Question 56

What is the smallest three-digit number which, when divided by 8 or 6, leaves a remainder of 1 in each case?

Solution

The LCM of 8 and 6 is 24.

So the required number = 24n+1 [Here 'n' is a integer number and 1 is added because it leaves the remainder.]

Here n = 1, 2, 3 and 4 will give a number that has two digits.

When n = 5, then the required number = $$24\times5+1$$ = 120+1 = 121

So 121 is the smallest three-digit number which, when divided by 8 or 6, leaves a remainder of 1 in each case.

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