1234
Q51: From a number M subtract 1. Take the reciprocal of the result to get the value of 'N'. Then which of the following is necessarily true?
A. $0\le M^{N}\le2$
B. $M^{N}>3$
C. $1
D. $1
Correct Answer: B
Solution:
Let the number be $M$.
We are given that $N = \frac{1}{M-1}$.
We need to find a relationship between $M$ and $N$ that is always true.
Let's analyze some examples: If $M = 2$, then $N = \frac{1}{2-1} = 1$.
$M = 2N$ is true for this case.
If $M = 3$, then $N = \frac{1}{3-1} = \frac{1}{2}$.
$M = 2N + 1$ is true for this case.
If $M = 4$, then $N = \frac{1}{4-1} = \frac{1}{3}$.
$M = 3N + 1$ is true for this case.
If $M = 5$, then $N = \frac{1}{5-1} = \frac{1}{4}$.
$M = 4N + 1$ is true for this case.
It appears that $M = (M-1)N + 1$ is true generally.
From $N = \frac{1}{M-1}$, we can write $N(M-1) = 1$.
$NM - N = 1$ $NM = N + 1$ $M = \frac{N+1}{N} = 1 + \frac{1}{N}$ Let's check: If $N=1$, $M=2$.
If $N=1/2$, $M=3$.
If $N=1/3$, $M=4$.
The relationship $M = 1 + \frac{1}{N}$ is always true.
There is no option relating to this, so let's re-examine.
$N(M-1) = 1$ $NM - N = 1$ $NM = N + 1$ $M = \frac{N+1}{N}$ This is always true.
However, we need to find an option that is necessarily true.
Let's consider the options.
The options are not provided, hence we cannot give a definite correct answer without knowing the options.
However, the relationship $M = \frac{N+1}{N}$ will always hold true.
Solution:
Let the number be $M$.
We are given that $N = \frac{1}{M-1}$.
We need to find a relationship between $M$ and $N$ that is always true.
Let's analyze some examples: If $M = 2$, then $N = \frac{1}{2-1} = 1$.
$M = 2N$ is true for this case.
If $M = 3$, then $N = \frac{1}{3-1} = \frac{1}{2}$.
$M = 2N + 1$ is true for this case.
If $M = 4$, then $N = \frac{1}{4-1} = \frac{1}{3}$.
$M = 3N + 1$ is true for this case.
If $M = 5$, then $N = \frac{1}{5-1} = \frac{1}{4}$.
$M = 4N + 1$ is true for this case.
It appears that $M = (M-1)N + 1$ is true generally.
From $N = \frac{1}{M-1}$, we can write $N(M-1) = 1$.
$NM - N = 1$ $NM = N + 1$ $M = \frac{N+1}{N} = 1 + \frac{1}{N}$ Let's check: If $N=1$, $M=2$.
If $N=1/2$, $M=3$.
If $N=1/3$, $M=4$.
The relationship $M = 1 + \frac{1}{N}$ is always true.
There is no option relating to this, so let's re-examine.
$N(M-1) = 1$ $NM - N = 1$ $NM = N + 1$ $M = \frac{N+1}{N}$ This is always true.
However, we need to find an option that is necessarily true.
Let's consider the options.
The options are not provided, hence we cannot give a definite correct answer without knowing the options.
However, the relationship $M = \frac{N+1}{N}$ will always hold true.
Q52: The cost of four mangoes, six guavas and sixteen watermelons is 500, while the cost of seven mangoes, nine guavas and nineteen watermelons is 620. What is the cost of one mango, one guava and one watermelon?
A. 120
B. 40
C. 150
D. Cannot be determined
Correct Answer: D
Solution:
Let $m$, $g$, and $w$ represent the cost of one mango, one guava, and one watermelon, respectively.
We can set up a system of two linear equations based on the given information: \begin{align} \label{eq:1} 4m + 6g + 16w &= 500 \\ 7m + 9g + 19w &= 620\end{align} We need to solve for $m+g+w$.
Notice that we don't need to solve for individual values of $m$, $g$, and $w$.
Let's try to manipulate the equations to eliminate some variables.
Multiplying the first equation by $\frac{7}{4}$ gives: $7m + \frac{21}{2}g + 28w = \frac{1750}{4} = 875$ Subtracting the second equation from this gives: $(\frac{21}{2}g - 9g) + (28w - 19w) = 875 - 620$ $\frac{3}{2}g + 9w = 255$ Multiplying the first equation by $\frac{9}{6} = \frac{3}{2}$ gives: $6m + 9g + 24w = 750$ Subtracting the second equation from this modified equation: $(6m - 7m) + (9g - 9g) + (24w - 19w) = 750 - 620$ $-m + 5w = 130$ However, this approach does not directly lead to $m+g+w$.
Let's try another approach.
Let's try to find a combination of the two equations that simplifies things.
Multiply the first equation by 3 and the second equation by 2: $12m + 18g + 48w = 1500$ $14m + 18g + 38w = 1240$ Subtracting the second equation from the first: $-2m + 10w = 260$ $-m + 5w = 130$ This doesn't help much.
Let's try a different approach.
It's unlikely that we can solve for individual $m$, $g$, $w$ and then sum them due to underdetermined system.
Let's observe that the question asks for the sum $m+g+w$.
Let's subtract the equations: $3m + 3g + 3w = 120$ Dividing by 3: $m + g + w = 40$
Solution:
Let $m$, $g$, and $w$ represent the cost of one mango, one guava, and one watermelon, respectively.
We can set up a system of two linear equations based on the given information: \begin{align} \label{eq:1} 4m + 6g + 16w &= 500 \\ 7m + 9g + 19w &= 620\end{align} We need to solve for $m+g+w$.
Notice that we don't need to solve for individual values of $m$, $g$, and $w$.
Let's try to manipulate the equations to eliminate some variables.
Multiplying the first equation by $\frac{7}{4}$ gives: $7m + \frac{21}{2}g + 28w = \frac{1750}{4} = 875$ Subtracting the second equation from this gives: $(\frac{21}{2}g - 9g) + (28w - 19w) = 875 - 620$ $\frac{3}{2}g + 9w = 255$ Multiplying the first equation by $\frac{9}{6} = \frac{3}{2}$ gives: $6m + 9g + 24w = 750$ Subtracting the second equation from this modified equation: $(6m - 7m) + (9g - 9g) + (24w - 19w) = 750 - 620$ $-m + 5w = 130$ However, this approach does not directly lead to $m+g+w$.
Let's try another approach.
Let's try to find a combination of the two equations that simplifies things.
Multiply the first equation by 3 and the second equation by 2: $12m + 18g + 48w = 1500$ $14m + 18g + 38w = 1240$ Subtracting the second equation from the first: $-2m + 10w = 260$ $-m + 5w = 130$ This doesn't help much.
Let's try a different approach.
It's unlikely that we can solve for individual $m$, $g$, $w$ and then sum them due to underdetermined system.
Let's observe that the question asks for the sum $m+g+w$.
Let's subtract the equations: $3m + 3g + 3w = 120$ Dividing by 3: $m + g + w = 40$
Q53: For the question above, what is the cost of a mango?
A. 20
B. 14
C. 15
D. Cannot be determined
Correct Answer: A
Solution:
The question asks for the cost of a mango.
To determine the subtopic, we need to analyze what information would be sufficient to answer the question.
Data sufficiency questions focus on whether given statements provide enough information to solve the problem, not on actually solving the problem.
Therefore, we would assess whether any provided statements (which are missing from the prompt) would allow us to determine the cost of a mango without further information.
We are not expected to find the actual cost; instead, we would determine if the given statements are sufficient to find the cost.
Solution:
The question asks for the cost of a mango.
To determine the subtopic, we need to analyze what information would be sufficient to answer the question.
Data sufficiency questions focus on whether given statements provide enough information to solve the problem, not on actually solving the problem.
Therefore, we would assess whether any provided statements (which are missing from the prompt) would allow us to determine the cost of a mango without further information.
We are not expected to find the actual cost; instead, we would determine if the given statements are sufficient to find the cost.
Q54: The following is known about three real numbers, x, y and z. $-4\le x\le4,-8\le y\le2$ and $-8\le z\le2$ Then the range of values that $M=xz/y$ can take is best represented by:
A. $ - \infty < x < \infty$
B. $-16
C. $-8\le x\le8$
D. $-16\le x\le16$
Correct Answer: A
Solution:
We are given that $-4 \le x \le 4$, $-8 \le y \le 2$, and $-8 \le z \le 2$.
We want to find the range of $M = \frac{xz}{y}$.
Let's consider the extreme cases: \begin{itemize} \item If $x=4$, $z=2$, and $y=-8$, then $M = \frac{4 \times 2}{-8} = -1$.
\item If $x=4$, $z=2$, and $y=2$, then $M = \frac{4 \times 2}{2} = 4$.
\item If $x=4$, $z=-2$, and $y=-8$, then $M = \frac{4 \times (-2)}{-8} = 1$.
\item If $x=4$, $z=-2$, and $y=2$, then $M = \frac{4 \times (-2)}{2} = -4$.
\item If $x=-4$, $z=2$, and $y=-8$, then $M = \frac{(-4) \times 2}{-8} = 1$.
\item If $x=-4$, $z=2$, and $y=2$, then $M = \frac{(-4) \times 2}{2} = -4$.
\item If $x=-4$, $z=-2$, and $y=-8$, then $M = \frac{(-4) \times (-2)}{-8} = -1$.
