Question

Find the smallest number which when divided by 48 and 60, leaves a remainder 7 in each case but when divided by 13 leaves no remainder. Find the sum of the digits in this required smallest number.
1. 13
2. 14
3. 15
4. 11

Answer 1

Correct Answer - Option 1 : 13

Given∶

A number when divided by 48 and 60, leaves a remainder 7 in each case.

But when divided by 13 leaves no remainder.

Formula Used∶

Concept of LCM and Hit and Trial method

Calculation∶

LCM of 48 and 60 = 2⁴ × 3 × 5 = 240

Let the required number be N

So, N = 240 × x + 7

Now use Hit and Trial method

When, x = 1

Then, N = 240 × 1 + 7 = 240 + 7 = 247

Hence, sum of the digits = 2 + 4 + 7 = 13

∴ 13 is the sum of the digits in the required smallest number which when divided by 48 and 60 leaves a remainder 7, but when divided by 13, the remainder is zero.