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The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

Asked Sep 07, 2023 Modified Sep 07, 2023 Viewed 0 times
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Mathematics
asked Sep 07, 2023 at 09:12
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L.C.M. of 6, 9, 15 and 18 is 90 Let required number be 90k + 4, which is multiple of 7 Least value of k for which (90k + 4) is divisible by 7 is k = 4 So Required number = (90 x 4) + 4 = 364
answered Sep 07, 2023 at 09:12
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