\item If $x=-4$, $z=-2$, and $y=2$, then $M = \frac{(-4) \times (-2)}{2} = 4$.
\end{itemize} Notice that $M$ can take both positive and negative values.
The smallest value of $M$ appears to be -4 and the largest is 4.
However, $y$ can be close to 0 which makes $M$ approach $\pm \infty$.
Since $y$ can be arbitrarily close to zero, the function is unbounded, hence the range is $(-\infty, \infty)$.
Solution:
We are given that $-4 \le x \le 4$, $-8 \le y \le 2$, and $-8 \le z \le 2$.
We want to find the range of $M = \frac{xz}{y}$.
Let's consider the extreme cases: \begin{itemize} \item If $x=4$, $z=2$, and $y=-8$, then $M = \frac{4 \times 2}{-8} = -1$.
\item If $x=4$, $z=2$, and $y=2$, then $M = \frac{4 \times 2}{2} = 4$.
\item If $x=4$, $z=-2$, and $y=-8$, then $M = \frac{4 \times (-2)}{-8} = 1$.
\item If $x=4$, $z=-2$, and $y=2$, then $M = \frac{4 \times (-2)}{2} = -4$.
\item If $x=-4$, $z=2$, and $y=-8$, then $M = \frac{(-4) \times 2}{-8} = 1$.
\item If $x=-4$, $z=2$, and $y=2$, then $M = \frac{(-4) \times 2}{2} = -4$.
\item If $x=-4$, $z=-2$, and $y=-8$, then $M = \frac{(-4) \times (-2)}{-8} = -1$.
\item If $x=-4$, $z=-2$, and $y=2$, then $M = \frac{(-4) \times (-2)}{2} = 4$.
\end{itemize} Notice that $M$ can take both positive and negative values.
The smallest value of $M$ appears to be -4 and the largest is 4.
However, $y$ can be close to 0 which makes $M$ approach $\pm \infty$.
Since $y$ can be arbitrarily close to zero, the function is unbounded, hence the range is $(-\infty, \infty)$.
Q55: A man sold 38 pieces of clothing (combined in the form of shirts, trousers and ties). If he sold at least 11 pieces of each item and he sold more shirts than trousers and more trousers than ties, then the number of ties that he must have sold is:1
A. Exactly 11
B. At least 11
C. At least 12
D. Cannot be determined
Correct Answer: B
Solution:
Let $s$, $t$, and $r$ represent the number of shirts, trousers, and ties sold, respectively.
We are given that $s + t + r = 38$, where $s, t, r \ge 11$.
Also, $s > t > r$.
Let $s' = s - 11$, $t' = t - 11$, $r' = r - 11$.
Then $s', t', r' \ge 0$.
Substituting, we get $(s' + 11) + (t' + 11) + (r' + 11) = 38$, which simplifies to $s' + t' + r' = 38 - 33 = 5$.
Since $s > t > r$, we have $s' > t' > r'$.
The possible non-negative integer solutions for $s', t', r'$ such that $s' + t' + r' = 5$ and $s' > t' > r'$ are: \begin{itemize} \item $s' = 3, t' = 2, r' = 0$ $\implies s = 14, t = 13, r = 11$ \item $s' = 4, t' = 1, r' = 0$ $\implies s = 15, t = 12, r = 11$ \item $s' = 2, t' = 1, r' = 2$ (This is invalid since $s'>t'>r'$ is not satisfied) \end{itemize} The only solutions that satisfy $s > t > r$ are $s = 14, t = 13, r = 11$ and $s = 15, t = 12, r = 11$.
In both cases, the number of ties sold is 11.
Solution:
Let $s$, $t$, and $r$ represent the number of shirts, trousers, and ties sold, respectively.
We are given that $s + t + r = 38$, where $s, t, r \ge 11$.
Also, $s > t > r$.
Let $s' = s - 11$, $t' = t - 11$, $r' = r - 11$.
Then $s', t', r' \ge 0$.
Substituting, we get $(s' + 11) + (t' + 11) + (r' + 11) = 38$, which simplifies to $s' + t' + r' = 38 - 33 = 5$.
Since $s > t > r$, we have $s' > t' > r'$.
The possible non-negative integer solutions for $s', t', r'$ such that $s' + t' + r' = 5$ and $s' > t' > r'$ are: \begin{itemize} \item $s' = 3, t' = 2, r' = 0$ $\implies s = 14, t = 13, r = 11$ \item $s' = 4, t' = 1, r' = 0$ $\implies s = 15, t = 12, r = 11$ \item $s' = 2, t' = 1, r' = 2$ (This is invalid since $s'>t'>r'$ is not satisfied) \end{itemize} The only solutions that satisfy $s > t > r$ are $s = 14, t = 13, r = 11$ and $s = 15, t = 12, r = 11$.
In both cases, the number of ties sold is 11.
Q56: The least number of 4 digits which is a perfect square is:
A. 1064
B. 1040
C. 1024
D. 1012
Correct Answer: C
Solution:
We need to find the least number of 4 digits which is a perfect square.
The smallest 4-digit number is 1000.
We need to find the smallest perfect square greater than or equal to 1000.
We can find the square root of 1000: $\sqrt{1000} \approx 31.62$.
The next integer is 32.
Therefore, $32^2 = 1024$.
This is the smallest 4-digit number that is a perfect square.
Solution:
We need to find the least number of 4 digits which is a perfect square.
The smallest 4-digit number is 1000.
We need to find the smallest perfect square greater than or equal to 1000.
We can find the square root of 1000: $\sqrt{1000} \approx 31.62$.
The next integer is 32.
Therefore, $32^2 = 1024$.
This is the smallest 4-digit number that is a perfect square.
Q57: The least multiple of 7 which leaves a remainder of 4 when divided by 6, 9, 15 and 18 is
A. 94
B. 184
C. 364
D. 74
Correct Answer: C
Solution:
Let the least multiple of 7 be $7x$, where $x$ is an integer.
We are given that when $7x$ is divided by 6, 9, 15, and 18, the remainder is 4 in each case.
This can be expressed as: $7x \equiv 4 \pmod{6}$ $7x \equiv 4 \pmod{9}$ $7x \equiv 4 \pmod{15}$ $7x \equiv 4 \pmod{18}$ This implies that $7x - 4$ is divisible by 6, 9, 15, and 18.
Therefore, $7x - 4$ is a multiple of the least common multiple (LCM) of 6, 9, 15, and 18.
First, find the LCM of 6, 9, 15, and 18: $6 = 2 \times 3$ $9 = 3^2$ $15 = 3 \times 5$ $18 = 2 \times 3^2$ LCM$(6, 9, 15, 18) = 2 \times 3^2 \times 5 = 90$ So, $7x - 4 = 90k$ for some integer $k$.
$7x = 90k + 4$ We need to find the smallest positive integer $x$ that satisfies this equation.
We can test values of $k$: If $k=1$, $7x = 94$, which is not divisible by 7.
If $k=2$, $7x = 184$, which is not divisible by 7.
If $k=3$, $7x = 274$, which is not divisible by 7.
If $k=4$, $7x = 364$, $x = 52$.
Therefore, the least multiple of 7 is $7x = 7(52) = 364$.
Let's check the remainders: $364 \div 6 = 60 R 4$ $364 \div 9 = 40 R 4$ $364 \div 15 = 24 R 4$ $364 \div 18 = 20 R 4$ The remainders are all 4.
Solution:
Let the least multiple of 7 be $7x$, where $x$ is an integer.
We are given that when $7x$ is divided by 6, 9, 15, and 18, the remainder is 4 in each case.
This can be expressed as: $7x \equiv 4 \pmod{6}$ $7x \equiv 4 \pmod{9}$ $7x \equiv 4 \pmod{15}$ $7x \equiv 4 \pmod{18}$ This implies that $7x - 4$ is divisible by 6, 9, 15, and 18.
Therefore, $7x - 4$ is a multiple of the least common multiple (LCM) of 6, 9, 15, and 18.
First, find the LCM of 6, 9, 15, and 18: $6 = 2 \times 3$ $9 = 3^2$ $15 = 3 \times 5$ $18 = 2 \times 3^2$ LCM$(6, 9, 15, 18) = 2 \times 3^2 \times 5 = 90$ So, $7x - 4 = 90k$ for some integer $k$.
$7x = 90k + 4$ We need to find the smallest positive integer $x$ that satisfies this equation.
We can test values of $k$: If $k=1$, $7x = 94$, which is not divisible by 7.
If $k=2$, $7x = 184$, which is not divisible by 7.
If $k=3$, $7x = 274$, which is not divisible by 7.
If $k=4$, $7x = 364$, $x = 52$.
Therefore, the least multiple of 7 is $7x = 7(52) = 364$.
Let's check the remainders: $364 \div 6 = 60 R 4$ $364 \div 9 = 40 R 4$ $364 \div 15 = 24 R 4$ $364 \div 18 = 20 R 4$ The remainders are all 4.
Q58: Find the least square number which is divisible by 6, 8 and 15
A. 2500
B. 3600
C. 4900
D. 4500
Correct Answer: B
Solution:
First, find the LCM of 6, 8, and 15.
The prime factorization of each number is: $6 = 2 \times 3$ $8 = 2^3$ $15 = 3 \times 5$ The LCM is $2^3 \times 3 \times 5 = 120$.
Now, we need to find the least square number divisible by 120.
This means we need to find the smallest number $n^2$ such that $120 | n^2$.
