Find the least number which when divided by 6, 15, 17 leaves a remainder 1, but when divided by 7 leaves no remainder.
Explanation:
Any number, which when divided by 6, 15 and 17 leaves a remainder of 1, is of the form {k × LCM(6, 15, 17) + 1}, where = 1, 2, 3… and so on.
k × LCM(6, 15, 17) + 1 = 510k + 1.
It is given that (510k + 1) is a multiple of 7.
When k = 1, we get 511, which is a multiple of 7.
Hence, 511.
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