Question

find the smallest number which when divided by 12 15 18 and 27 leaves as reminder 8 11 14 and 23 respectively​

Answer 1

Answer:

536

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

Answer 2

Answer:

The number is 536.

Explanation:

Given :

Divisors = 12, 15, 18, 27

Remainder = 8, 11, 14, 23

To find :

smallest possible number divisible by given divisors with respective remainders.

Solution :

For 12,

12 = 2 x 2 x 3

For 15,

15 = 3 x 5

For 18,

18 = 2 x 3 x 3

For 27,

27 = 3 x 3 x 3

Now,

LCM of the numbers 12, 15, 18, 27 = 540

Also,

Difference between divisor and remainder = 4

∴ 540 - 4 = 536

∴ The number is 536.

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