The prime factorization of 120 is $2^3 \times 3 \times 5$.
For $n^2$ to be divisible by 120, it must have at least the following prime factors: $2^3$, $3^1$, $5^1$.
To make it a perfect square, we need to have even powers for each prime factor.
Thus, we need to multiply 120 by $2 \times 3 \times 5 = 30$ to get: $120 \times 30 = 3600 = 60^2$ Therefore, the least square number divisible by 6, 8, and 15 is 3600.
Solution:
First, find the LCM of 6, 8, and 15.
The prime factorization of each number is: $6 = 2 \times 3$ $8 = 2^3$ $15 = 3 \times 5$ The LCM is $2^3 \times 3 \times 5 = 120$.
Now, we need to find the least square number divisible by 120.
This means we need to find the smallest number $n^2$ such that $120 | n^2$.
The prime factorization of 120 is $2^3 \times 3 \times 5$.
For $n^2$ to be divisible by 120, it must have at least the following prime factors: $2^3$, $3^1$, $5^1$.
To make it a perfect square, we need to have even powers for each prime factor.
Thus, we need to multiply 120 by $2 \times 3 \times 5 = 30$ to get: $120 \times 30 = 3600 = 60^2$ Therefore, the least square number divisible by 6, 8, and 15 is 3600.
Q59: By how much is three fourth of 116 greater than four fifth of 45?
A. 31
B. 41
C. 46
D. None of these
Correct Answer: D
Solution:
First, we calculate three-fourth of 116: $$ \frac{3}{4} \times 116 = 3 \times 29 = 87 $$ Next, we calculate four-fifth of 45: $$ \frac{4}{5} \times 45 = 4 \times 9 = 36 $$ Finally, we find the difference between the two results: $$ 87 - 36 = 51 $$
Solution:
First, we calculate three-fourth of 116: $$ \frac{3}{4} \times 116 = 3 \times 29 = 87 $$ Next, we calculate four-fifth of 45: $$ \frac{4}{5} \times 45 = 4 \times 9 = 36 $$ Finally, we find the difference between the two results: $$ 87 - 36 = 51 $$
Q60: Two friends were discussing their marks in an examination. While doing so they realized that both the numbers had the same prime factors, although Raveesh got a score which had two more factors than Harish. If their marks are represented by one of the options as given below, which of the following options would correctly represent the number of marks they got?
A. 30,60
B. 20,80
C. 40,80
D. 20,60
Correct Answer: B
Solution:
Let the marks obtained by Harish be $H$ and the marks obtained by Raveesh be $R$.
The problem states that $H$ and $R$ have the same prime factors.
Let's denote the prime factors as $p_1, p_2, ..., p_n$.
Then we can represent $H$ and $R$ as: $H = p_1^{a_1} p_2^{a_2} ...
p_n^{a_n}$ $R = p_1^{b_1} p_2^{b_2} ...
p_n^{b_n}$ where $a_i$ and $b_i$ are non-negative integers.
The number of factors of $H$ is given by: $N_H = (a_1 + 1)(a_2 + 1)...(a_n + 1)$ Similarly, the number of factors of $R$ is: $N_R = (b_1 + 1)(b_2 + 1)...(b_n + 1)$ The problem states that $R$ has two more factors than $H$, so $N_R = N_H + 2$.
We need to find a pair of numbers that satisfies this condition and have the same prime factors.
This requires checking the given options (not provided in the prompt).
Without the options, a definitive answer cannot be given.
However, a suitable example would be if Harish got 12 marks ($2^2 \times 3$, 6 factors) and Raveesh got 18 marks ($2 \times 3^2$, 6 factors).
This doesn't meet the condition.
Another example could be if Harish scored 16 (factors:1,2,4,8,16, total 5) and Raveesh scored 24 (factors 1,2,3,4,6,8,12,24, total 8) .
That fulfills the condition of same prime factors and a difference of 2 factors.
More examples must be tested against the provided options.
Solution:
Let the marks obtained by Harish be $H$ and the marks obtained by Raveesh be $R$.
The problem states that $H$ and $R$ have the same prime factors.
Let's denote the prime factors as $p_1, p_2, ..., p_n$.
Then we can represent $H$ and $R$ as: $H = p_1^{a_1} p_2^{a_2} ...
p_n^{a_n}$ $R = p_1^{b_1} p_2^{b_2} ...
p_n^{b_n}$ where $a_i$ and $b_i$ are non-negative integers.
The number of factors of $H$ is given by: $N_H = (a_1 + 1)(a_2 + 1)...(a_n + 1)$ Similarly, the number of factors of $R$ is: $N_R = (b_1 + 1)(b_2 + 1)...(b_n + 1)$ The problem states that $R$ has two more factors than $H$, so $N_R = N_H + 2$.
We need to find a pair of numbers that satisfies this condition and have the same prime factors.
This requires checking the given options (not provided in the prompt).
Without the options, a definitive answer cannot be given.
However, a suitable example would be if Harish got 12 marks ($2^2 \times 3$, 6 factors) and Raveesh got 18 marks ($2 \times 3^2$, 6 factors).
This doesn't meet the condition.
Another example could be if Harish scored 16 (factors:1,2,4,8,16, total 5) and Raveesh scored 24 (factors 1,2,3,4,6,8,12,24, total 8) .
That fulfills the condition of same prime factors and a difference of 2 factors.
More examples must be tested against the provided options.
Q61: A number is such that when divided by 3, 5, 6, or 7 it leaves the remainder 1, 3, 4, or 5 respectively. Which is the largest number below 4000 that satisfies this property?
A. 3358
B. 3988
C. 3778
D. 2938
Correct Answer: B
Solution:
Let the number be $n$.
According to the problem, we have: $n \equiv 1 \pmod{3}$ $n \equiv 3 \pmod{5}$ $n \equiv 4 \pmod{6}$ $n \equiv 5 \pmod{7}$ Notice that in each case, the remainder is one less than the divisor.
This means that $n+2$ is divisible by 3, 5, 6, and 7.
Therefore, $n+2$ is a multiple of the least common multiple (LCM) of 3, 5, 6, and 7.
We find the LCM of 3, 5, 6, and 7: $3 = 3$ $5 = 5$ $6 = 2 \times 3$ $7 = 7$ LCM$(3, 5, 6, 7) = 2 \times 3 \times 5 \times 7 = 210$ So, $n+2$ is a multiple of 210.
We can write $n+2 = 210k$ for some integer $k$.
Then $n = 210k - 2$.
We want the largest number $n$ below 4000.
We find the largest integer $k$ such that $210k - 2 < 4000$: $210k < 4002$ $k < \frac{4002}{210} \approx 19.057$ The largest integer $k$ is 19.
Therefore, the largest number $n$ is $210(19) - 2 = 3990 - 2 = 3988$.
Solution:
Let the number be $n$.
According to the problem, we have: $n \equiv 1 \pmod{3}$ $n \equiv 3 \pmod{5}$ $n \equiv 4 \pmod{6}$ $n \equiv 5 \pmod{7}$ Notice that in each case, the remainder is one less than the divisor.
This means that $n+2$ is divisible by 3, 5, 6, and 7.
Therefore, $n+2$ is a multiple of the least common multiple (LCM) of 3, 5, 6, and 7.
We find the LCM of 3, 5, 6, and 7: $3 = 3$ $5 = 5$ $6 = 2 \times 3$ $7 = 7$ LCM$(3, 5, 6, 7) = 2 \times 3 \times 5 \times 7 = 210$ So, $n+2$ is a multiple of 210.
We can write $n+2 = 210k$ for some integer $k$.
Then $n = 210k - 2$.
We want the largest number $n$ below 4000.
We find the largest integer $k$ such that $210k - 2 < 4000$: $210k < 4002$ $k < \frac{4002}{210} \approx 19.057$ The largest integer $k$ is 19.
Therefore, the largest number $n$ is $210(19) - 2 = 3990 - 2 = 3988$.
Q62: A number when divided by 2, 3 and 4 leaves a remainder of 1. Find the highest 2 digit number that satisfies this requirement.
A. 91
B. 93
C. 97
D. 95
Correct Answer: C
Solution:
Let the number be $n$.
According to the problem, when $n$ is divided by 2, 3, and 4, the remainder is 1 in each case.
This can be expressed as: $n \equiv 1 \pmod{2}$ $n \equiv 1 \pmod{3}$ $n \equiv 1 \pmod{4}$ This means that $n-1$ is divisible by 2, 3, and 4.
Therefore, $n-1$ must be a multiple of the least common multiple (LCM) of 2, 3, and 4.
The LCM of 2, 3, and 4 is $LCM(2,3,4) = 12$.
So, $n-1$ is a multiple of 12.
We can write this as $n-1 = 12k$, where $k$ is an integer.
Then $n = 12k + 1$.
We are looking for the highest 2-digit number that satisfies this condition.
Let's find the largest integer $k$ such that $12k + 1 \le 99$.
$12k \le 98$ $k \le \frac{98}{12} = 8.166...$ The largest integer $k$ is 8.
Therefore, the highest 2-digit number is $n = 12(8) + 1 = 96 + 1 = 97$.
Solution:
Let the number be $n$.
According to the problem, when $n$ is divided by 2, 3, and 4, the remainder is 1 in each case.
This can be expressed as: $n \equiv 1 \pmod{2}$ $n \equiv 1 \pmod{3}$ $n \equiv 1 \pmod{4}$ This means that $n-1$ is divisible by 2, 3, and 4.
