Problem #88 · October 23, 2026

EKUB greatest common divisor and EKUK least common multiple

Find the smallest three-digit number which, when divided by both 12 and 15, leaves a remainder of 7 in each case.

WORKED EXAMPLE

Solution

If we subtract 7 from the required number, the result is divisible by both 12 and 15 without a remainder, that is, it is a multiple of

EKUK(12; 15) = 60

. So the number has the form

60k + 7

; the smallest three-digit one occurs at

k = 2

120 + 7 = 127

. Check:

127 = 10 \cdot 12 + 7 = 8 \cdot 15 + 7