CAT Remainders Questions

Master CAT Remainders Questions with practice questions and detailed solutions.

Question 1.

When $10^{100}$ is divided by 7, the remainder is

Video Explanation

Text Explanation:

To find the value of $10^{100}mod\left(7\right)$

When 10 is divided by 7, it leaves a remainder 3, so the above equation can be written as,

$3^{100}mod\left(7\right)$

Now looking at the cyclicality of powers of 3 when divided by 7,

$3^1$ mod $7=3$

$3^2$ mod $7=2$

$3^3$ mod $7=6$

$3^4$ mod $7=4$

$3^5$ mod $7=5$

$3^6$ mod $7=1$

From this calculation, it is evident that the powers of 3 modulo 7 repeat every 6 steps. This forms a cycle: 3, 2, 6, 4, 5, 1

$3^{100}=\left(3^6\right)^{16}\times\ \left(3^4\right)$

Since $3^6$ mod $7=1$

We just need to consider $3^4$ mod $7$ which equals 4

Hence the answer is 4.

Question 2.

When $3^{333}$ is divided by 11, the remainder is

Video Explanation

Text Explanation:

There are multiple ways of solving these sorts of questions. One method is to look for powers of the term in the numerator that leave a remainder of 1 or -1 when divided by the denominator.

Noting down the powers of 3, 3, 9, 27, 81, 243

243 is one such number, 242 is multiple of 11 (11 times 22), hence 243 will leave a remainder of 1 when divided by 11.

243 is 3 raised to power 5; we can rewrite the given term as $\dfrac{3^{330}\times\ 3^3}{11}$

The overall remainder will be $\left[\dfrac{3^{330}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$\left[\dfrac{3^{5\times\ 66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$\left[\dfrac{243^{66}}{11}\right]_R\times\ \left[\dfrac{3^3}{11}\right]_R$

$1^{66}\times\ \left[\dfrac{27}{11}\right]_R$

$1\times\ 5$

$5$

Therefore, Option A is the correct answer.

Question 3.

If $10^{68}$ is divided by 13, the remainder is

Video Explanation

Text Explanation:

There are multiple ways of solving such questions involving remainders; one easy way is to look for a power of numerator that leaves a remainder of 1 or -1 when divided by the denominator.

In this instance, 1000, when divided by 13, leaves a remainder of -1

We can rewrite the numerator as $\frac{10^{66}\times\ 100}{13}$

The remainder would be $\left[\frac{10^{66}}{13}\right]_R\times\ \left[\frac{100}{13}\right]_R$

$\left(-1\right)^{22}\times\ 9$

Therefore, Option C is the correct answer.

Question 4.

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

Video Explanation

Question 5.

Text Explanation:

At $x = 0, 2^x = 1$ which is in the given range [0.25, 200]
$2^x + 2$ = 1 + 2 = 3 Which is divisible by 3. Hence, x = 0 is one possible solution.
At $x = 1, 2^x = 2$ which is in the given range [0.25, 200]
$2^x + 2$ = 2 + 2 = 3 Which is divisible by 4. Hence, x = 1 is one possible solution.
At $x = 2, 2^x = 4$ which is in the given range [0.25, 200]
$2^x + 2$ = 4 + 2 = 6 Which is divisible by 3. Hence, x = 2 is one possible solution.
At $x = 3, 2^x = 8$ which is in the given range [0.25, 200]
$2^x + 2$ = 8 + 2 = 3 Which is not divisible by 3 or 4. Hence, x = 3 can't be a solution.
At $x = 4, 2^x = 16$ which is in the given range [0.25, 200]
$2^x + 2$ = 16 + 2 = 18 Which is divisible by 3. Hence, x = 4 is one possible solution.
At $x = 5, 2^x = 32$ which is in the given range [0.25, 200]
$2^x + 2$ = 32 + 2 = 34 Which is not divisible by 3 or 4. Hence, x = 5 can't be a solution.
At $x = 6, 2^x = 64$ which is in the given range [0.25, 200]
$2^x + 2$ = 64 + 2 = 66 Which is divisible by 3. Hence, x = 6 is one possible solution.
At $x = 7, 2^x = 128$ which is in the given range [0.25, 200]
$2^x + 2$ = 128 + 2 = 130 Which is not divisible by 3 or 4. Hence, x = 7 can't be a solution.
At $x = 8, 2^x = 256$ which is not in the given range [0.25, 200]. Hence, x can't take any value greater than 7.
Therefore, all possible values of x = {0,1,2,4,6}. Hence, we can say that 'x' can take 5 different integer values.

Question 6.

Instructions

Instructions
For the following questions answer them individually

Question 7.