Therefore, $n-1$ must be a multiple of the least common multiple (LCM) of 2, 3, and 4.
The LCM of 2, 3, and 4 is $LCM(2,3,4) = 12$.
So, $n-1$ is a multiple of 12.
We can write this as $n-1 = 12k$, where $k$ is an integer.
Then $n = 12k + 1$.
We are looking for the highest 2-digit number that satisfies this condition.
Let's find the largest integer $k$ such that $12k + 1 \le 99$.
$12k \le 98$ $k \le \frac{98}{12} = 8.166...$ The largest integer $k$ is 8.
Therefore, the highest 2-digit number is $n = 12(8) + 1 = 96 + 1 = 97$.
Q63: For the question above, how many times would they ring, together in the next 1 hour?
A. 17
B. 18
C. 19
D. None of these
Correct Answer: C
Solution:
This question is incomplete; it lacks crucial information about the individual ringing frequencies of the objects.
To solve it, we need to know how frequently each object rings.
Let's assume we have two objects ringing at intervals $x$ and $y$ minutes, respectively.
To find out when they ring together, we need to find the least common multiple (LCM) of $x$ and $y$.
The number of times they ring together in one hour (60 minutes) would be $\left\lfloor \frac{60}{LCM(x,y)} \right\rfloor$.
The floor function ($\lfloor \rfloor$) is used because any remainder implies the next simultaneous ring would occur after the one-hour mark.
Without knowing $x$ and $y$, we cannot calculate the answer.
The question needs additional data about the timing of the rings.
Solution:
This question is incomplete; it lacks crucial information about the individual ringing frequencies of the objects.
To solve it, we need to know how frequently each object rings.
Let's assume we have two objects ringing at intervals $x$ and $y$ minutes, respectively.
To find out when they ring together, we need to find the least common multiple (LCM) of $x$ and $y$.
The number of times they ring together in one hour (60 minutes) would be $\left\lfloor \frac{60}{LCM(x,y)} \right\rfloor$.
The floor function ($\lfloor \rfloor$) is used because any remainder implies the next simultaneous ring would occur after the one-hour mark.
Without knowing $x$ and $y$, we cannot calculate the answer.
The question needs additional data about the timing of the rings.
Q64: For the question above, what are the minimum number of rows that would be required to be formed?
A. 11
B. 19
C. 18
D. None of these
Correct Answer: C
Solution:
The question asks for the minimum number of rows.
This requires analyzing the sufficiency of information to answer the question, not actually calculating the number of rows.
The question itself doesn't provide any data or context about rows (e.g., in a table, matrix, etc.).
Therefore, without additional information, it is impossible to determine the minimum number of rows required.
The question lacks sufficient information to be answered.
The focus is on determining if sufficient information is present, a core aspect of data sufficiency problems, not on solving for an answer.
Solution:
The question asks for the minimum number of rows.
This requires analyzing the sufficiency of information to answer the question, not actually calculating the number of rows.
The question itself doesn't provide any data or context about rows (e.g., in a table, matrix, etc.).
Therefore, without additional information, it is impossible to determine the minimum number of rows required.
The question lacks sufficient information to be answered.
The focus is on determining if sufficient information is present, a core aspect of data sufficiency problems, not on solving for an answer.
Q65: For the above question, what is the size of the largest bottle which can be used?
A. 1
B. 2
C. 17
D. 34
Correct Answer: B
Solution:
The question is incomplete.
It lacks crucial information: the sizes of the bottles available and the context of how these bottles are being used (e.g., to fill a larger container, to pour into smaller containers without waste).
Without this information, determining the size of the largest bottle that can be used is impossible.
The problem requires knowing the quantities and possibly the constraints regarding the volumes of the liquids involved.
To solve a similar problem, one would find the greatest common divisor (GCD) of the quantities involved if the goal is to fill the same amount in the largest possible bottle size.
If the goal is to use bottles of the same size to fill several quantities, the LCM will be involved.
Therefore, more context is needed.
Solution:
The question is incomplete.
It lacks crucial information: the sizes of the bottles available and the context of how these bottles are being used (e.g., to fill a larger container, to pour into smaller containers without waste).
Without this information, determining the size of the largest bottle that can be used is impossible.
The problem requires knowing the quantities and possibly the constraints regarding the volumes of the liquids involved.
To solve a similar problem, one would find the greatest common divisor (GCD) of the quantities involved if the goal is to fill the same amount in the largest possible bottle size.
If the goal is to use bottles of the same size to fill several quantities, the LCM will be involved.
Therefore, more context is needed.
Q66: For Question 105, what are the minimum number of bottles that would be required?
A. 11
B. 19
C. 18
D. None of these
Correct Answer: C
Solution:
The question "For Question 105, what are the minimum number of bottles that would be required?" is a data sufficiency question because it doesn't provide enough information to answer.
To determine the minimum number of bottles, we need the context of Question 105.
The question itself only asks about the sufficiency of information, not the actual calculation.
We need the details of Question 105 (like the amount of liquid and the capacity of each bottle) to judge if enough information is provided to determine the minimum number of bottles needed.
Therefore, this question highlights the core concept of data sufficiency: assessing whether the given data is sufficient to answer a question without necessarily finding the complete solution.
The statement is insufficient on its own.
More information about Question 105 is needed.
Solution:
The question "For Question 105, what are the minimum number of bottles that would be required?" is a data sufficiency question because it doesn't provide enough information to answer.
To determine the minimum number of bottles, we need the context of Question 105.
The question itself only asks about the sufficiency of information, not the actual calculation.
We need the details of Question 105 (like the amount of liquid and the capacity of each bottle) to judge if enough information is provided to determine the minimum number of bottles needed.
Therefore, this question highlights the core concept of data sufficiency: assessing whether the given data is sufficient to answer a question without necessarily finding the complete solution.
The statement is insufficient on its own.
More information about Question 105 is needed.
Q67: Find the number of zeroes at the end of 1400!
A. 347
B. 336
C. 349
D. 348
Correct Answer: C
Solution:
The number of zeroes at the end of a factorial is determined by the number of times 10 is a factor.
Since $10 = 2 \times 5$, and there are always more factors of 2 than 5 in a factorial, we need to count the number of factors of 5 in 1400!.
We can use Legendre's formula for this: $$ \sum_{i=1}^{\infty} \left\lfloor \frac{1400}{5^i} \right\rfloor $$ where $\lfloor x \rfloor$ denotes the floor function (the greatest integer less than or equal to $x$).
Let's calculate the terms: $i = 1$: $\left\lfloor \frac{1400}{5} \right\rfloor = 280$ $i = 2$: $\left\lfloor \frac{1400}{25} \right\rfloor = 56$ $i = 3$: $\left\lfloor \frac{1400}{125} \right\rfloor = 11$ $i = 4$: $\left\lfloor \frac{1400}{625} \right\rfloor = 2$ $i = 5$: $\left\lfloor \frac{1400}{3125} \right\rfloor = 0$ The sum is $280 + 56 + 11 + 2 + 0 = 349$.
Therefore, there are 349 factors of 5 in 1400!.
Since there are more factors of 2 than 5, there are 349 zeroes at the end of 1400!.
Solution:
The number of zeroes at the end of a factorial is determined by the number of times 10 is a factor.
Since $10 = 2 \times 5$, and there are always more factors of 2 than 5 in a factorial, we need to count the number of factors of 5 in 1400!.
We can use Legendre's formula for this: $$ \sum_{i=1}^{\infty} \left\lfloor \frac{1400}{5^i} \right\rfloor $$ where $\lfloor x \rfloor$ denotes the floor function (the greatest integer less than or equal to $x$).
Let's calculate the terms: $i = 1$: $\left\lfloor \frac{1400}{5} \right\rfloor = 280$ $i = 2$: $\left\lfloor \frac{1400}{25} \right\rfloor = 56$ $i = 3$: $\left\lfloor \frac{1400}{125} \right\rfloor = 11$ $i = 4$: $\left\lfloor \frac{1400}{625} \right\rfloor = 2$ $i = 5$: $\left\lfloor \frac{1400}{3125} \right\rfloor = 0$ The sum is $280 + 56 + 11 + 2 + 0 = 349$.
Therefore, there are 349 factors of 5 in 1400!.
Since there are more factors of 2 than 5, there are 349 zeroes at the end of 1400!.
Q68: Find the number of zeroes at the end of 380!
A. 90
B. 91
C. 94
D. 95
Correct Answer: C
Solution:
The number of zeroes at the end of $380!$ is determined by the number of times 10 is a factor in $380!$.
Since $10 = 2 \times 5$, and there will always be more factors of 2 than 5 in a factorial, we need to count the number of factors of 5 in $380!$.
We can use Legendre's formula for this: The number of factors of 5 in $380!$ is given by: $$ \left\lfloor \frac{380}{5} \right\rfloor + \left\lfloor \frac{380}{5^2} \right\rfloor + \left\lfloor \frac{380}{5^3} \right\rfloor + \left\lfloor \frac{380}{5^4} \right\rfloor + ...
$$ where $\lfloor x \rfloor$ denotes the floor function (the greatest integer less than or equal to $x$).
Calculating this: $$ \left\lfloor \frac{380}{5} \right\rfloor = 76 $$ $$ \left\lfloor \frac{380}{25} \right\rfloor = 15 $$ $$ \left\lfloor \frac{380}{125} \right\rfloor = 3 $$ $$ \left\lfloor \frac{380}{625} \right\rfloor = 0 $$ The terms become 0 from here onwards.
