Question

The leaste mutiple of 7 which leaves a remainder of 4 when divided by 6 9 15 18​

Answer 1

Answer:

364

Step-by-step explanation:

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.

∴ Required number =(90×4)+4=364

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