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NIMCET Previous Year Questions (PYQs)

NIMCET Remainder PYQ

NIMCET PYQ
The remainder when 231 is divided by 5 is





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NIMCET Previous Year PYQNIMCET 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

NIMCET PYQ
The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:





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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ

Solution

Required number
 = H.C.F. of (1657 - 6) and (2037 - 5)   
 = H.C.F. of 1651 and 2032 = 127.
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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NIMCET Previous Year PYQNIMCET 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. 

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