Adding these up: $76 + 15 + 3 = 94$ Therefore, there are 94 factors of 5 in $380!$.
Since there are more factors of 2, there are 94 factors of 10, which means there are 94 zeroes at the end of $380!$.
Solution:
The number of zeroes at the end of $380!$ is determined by the number of times 10 is a factor in $380!$.
Since $10 = 2 \times 5$, and there will always be more factors of 2 than 5 in a factorial, we need to count the number of factors of 5 in $380!$.
We can use Legendre's formula for this: The number of factors of 5 in $380!$ is given by: $$ \left\lfloor \frac{380}{5} \right\rfloor + \left\lfloor \frac{380}{5^2} \right\rfloor + \left\lfloor \frac{380}{5^3} \right\rfloor + \left\lfloor \frac{380}{5^4} \right\rfloor + ...
$$ where $\lfloor x \rfloor$ denotes the floor function (the greatest integer less than or equal to $x$).
Calculating this: $$ \left\lfloor \frac{380}{5} \right\rfloor = 76 $$ $$ \left\lfloor \frac{380}{25} \right\rfloor = 15 $$ $$ \left\lfloor \frac{380}{125} \right\rfloor = 3 $$ $$ \left\lfloor \frac{380}{625} \right\rfloor = 0 $$ The terms become 0 from here onwards.
Adding these up: $76 + 15 + 3 = 94$ Therefore, there are 94 factors of 5 in $380!$.
Since there are more factors of 2, there are 94 factors of 10, which means there are 94 zeroes at the end of $380!$.
Q69: $115!/7^{n}$ is an integer, Find the highest possible value of n for this to be true.
A. 15
B. 17
C. 16
D. 18
Correct Answer: D
Solution:
We need to find the highest power of 7 that divides $115!$.
We can use Legendre's formula for this: The exponent of a prime $p$ in the prime factorization of $n!$ is given by: $$v_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$$ In our case, $n = 115$ and $p = 7$.
We calculate: $$v_7(115!) = \left\lfloor \frac{115}{7} \right\rfloor + \left\lfloor \frac{115}{49} \right\rfloor + \left\lfloor \frac{115}{343} \right\rfloor + \dots$$ $$v_7(115!) = \left\lfloor 16.42 \right\rfloor + \left\lfloor 2.34 \right\rfloor + \left\lfloor 0.33 \right\rfloor + \dots$$ $$v_7(115!) = 16 + 2 + 0 + \dots = 18$$ Therefore, the highest power of 7 that divides $115!$ is $7^{18}$.
Thus, the highest possible value of $n$ for which $115!/7^n$ is an integer is 18.
Solution:
We need to find the highest power of 7 that divides $115!$.
We can use Legendre's formula for this: The exponent of a prime $p$ in the prime factorization of $n!$ is given by: $$v_p(n!) = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor$$ In our case, $n = 115$ and $p = 7$.
We calculate: $$v_7(115!) = \left\lfloor \frac{115}{7} \right\rfloor + \left\lfloor \frac{115}{49} \right\rfloor + \left\lfloor \frac{115}{343} \right\rfloor + \dots$$ $$v_7(115!) = \left\lfloor 16.42 \right\rfloor + \left\lfloor 2.34 \right\rfloor + \left\lfloor 0.33 \right\rfloor + \dots$$ $$v_7(115!) = 16 + 2 + 0 + \dots = 18$$ Therefore, the highest power of 7 that divides $115!$ is $7^{18}$.
Thus, the highest possible value of $n$ for which $115!/7^n$ is an integer is 18.
Q70: $155!/20^{n}$ is an integer. Find the highest possible value of n for this to be true.
A. 77
B. 38
C. 75
D. 37
Correct Answer: B
Solution:
We need to find the highest power of 20 that divides 155!.
Since $20 = 2^2 \times 5$, we need to find the highest power of 2 and 5 that divides 155!.
The highest power of 5 that divides 155!
is given by Legendre's formula: $$ \sum_{k=1}^{\infty} \left\lfloor \frac{155}{5^k} \right\rfloor = \left\lfloor \frac{155}{5} \right\rfloor + \left\lfloor \frac{155}{25} \right\rfloor + \left\lfloor \frac{155}{125} \right\rfloor + \dots = 31 + 6 + 1 = 38 $$ Thus, the highest power of 5 that divides 155!
is $5^{38}$.
The highest power of 2 that divides 155!
is given by Legendre's formula: $$ \sum_{k=1}^{\infty} \left\lfloor \frac{155}{2^k} \right\rfloor = \left\lfloor \frac{155}{2} \right\rfloor + \left\lfloor \frac{155}{4} \right\rfloor + \left\lfloor \frac{155}{8} \right\rfloor + \left\lfloor \frac{155}{16} \right\rfloor + \left\lfloor \frac{155}{32} \right\rfloor + \left\lfloor \frac{155}{64} \right\rfloor + \left\lfloor \frac{155}{128} \right\rfloor + \dots $$ $$ = 77 + 38 + 19 + 9 + 4 + 2 + 1 = 150 $$ Thus, the highest power of 2 that divides 155!
is $2^{150}$.
Since $20 = 2^2 \times 5$, we have $20^n = (2^2 \times 5)^n = 2^{2n} \times 5^n$.
For $20^n$ to divide 155!, we must have $2n \le 150$ and $n \le 38$.
From $2n \le 150$, we get $n \le 75$.
Since both conditions must be satisfied, we must have $n \le \min(75, 38) = 38$.
Therefore, the highest possible value of $n$ is 38.
Solution:
We need to find the highest power of 20 that divides 155!.
Since $20 = 2^2 \times 5$, we need to find the highest power of 2 and 5 that divides 155!.
The highest power of 5 that divides 155!
is given by Legendre's formula: $$ \sum_{k=1}^{\infty} \left\lfloor \frac{155}{5^k} \right\rfloor = \left\lfloor \frac{155}{5} \right\rfloor + \left\lfloor \frac{155}{25} \right\rfloor + \left\lfloor \frac{155}{125} \right\rfloor + \dots = 31 + 6 + 1 = 38 $$ Thus, the highest power of 5 that divides 155!
is $5^{38}$.
The highest power of 2 that divides 155!
is given by Legendre's formula: $$ \sum_{k=1}^{\infty} \left\lfloor \frac{155}{2^k} \right\rfloor = \left\lfloor \frac{155}{2} \right\rfloor + \left\lfloor \frac{155}{4} \right\rfloor + \left\lfloor \frac{155}{8} \right\rfloor + \left\lfloor \frac{155}{16} \right\rfloor + \left\lfloor \frac{155}{32} \right\rfloor + \left\lfloor \frac{155}{64} \right\rfloor + \left\lfloor \frac{155}{128} \right\rfloor + \dots $$ $$ = 77 + 38 + 19 + 9 + 4 + 2 + 1 = 150 $$ Thus, the highest power of 2 that divides 155!
is $2^{150}$.
Since $20 = 2^2 \times 5$, we have $20^n = (2^2 \times 5)^n = 2^{2n} \times 5^n$.
For $20^n$ to divide 155!, we must have $2n \le 150$ and $n \le 38$.
From $2n \le 150$, we get $n \le 75$.
Since both conditions must be satisfied, we must have $n \le \min(75, 38) = 38$.
Therefore, the highest possible value of $n$ is 38.
Q71: How many numbers between 1 and 200 are exactly divisible by exactly two of 3, 9 and 27?
A. 14
B. 15
C. 16
D. 17
Correct Answer: B
Solution:
Let $A$ be the set of numbers divisible by 3, $B$ be the set of numbers divisible by 9, and $C$ be the set of numbers divisible by 27.
We want to find the number of integers between 1 and 200 that are divisible by exactly two of 3, 9, and 27.
Numbers divisible by exactly two of 3, 9, and 27 are those divisible by: \begin{itemize} \item 3 and 9, but not 27: Multiples of 9 that are not multiples of 27.
These are $9, 18, 36, \dots, 198$.
The number of such integers is $\left\lfloor \frac{198}{9} \right\rfloor - \left\lfloor \frac{198}{27} \right\rfloor = 22 - 7 = 15$.
\item 3 and 27, but not 9: This is impossible since if a number is divisible by 27, it must be divisible by 9 and 3.
\item 9 and 27, but not 3: This is also impossible since if a number is divisible by 9 and 27, it must be divisible by 3.
\end{itemize} Therefore, the only possibility is that the numbers are divisible by 3 and 9 but not 27.
There are 15 such numbers.
Solution:
Let $A$ be the set of numbers divisible by 3, $B$ be the set of numbers divisible by 9, and $C$ be the set of numbers divisible by 27.
We want to find the number of integers between 1 and 200 that are divisible by exactly two of 3, 9, and 27.
Numbers divisible by exactly two of 3, 9, and 27 are those divisible by: \begin{itemize} \item 3 and 9, but not 27: Multiples of 9 that are not multiples of 27.
These are $9, 18, 36, \dots, 198$.
The number of such integers is $\left\lfloor \frac{198}{9} \right\rfloor - \left\lfloor \frac{198}{27} \right\rfloor = 22 - 7 = 15$.
\item 3 and 27, but not 9: This is impossible since if a number is divisible by 27, it must be divisible by 9 and 3.
\item 9 and 27, but not 3: This is also impossible since if a number is divisible by 9 and 27, it must be divisible by 3.
