SSC JE Mechanical Engineering 25th May 2014 Shift-2 Question 18

Question 18

Find the smallest number which when divided by 25, 40, or 56 has in each case 13 as remainder.

Solution

Smallest number = (LCM of 25, 40 and 56) + remainder

Factor of 25 = $$5^2$$

Factor of 40 = $$2^3.5$$

Factor of 56 = $$2^3.7$$

LCM of 25, 40 and 56 = $$2^3.5^2.7$$ = 1400

Smallest number = 1400 + 13 = 1413



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