Which is the smallest multiple of 7, which leaves 5 as the remainder in each case, when divided by 8,9,12, and 15?
Show answer & explanation
Correct answer: B
Explanation
The correct answer is option 2 i.e. 1085.
LCM of 8, 9, 12, and 15 = 1080
Let, the required number is of form 1080k + 5
When divided by 7, 1080k + 5 leaves a remainder of 5
⇒ (1080k + 5)/7 = 1078k/7 + (2k + 5)/7
When k = 1, (2k + 5)/7 is divisible by 7
Thus, the smallest such number is (1080 × 1 + 5) = 1085