\end{itemize} Therefore, the only possibility is that the numbers are divisible by 3 and 9 but not 27.
There are 15 such numbers.
Q72: A number N is squared to give a value of S. The minimum value of N+S would happen when N is
A. -0.3
B. -0.5
C. -0.7
D. None of these
Correct Answer: D
Solution:
Let $N$ be the number.
We are given that $N^2 = S$.
We want to minimize $N + S = N + N^2$.
Let $f(N) = N + N^2$.
To find the minimum value, we can complete the square or use calculus.
Completing the square: $f(N) = N^2 + N = (N^2 + N + \frac{1}{4}) - \frac{1}{4} = (N + \frac{1}{2})^2 - \frac{1}{4}$.
Since $(N + \frac{1}{2})^2 \ge 0$, the minimum value occurs when $(N + \frac{1}{2})^2 = 0$, which means $N = -\frac{1}{2}$.
Using calculus: $f(N) = N^2 + N$ $f'(N) = 2N + 1$ Setting $f'(N) = 0$, we get $2N + 1 = 0$, so $N = -\frac{1}{2}$.
$f''(N) = 2 > 0$, indicating a minimum.
Thus, the minimum value of $N + S$ occurs when $N = -\frac{1}{2}$.
Solution:
Let $N$ be the number.
We are given that $N^2 = S$.
We want to minimize $N + S = N + N^2$.
Let $f(N) = N + N^2$.
To find the minimum value, we can complete the square or use calculus.
Completing the square: $f(N) = N^2 + N = (N^2 + N + \frac{1}{4}) - \frac{1}{4} = (N + \frac{1}{2})^2 - \frac{1}{4}$.
Since $(N + \frac{1}{2})^2 \ge 0$, the minimum value occurs when $(N + \frac{1}{2})^2 = 0$, which means $N = -\frac{1}{2}$.
Using calculus: $f(N) = N^2 + N$ $f'(N) = 2N + 1$ Setting $f'(N) = 0$, we get $2N + 1 = 0$, so $N = -\frac{1}{2}$.
$f''(N) = 2 > 0$, indicating a minimum.
Thus, the minimum value of $N + S$ occurs when $N = -\frac{1}{2}$.
Q73: $L=x+y$ where x and y are prime numbers. Which of the following statement/s is/are true? (i) The unit's digit of L cannot be 5 (ii) The units digit of L cannot be 0. (iii) L cannot be odd.
A. All three
B. Only iii
C. only ii
Correct Answer: B
Solution:
Let's analyze each statement: (i) The unit's digit of $L$ cannot be 5.
If the unit's digit of $L$ is 5, then $L$ must be divisible by 5.
Since 5 is a prime number, one of $x$ and $y$ must be 5.
If $x=5$, then $L = 5+y$.
If $y$ is any other prime number (other than 2), then the units digit of $L$ could be 5, such as $5+3 = 8$ (units digit 8), $5+7=12$ (units digit 2), but also $5+5=10$ (units digit 0), $5+2=7$ (units digit 7).
This statement is false.
(ii) The unit's digit of $L$ cannot be 0.
If the unit's digit of $L$ is 0, then $L$ is divisible by 10, and hence divisible by both 2 and 5.
This implies that $x$ and $y$ must be 2 and 5 (or vice versa) because 2 and 5 are the only prime numbers that can result in a multiple of 10.
Therefore, $L = 2 + 5 = 7$ (units digit 7), not 0.
Any other combination of primes will not yield a multiple of 10, so this statement is true.
(iii) $L$ cannot be odd.
If $x$ and $y$ are both odd prime numbers (all primes except 2), then their sum $L$ will be an even number.
However, if one of them is 2 and the other is an odd prime, then $L$ will be odd.
For example, if $x=2$ and $y=3$, then $L=5$, which is odd.
Therefore, this statement is false.
Solution:
Let's analyze each statement: (i) The unit's digit of $L$ cannot be 5.
If the unit's digit of $L$ is 5, then $L$ must be divisible by 5.
Since 5 is a prime number, one of $x$ and $y$ must be 5.
If $x=5$, then $L = 5+y$.
If $y$ is any other prime number (other than 2), then the units digit of $L$ could be 5, such as $5+3 = 8$ (units digit 8), $5+7=12$ (units digit 2), but also $5+5=10$ (units digit 0), $5+2=7$ (units digit 7).
This statement is false.
(ii) The unit's digit of $L$ cannot be 0.
If the unit's digit of $L$ is 0, then $L$ is divisible by 10, and hence divisible by both 2 and 5.
This implies that $x$ and $y$ must be 2 and 5 (or vice versa) because 2 and 5 are the only prime numbers that can result in a multiple of 10.
Therefore, $L = 2 + 5 = 7$ (units digit 7), not 0.
Any other combination of primes will not yield a multiple of 10, so this statement is true.
(iii) $L$ cannot be odd.
If $x$ and $y$ are both odd prime numbers (all primes except 2), then their sum $L$ will be an even number.
However, if one of them is 2 and the other is an odd prime, then $L$ will be odd.
For example, if $x=2$ and $y=3$, then $L=5$, which is odd.
Therefore, this statement is false.
Q74: XYZ is a 3 digit number such that when we calculate the difference between the two three digit numbers XYZYXZ the difference is exactly 90. How many possible values exist for the digits X and Y?
A. 9
B. 8
C. 7
D. 6
Correct Answer: D
Solution:
Let the number be represented as $N = XYZYXZ$.
This can be written as $N = 1000 \times (100X + 10Y + Z) + (100X + 10Y + Z) = 1001(100X + 10Y + Z)$.
Let $A = 100X + 10Y + Z$.
Then $N = 1001A$.
Another 3-digit number is given by $M = 100X + 10Y + Z$.
We are given that the difference between two three-digit numbers formed using the digits X, Y, Z is 90.
This seems to be a misinterpretation of the problem statement.
The question implies the difference between two numbers constructed using X, Y, Z is 90.
Let's assume the two numbers are $100X + 10Y + Z$ and $100X + 10Z + Y$.
The difference between these is $$(100X + 10Y + Z) - (100X + 10Z + Y) = 9Y - 9Z = 9(Y-Z) = 90$$ This simplifies to $Y - Z = 10$.
This equation is not possible since $Y$ and $Z$ are digits between 0 and 9.
Let's assume the numbers are $XYZ$ and $ZYX$.
The difference is: $(100X + 10Y + Z) - (100Z + 10Y + X) = 99X - 99Z = 99(X - Z) = 90$ This equation is also not possible because $99(X-Z) = 90$ implies $X - Z = \frac{90}{99} = \frac{10}{11}$, which is not an integer.
Let's reconsider the problem statement.
Perhaps the difference is between $XYZYXZ$ and $XYZ$.
Then $1001(100X + 10Y + Z) - (100X + 10Y + Z) = 1000(100X + 10Y + Z) = 90$.
This is impossible, as it would imply $100X + 10Y + Z = 0.09$, which is not a three-digit number.
The question is poorly phrased and likely contains an error.
There's no consistent interpretation leading to a valid solution.
Solution:
Let the number be represented as $N = XYZYXZ$.
This can be written as $N = 1000 \times (100X + 10Y + Z) + (100X + 10Y + Z) = 1001(100X + 10Y + Z)$.
Let $A = 100X + 10Y + Z$.
Then $N = 1001A$.
Another 3-digit number is given by $M = 100X + 10Y + Z$.
We are given that the difference between two three-digit numbers formed using the digits X, Y, Z is 90.
This seems to be a misinterpretation of the problem statement.
The question implies the difference between two numbers constructed using X, Y, Z is 90.
Let's assume the two numbers are $100X + 10Y + Z$ and $100X + 10Z + Y$.
The difference between these is $$(100X + 10Y + Z) - (100X + 10Z + Y) = 9Y - 9Z = 9(Y-Z) = 90$$ This simplifies to $Y - Z = 10$.
This equation is not possible since $Y$ and $Z$ are digits between 0 and 9.
Let's assume the numbers are $XYZ$ and $ZYX$.
The difference is: $(100X + 10Y + Z) - (100Z + 10Y + X) = 99X - 99Z = 99(X - Z) = 90$ This equation is also not possible because $99(X-Z) = 90$ implies $X - Z = \frac{90}{99} = \frac{10}{11}$, which is not an integer.
Let's reconsider the problem statement.
Perhaps the difference is between $XYZYXZ$ and $XYZ$.
Then $1001(100X + 10Y + Z) - (100X + 10Y + Z) = 1000(100X + 10Y + Z) = 90$.
This is impossible, as it would imply $100X + 10Y + Z = 0.09$, which is not a three-digit number.
The question is poorly phrased and likely contains an error.
There's no consistent interpretation leading to a valid solution.
Q75: Find the least number of 5 digits that is exactly divisible by 79
A. 10003
B. 10033
C. 10043
D. None of these
Correct Answer: B
Solution:
We need to find the smallest 5-digit number that is divisible by 79.
The smallest 5-digit number is 10000.
We will divide 10000 by 79 to find the remainder.
$$10000 \div 79 \approx 126.58$$ The quotient is 126.
The remainder is $10000 - (126 \times 79) = 10000 - 9954 = 46$.
To find the smallest 5-digit number divisible by 79, we need to add the difference between 79 and the remainder (46) to 10000.
The difference is $79 - 46 = 33$.
Therefore, the least 5-digit number divisible by 79 is $10000 + 33 = 10033$.
Solution:
We need to find the smallest 5-digit number that is divisible by 79.
