NIMCET 2027 Remainder Previous Year Questions (PYQs)
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Showing 5 questions
The remainder when
231
is divided by 5 is
1
2
3
4
Solution
Remainder of 231 Divided by 5
To find the remainder of 231 divided by 5, we look at the cycle of powers of 2 mod 5: 2, 4, 3, 1 (repeats every 4 terms).
31 mod 4 = 3, so 231 corresponds to the 3rd term, which is 3.
Answer: 3
Find the remainder when
$X = 1! + 2! + 3! + \cdots + 100!$
is divided by $240$.
1
2
3
4
Solution
For $n \ge 6$,
$n!$ is divisible by $240 = 2^4 \cdot 3 \cdot 5$.
Thus only first 5 factorials matter:
$1! + 2! + 3! + 4! + 5!$
$= 1 + 2 + 6 + 24 + 120 = 153$
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
1
2
3
4
Solution
Required number
= H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
The integers 34041 and 32506, when divided by a 3 - digit integer n, leave the same remainder. What can be the value of n?
1
2
3
4
Solution
Find the difference between both 34041 and 32506.
34041 - 32506 = 1535
So n should be a factor of 1535.
Factors of 1535:-
1, 5, 307, 1535
Here, 307 is a three digit factor of 1535. So n = 307
34041 and 32506 divided by 307 leaves remainder 271.
So 307 is the answer.
If both $7^2$ and $3^3$ are factors of
$a\times11^3\times6^2\times13^{11}$,
then the smallest possible value of $a$ is
1
2
3
4
Solution
$6^2=(2^2)(3^2)$ gives only $3^2$.
To make $3^3$, one more $3$ is needed, and $7^2$ is missing completely.
So $a=3\times7^2=147$.
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