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?

3601

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1469

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2091

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4879

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Solution

The correct option is A 3601
LCM of 5, 6, 8, 9 and 12 $= 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360$
$∴ R e q u i r e d n u m b e r = (360 k + 1)$
(360k+1) is divisible by 13 then k=10
Required number $= 360 \times 10 + 1 = 3601$

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