The smallest 5-digit number is 10000.
We will divide 10000 by 79 to find the remainder.
$$10000 \div 79 \approx 126.58$$ The quotient is 126.
The remainder is $10000 - (126 \times 79) = 10000 - 9954 = 46$.
To find the smallest 5-digit number divisible by 79, we need to add the difference between 79 and the remainder (46) to 10000.
The difference is $79 - 46 = 33$.
Therefore, the least 5-digit number divisible by 79 is $10000 + 33 = 10033$.
Q76: A number when divided by 84 leaves a remainder of 57. What is the remainder when the same number is divided by 12?
A. 7
B. 8
C. 9
D. Cannot be determined
Correct Answer: D
Solution:
Let the number be $N$.
We are given that when $N$ is divided by 84, the remainder is 57.
This can be expressed as: $N = 84k + 57$, where $k$ is the quotient.
We want to find the remainder when $N$ is divided by 12.
We can rewrite the equation as: $N = 12(7k) + 12(4) + 9$ $N = 12(7k + 4) + 9$ Since $7k + 4$ is an integer, let's call it $m$.
Then: $N = 12m + 9$ This is in the form of dividend = divisor × quotient + remainder.
Therefore, when $N$ is divided by 12, the remainder is 9.
Solution:
Let the number be $N$.
We are given that when $N$ is divided by 84, the remainder is 57.
This can be expressed as: $N = 84k + 57$, where $k$ is the quotient.
We want to find the remainder when $N$ is divided by 12.
We can rewrite the equation as: $N = 12(7k) + 12(4) + 9$ $N = 12(7k + 4) + 9$ Since $7k + 4$ is an integer, let's call it $m$.
Then: $N = 12m + 9$ This is in the form of dividend = divisor × quotient + remainder.
Therefore, when $N$ is divided by 12, the remainder is 9.
Q77: 511 and 667 when divided by the same number, leave the same remainder. How many numbers can be used as the divisor in order to make this occur?
A. 14
B. 12
C. 10
D. 8
Correct Answer: B
Solution:
Let the same number be $x$.
When 511 is divided by $x$, the remainder is $r$.
When 667 is divided by $x$, the remainder is $r$.
This means that $511 = qx + r$ and $667 = px + r$ for some integers $p$ and $q$.
Subtracting the two equations, we get: $667 - 511 = px - qx$ $156 = x(p - q)$ This means that $x$ is a divisor of 156.
The divisors of 156 are 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156.
However, since the remainder must be the same and less than the divisor, the divisor must be greater than the remainder.
Also, the remainder must be less than the smaller number (511).
Let's find the divisors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156.
Since the remainder must be less than the divisor, we eliminate any divisors greater than 511.
All divisors of 156 are less than 511.
The number of divisors of 156 is the number of possible values of $x$.
To find the number of divisors, we find the prime factorization of 156: $156 = 2^2 \times 3 \times 13$.
The number of divisors is $(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12$.
Therefore, there are 12 numbers that can be used as the divisor.
Solution:
Let the same number be $x$.
When 511 is divided by $x$, the remainder is $r$.
When 667 is divided by $x$, the remainder is $r$.
This means that $511 = qx + r$ and $667 = px + r$ for some integers $p$ and $q$.
Subtracting the two equations, we get: $667 - 511 = px - qx$ $156 = x(p - q)$ This means that $x$ is a divisor of 156.
The divisors of 156 are 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156.
However, since the remainder must be the same and less than the divisor, the divisor must be greater than the remainder.
Also, the remainder must be less than the smaller number (511).
Let's find the divisors of 156: 1, 2, 3, 4, 6, 12, 13, 26, 39, 52, 78, 156.
Since the remainder must be less than the divisor, we eliminate any divisors greater than 511.
All divisors of 156 are less than 511.
The number of divisors of 156 is the number of possible values of $x$.
To find the number of divisors, we find the prime factorization of 156: $156 = 2^2 \times 3 \times 13$.
The number of divisors is $(2+1)(1+1)(1+1) = 3 \times 2 \times 2 = 12$.
Therefore, there are 12 numbers that can be used as the divisor.
Q78: The sum of two numbers is equal to thrice their difference. If the smaller of the numbers is 10 find the other number.
A. 15
B. 20
C. 40
D. None of these
Correct Answer: D
Solution:
Let the two numbers be $x$ and $y$, with $x > y$.
We are given that the sum of the two numbers is equal to thrice their difference.
This can be written as an equation: $x + y = 3(x - y)$ We are also given that the smaller number is 10, so $y = 10$.
Substituting this into the equation: $x + 10 = 3(x - 10)$ $x + 10 = 3x - 30$ $2x = 40$ $x = 20$ Therefore, the other number is 20.
Solution:
Let the two numbers be $x$ and $y$, with $x > y$.
We are given that the sum of the two numbers is equal to thrice their difference.
This can be written as an equation: $x + y = 3(x - y)$ We are also given that the smaller number is 10, so $y = 10$.
Substituting this into the equation: $x + 10 = 3(x - 10)$ $x + 10 = 3x - 30$ $2x = 40$ $x = 20$ Therefore, the other number is 20.
Q79: HCF of $3^{15}-1$ & $3^{25}-1$ is
Correct Answer: Cannot
Solution:
Let $A = 3^{15} - 1$ and $B = 3^{25} - 1$.
We use the identity $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + ...
+ xy^{n-2} + y^{n-1})$.
Then $A = 3^{15} - 1 = (3^5 - 1)(3^{10} + 3^5 + 1) = (3^5 - 1)(3^{10} + 3^5 + 1) = (243 - 1)(59049 + 243 + 1) = 242(59293) = 2 \times 11^2 \times 59293$.
Also, $B = 3^{25} - 1 = (3^5 - 1)(3^{20} + 3^{15} + 3^{10} + 3^5 + 1) = 242 (3^{20} + 3^{15} + 3^{10} + 3^5 + 1)$.
We have $3^5 - 1 = 242 = 2 \times 11^2$.
Let's find the HCF of $3^{15} - 1$ and $3^{25} - 1$.
We can use the property that $HCF(x^a - 1, x^b - 1) = x^{HCF(a, b)} - 1$.
Therefore, $HCF(3^{15} - 1, 3^{25} - 1) = 3^{HCF(15, 25)} - 1 = 3^5 - 1 = 243 - 1 = 242$.
Solution:
Let $A = 3^{15} - 1$ and $B = 3^{25} - 1$.
We use the identity $x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + ...
+ xy^{n-2} + y^{n-1})$.
Then $A = 3^{15} - 1 = (3^5 - 1)(3^{10} + 3^5 + 1) = (3^5 - 1)(3^{10} + 3^5 + 1) = (243 - 1)(59049 + 243 + 1) = 242(59293) = 2 \times 11^2 \times 59293$.
Also, $B = 3^{25} - 1 = (3^5 - 1)(3^{20} + 3^{15} + 3^{10} + 3^5 + 1) = 242 (3^{20} + 3^{15} + 3^{10} + 3^5 + 1)$.
We have $3^5 - 1 = 242 = 2 \times 11^2$.
Let's find the HCF of $3^{15} - 1$ and $3^{25} - 1$.
We can use the property that $HCF(x^a - 1, x^b - 1) = x^{HCF(a, b)} - 1$.
Therefore, $HCF(3^{15} - 1, 3^{25} - 1) = 3^{HCF(15, 25)} - 1 = 3^5 - 1 = 243 - 1 = 242$.
Q80: The LCM of two numbers is 421. What is the HCF of these two numbers?
Correct Answer: 440
Solution:
The question is insufficient to determine the HCF.
The relationship between the HCF and LCM of two numbers $a$ and $b$ is given by: $HCF(a, b) \times LCM(a, b) = a \times b$ We are given that $LCM(a, b) = 421$.
However, we don't know the values of $a$ and $b$, or their product.
Therefore, we cannot determine the HCF.
The question needs additional information, such as the value of either $a$ or $b$, or their product, to solve for the HCF.
421 is a prime number ($421 = 1 \times 421$), meaning its only divisors are 1 and itself.
This limits the possible pairs of numbers whose LCM is 421 to (1, 421) and (421, 1).
In both cases, the HCF is 1.
Solution:
The question is insufficient to determine the HCF.
The relationship between the HCF and LCM of two numbers $a$ and $b$ is given by: $HCF(a, b) \times LCM(a, b) = a \times b$ We are given that $LCM(a, b) = 421$.
However, we don't know the values of $a$ and $b$, or their product.
Therefore, we cannot determine the HCF.
The question needs additional information, such as the value of either $a$ or $b$, or their product, to solve for the HCF.
421 is a prime number ($421 = 1 \times 421$), meaning its only divisors are 1 and itself.
This limits the possible pairs of numbers whose LCM is 421 to (1, 421) and (421, 1).
In both cases, the HCF is 1.
Q81: If X is a natural number and X! ends with Y zeros then number of zeros at the end of (5X) is
Correct Answer: 5 173
Solution:
The number of zeros at the end of $X!$ is determined by the number of times 10 is a factor in $X!$.
Since $10 = 2 \times 5$, and there are always more factors of 2 than 5, we need to count the number of factors of 5 in $X!$.
This is given by Legendre's formula: $$ Z(X) = \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^i} \right\rfloor $$ where $Z(X)$ is the number of trailing zeros in $X!$ and $\lfloor x \rfloor$ denotes the floor function.
We are given that $Z(X) = Y$.
