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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Remainder PYQ

MCA NIMCET PYQ
The remainder when 231 is divided by 5 is





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

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

MCA NIMCET PYQ
Find the remainder when $X = 1! + 2! + 3! + \cdots + 100!$ is divided by $240$.





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2009 PYQ

Solution

For $n \ge 6$, 
$n!$ is divisible by $240 = 2^4 \cdot 3 \cdot 5$. 
 Thus only first 5 factorials matter: 
 $1! + 2! + 3! + 4! + 5!$ 
 $= 1 + 2 + 6 + 24 + 120 = 153$
MCA NIMCET PYQ
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Required number
 = H.C.F. of (1657 - 6) and (2037 - 5)   
 = H.C.F. of 1651 and 2032 = 127.
MCA NIMCET PYQ
The integers 34041 and 32506, when divided by a 3 - digit integer n, leave the same remainder. What can be the value of n?





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

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. 
MCA NIMCET PYQ
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





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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2008 PYQ

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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