Question

6. Find the smallest number which when divided by 12, 15, 18 and 27 leaves as
remainder 8, 11, 14 and 23 respectively,​

Answer 1

Answer:

hey

Step-by-step explanation:

Prime factorizations first.

12 = 2 x 2 x 3

15 = 3 x 5

18 = 2 x 3 x 3

27 = 3 x 3 x 3

LCM = 2 x 2 x 3 x 3 x 3 x 5 = 540

Since all of the remainders are 4 less than the divisors, the number will be 540 - 4 = 536

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