RRB NTPC 27 April 2016 - Shift 2

Question 45

Find the greatest number which on dividing 391 and 318 leaves remainders 7 and 6 respectively.

Solution

Since on dividing 391 by the required no. the remainder is 7 
then 391-7 ie. 384 will be exactly divisible by the required no.
Similarly,318-6 ie. 312 will be exactly divisible by the required no.

Now we will do prime factorisation of 384 and 312,

384 = $$2^2\times\ 3$$

312 = $$2^3\times\ 3\times\ 13$$

H.C.F = $$2^3\times\ 3=24$$

Required number is 24.

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