1.

What is the least number which when divided by 8, 12 and 16 leaves 3 as the remainder in each case, but when divided by 7 leaves no any remainder?

Answer»

SOLUTION :The POSSIBLE value = m (LCM of 8 , 12 and 16) + 3 = m (48) + 3
Now put the value of m such that "m (48) = 3" becomes DIVISIBLE by 7 . So `3 xx 48 + 3 = 147` which is the least possible required number ?


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