The least number which when divided by 2, 3, 4, 5 and 6 leaves the remainder 1 in each case. If the same number is divided by 7 it leaves no remainder. The number is

231

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301

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371

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441

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Solution

The correct option is A 301
The least common multiple of 2, 3, 4, 5, 6 is their LCM $= 60$ .
So, the answer is the least number such that it is a multiple of 7 and a multiple of 60 leaving remainder 1.
$∴$ The required answer is 301.

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Find the least number which when divided by 2, 3, 4, 5 and 6 leaves a reminder of 1 in each case but when divided by 7 leaves no remainder.

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