Question

Find the least number which on being divided by 5, 6, 8, 9, 12 leaves in each case a remainder 1 but when divided by 13 leaves no remainder?

• A. 3601
• B. 1469
• C. 2091
• D. 4879

Answer 1

The correct option is A 3601

LCM of 5, 6, 8, 9 and 12$=2\times 2\times 2\times 3\times 3\times 5=360$
$\therefore Requirednumber=(360k+1)$
(360k+1) is divisible by 13 then k=10
Required number $=360\times 10+1=3601$