Now, we need to find the number of trailing zeros in $(5X)!$.
This is given by: $$ Z(5X) = \sum_{i=1}^{\infty} \left\lfloor \frac{5X}{5^i} \right\rfloor = \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^{i-1}} \right\rfloor = \left\lfloor \frac{X}{1} \right\rfloor + \left\lfloor \frac{X}{5} \right\rfloor + \left\lfloor \frac{X}{25} \right\rfloor + \dots $$ We can rewrite this as: $$ Z(5X) = X + \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^i} \right\rfloor = X + Z(X) $$ Since $Z(X) = Y$, the number of zeros at the end of $(5X)!$ is $X + Y$.
However, we don't know the value of $X$.
The question is flawed as it requires the value of X to give a numerical answer, or it needs to be rephrased to express the answer in terms of X and Y.
The number of zeros at the end of $(5X)!$ is $X + Y$.
Solution:
The number of zeros at the end of $X!$ is determined by the number of times 10 is a factor in $X!$.
Since $10 = 2 \times 5$, and there are always more factors of 2 than 5, we need to count the number of factors of 5 in $X!$.
This is given by Legendre's formula: $$ Z(X) = \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^i} \right\rfloor $$ where $Z(X)$ is the number of trailing zeros in $X!$ and $\lfloor x \rfloor$ denotes the floor function.
We are given that $Z(X) = Y$.
Now, we need to find the number of trailing zeros in $(5X)!$.
This is given by: $$ Z(5X) = \sum_{i=1}^{\infty} \left\lfloor \frac{5X}{5^i} \right\rfloor = \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^{i-1}} \right\rfloor = \left\lfloor \frac{X}{1} \right\rfloor + \left\lfloor \frac{X}{5} \right\rfloor + \left\lfloor \frac{X}{25} \right\rfloor + \dots $$ We can rewrite this as: $$ Z(5X) = X + \sum_{i=1}^{\infty} \left\lfloor \frac{X}{5^i} \right\rfloor = X + Z(X) $$ Since $Z(X) = Y$, the number of zeros at the end of $(5X)!$ is $X + Y$.
However, we don't know the value of $X$.
The question is flawed as it requires the value of X to give a numerical answer, or it needs to be rephrased to express the answer in terms of X and Y.
The number of zeros at the end of $(5X)!$ is $X + Y$.
Q82: If $A=n^{2n^{n}}$, $B=n^{n^{2n}}$, $C=(n^{2n})^{n}$, $D=(n^{n})^{n^{2}}$ when n is a natural number & $n\ne1$. Then arrange them in terms of their values.
Correct Answer: 4 175
Solution:
We need to simplify A, B, C, and D using the laws of indices.
$A = n^{2n^n}$ $B = n^{n^{2n}} = n^{(n^{2n})}$ $C = (n^{2n})^n = n^{2n \times n} = n^{2n^2}$ $D = (n^n)^{n^2} = n^{n \times n^2} = n^{n^3}$ Now let's compare the exponents: $2n^n$, $n^{2n}$, $2n^2$, $n^3$ Let's consider $n=2$ as an example: $A = 2^{2(2^2)} = 2^{2(4)} = 2^8 = 256$ $B = 2^{2^{2(2)}} = 2^{2^4} = 2^{16} = 65536$ $C = 2^{2(2^2)} = 2^8 = 256$ $D = 2^{2^3} = 2^8 = 256$ For $n=2$, $B > A = C = D$.
Let's consider $n=3$: $A = 3^{2(3^3)} = 3^{2(27)} = 3^{54}$ $B = 3^{3^{2(3)}} = 3^{3^6} = 3^{729}$ $C = 3^{2(3^2)} = 3^{18}$ $D = 3^{3^3} = 3^{27}$ For $n=3$, $B > A > D > C$.
In general, since the exponential tower grows very fast, the order is $B > A > D > C$ for $n>1$.
This is because the exponent in B grows fastest.
A and C have similar growth but A is always larger.
D grows relatively slower.
Solution:
We need to simplify A, B, C, and D using the laws of indices.
$A = n^{2n^n}$ $B = n^{n^{2n}} = n^{(n^{2n})}$ $C = (n^{2n})^n = n^{2n \times n} = n^{2n^2}$ $D = (n^n)^{n^2} = n^{n \times n^2} = n^{n^3}$ Now let's compare the exponents: $2n^n$, $n^{2n}$, $2n^2$, $n^3$ Let's consider $n=2$ as an example: $A = 2^{2(2^2)} = 2^{2(4)} = 2^8 = 256$ $B = 2^{2^{2(2)}} = 2^{2^4} = 2^{16} = 65536$ $C = 2^{2(2^2)} = 2^8 = 256$ $D = 2^{2^3} = 2^8 = 256$ For $n=2$, $B > A = C = D$.
Let's consider $n=3$: $A = 3^{2(3^3)} = 3^{2(27)} = 3^{54}$ $B = 3^{3^{2(3)}} = 3^{3^6} = 3^{729}$ $C = 3^{2(3^2)} = 3^{18}$ $D = 3^{3^3} = 3^{27}$ For $n=3$, $B > A > D > C$.
In general, since the exponential tower grows very fast, the order is $B > A > D > C$ for $n>1$.
This is because the exponent in B grows fastest.
A and C have similar growth but A is always larger.
D grows relatively slower.
Q83: How many numbers in the form of $2^{n}-1$, which are less than 5000 are prime?
Correct Answer:
Solution:
We are looking for prime numbers of the form $2^n - 1$ which are less than 5000.
Numbers of this form are called Mersenne numbers.
A Mersenne prime is a prime number that is one less than a power of two.
That is, it is a prime number of the form $M_n = 2^n - 1$ for some integer $n$.
Let's test values of $n$: \begin{itemize} \item If $n=1$, $2^1 - 1 = 1$, which is not prime.
\item If $n=2$, $2^2 - 1 = 3$, which is prime.
\item If $n=3$, $2^3 - 1 = 7$, which is prime.
\item If $n=4$, $2^4 - 1 = 15 = 3 \times 5$, which is not prime.
\item If $n=5$, $2^5 - 1 = 31$, which is prime.
\item If $n=6$, $2^6 - 1 = 63 = 3^2 \times 7$, which is not prime.
\item If $n=7$, $2^7 - 1 = 127$, which is prime.
\item If $n=8$, $2^8 - 1 = 255 = 3 \times 5 \times 17$, which is not prime.
\item If $n=9$, $2^9 - 1 = 511 = 7 \times 73$, which is not prime.
\item If $n=10$, $2^{10} - 1 = 1023 = 3 \times 11 \times 31$, which is not prime.
\item If $n=11$, $2^{11} - 1 = 2047 = 23 \times 89$, which is not prime.
\item If $n=12$, $2^{12} - 1 = 4095 = 3^2 \times 5 \times 7 \times 13$, which is not prime.
\end{itemize} We can observe that if $n$ is composite, then $2^n - 1$ is also composite.
If $n$ is a composite number $ab$ ($a, b > 1$), then $2^n - 1 = 2^{ab} - 1 = (2^a)^b - 1$, which is divisible by $2^a - 1$.
Therefore, $n$ must be prime.
Let's check prime values of $n$: For $n=13$, $2^{13}-1 = 8191$, which is prime.
For $n=17$, $2^{17}-1 = 131071$, which is prime, but > 5000.
The primes are 3, 7, 31, 127, 8191.
Only 3, 7, 31, 127 are less than 5000.
Solution:
We are looking for prime numbers of the form $2^n - 1$ which are less than 5000.
Numbers of this form are called Mersenne numbers.
A Mersenne prime is a prime number that is one less than a power of two.
That is, it is a prime number of the form $M_n = 2^n - 1$ for some integer $n$.
Let's test values of $n$: \begin{itemize} \item If $n=1$, $2^1 - 1 = 1$, which is not prime.
\item If $n=2$, $2^2 - 1 = 3$, which is prime.
\item If $n=3$, $2^3 - 1 = 7$, which is prime.
\item If $n=4$, $2^4 - 1 = 15 = 3 \times 5$, which is not prime.
\item If $n=5$, $2^5 - 1 = 31$, which is prime.
\item If $n=6$, $2^6 - 1 = 63 = 3^2 \times 7$, which is not prime.
\item If $n=7$, $2^7 - 1 = 127$, which is prime.
\item If $n=8$, $2^8 - 1 = 255 = 3 \times 5 \times 17$, which is not prime.
\item If $n=9$, $2^9 - 1 = 511 = 7 \times 73$, which is not prime.
\item If $n=10$, $2^{10} - 1 = 1023 = 3 \times 11 \times 31$, which is not prime.
\item If $n=11$, $2^{11} - 1 = 2047 = 23 \times 89$, which is not prime.
\item If $n=12$, $2^{12} - 1 = 4095 = 3^2 \times 5 \times 7 \times 13$, which is not prime.
\end{itemize} We can observe that if $n$ is composite, then $2^n - 1$ is also composite.
If $n$ is a composite number $ab$ ($a, b > 1$), then $2^n - 1 = 2^{ab} - 1 = (2^a)^b - 1$, which is divisible by $2^a - 1$.
Therefore, $n$ must be prime.
Let's check prime values of $n$: For $n=13$, $2^{13}-1 = 8191$, which is prime.
For $n=17$, $2^{17}-1 = 131071$, which is prime, but > 5000.
The primes are 3, 7, 31, 127, 8191.
Only 3, 7, 31, 127 are less than 5000.
1234
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