Let $n!=123* ...n$ for integer $n \geq 1$.
If $p = 1!+(22!)+(33!)+... +(1010!)$, then $p+2$ when divided by 11! leaves a remainder of

Question 8.

Question 9.

Text Explanation:

Since after division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively, the number is of form ((((4*4)+1)*3)+2)k = 53K
Let k = 1; the number becomes 53
If it is divided by 84, the remainder is 53.
Option d) is the correct answer.
Alternative Solution.
Consider only for 3 and 4 and the remainders are 2 and 1 respectively.
So 5 is the first number to satisfy both the conditions. The number will be of the form 12k+5. Put different integral values of k to find whether it will leave remainder 5 when divided by 7. So the first number to satisfy such condition is 48x4+5= 53

Question 10.

Question 11.

Text Explanation:

We know that a=$b^2-b$.
So$a^2-a$ = b($b^3-2b^2-b+2$) . = (b - 2)(b - 1)( b)(b + 1)
The above given is a product of 4 consecutive numbers with the lowest number of the product being 2(given b >= 4)
In any set of four consecutive numbers, one of the numbers would be divisible by 3 and there would be two even numbers with the minimum value of the pair being (2,4).
Thus, for any value of b >=4, $a^2-4$ would be divisible by 3 x 2 x 4 = 24.
Thus, option C is the right choice. Options A and B are definitely wrong as a set of four consecutive numbers need not always include a multiple of 5 eg:(6,7,8,9)

Question 12.

Question 13.

Text Explanation:

Between 100 and 200 both included there are 51 even nos. There are 7 even nos which are divisible by 7 and 6 nos which are divisible by 9 and 1 no divisible by both. hence in total 51 - (7+6-1) = 39
There is one more method through which we can find the answer. Since we have to find even numbers, consider the numbers which are divisible by 14, 18 and 126 between 100 and 200. These are 7, 6 and 1 respectively.

Question 14.

Question 15.

Question 16.

Question 17.

Question 18.

Question 19.

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CAT Remainders Questions

Question 1.

When $10^{100}$ is divided by 7, the remainder is

Question 2.

When $3^{333}$ is divided by 11, the remainder is

Question 3.

If $10^{68}$ is divided by 13, the remainder is

Question 4.

How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7?

Question 5.

<div>The number of integers x such that $0.25 \leq 2^x \leq 200$ and $2^x + 2$ is perfectly divisible by either 3 or 4, is</div>

Question 6.

<div>If x = $(16^3 + 17^3+ 18^3+ 19^3 )$, then x divided by 70 leaves a remainder of</div>

Instructions

Instructions For the following questions answer them individually

Question 7.

Let $n!=1*2*3* ...*n$ for integer $n \geq 1$.If $p = 1!+(2*2!)+(3*3!)+... +(10*10!)$, then $p+2$ when divided by 11! leaves a remainder of

Question 8.

<div>When $2^{256}$ is divided by 17, the remainder would be</div>

Question 9.

<div>After the division of a number successively by 3, 4 and 7, the remainders obtained are 2, 1 and 4 respectively. What will be the remainder if 84 divides the same number?</div>

Question 10.

<div>$7^{6n} - 6^{6n}$, where n is an integer &gt; 0, is divisible by</div>

Question 11.

<div>Let $b$ be a positive integer and $a = b^2 - b$. If $b \geq 4$ , then $a^2 - 2a$ is divisible by</div>

Question 12.

<div>Let n be the number of different five-digit numbers, divisible by 4 with the digits 1, 2, 3, 4, 5 and 6, no digit being repeated in the numbers. What is the value of n?</div>

Question 13.

<div>How many even integers n, where $100 \leq n \leq 200$ , are divisible neither by seven nor by nine?</div>

Question 14.

<div>The number of positive integers n in the range $12 \leq n \leq 40$ such that the product (n -1)*(n - 2)*…*3*2*1 is not divisible by n is</div>

Question 15.

<div>Let N = 1421 * 1423 * 1425. What is the&nbsp;remainder when N is divided by 12?</div>

Question 16.

<div>The integers 34041 and 32506 when divided by a three-digit integer n leave the same remainder. What is n?</div>

Question 17.

<div>Let N = $55^3 + 17^3 - 72^3$. N is divisible by:</div>

Question 18.

<div>The remainder when $7^{84}$ is divided by $342$&nbsp;is :</div>

Question 19.

<div>A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.</div>

Free CAT Quant Questions

Instructions
For the following questions answer them individually

Let $n!=123* ...n$ for integer $n \geq 1$.
If $p = 1!+(22!)+(33!)+... +(1010!)$, then $p+2$ when divided by 11! leaves a remainder of