Find the least number which when divided by 12, 24, 36 and 40 leaves a remainder 1 but when divided by 7 leaves no remainder

361

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1080

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721

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371

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Solution

The correct option is C 721
LCM of 12, 24, 36 and 40 $= 360$
$∴$ The required number is of the form $= 360 k + 1$
Last value of k for which (360k+1) is divisible by 7 is $k = 2$
$∴ R e q u i r e d n u m b e r = 360 \times 2 + 1 = 721$

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