Question

find smallest number which when divided by 12 18 and 32 leaves remainder 5 in each case​

Answer 1

Answer:

Step-by-step explanation:

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Answer 2

The smallest number which when divided by 12 18 and 32 leaves remainder 5 in each case​ is 293

Solution:

Given that,

We have to find smallest number which when divided by 12 18 and 32 leaves remainder 5 in each case​

Thus,

smallest number = (LCM of 12, 18, 32 ) + 5

Find LCM of 12, 18, 32

List all prime factors for each number.

Prime Factorization of 12 is:  2 x 2 x 3

Prime Factorization of 18 is:  2 x 3 x 3  

Prime Factorization of 32 is:  2 x 2 x 2 x 2 x 2  

For each prime factor, find where it occurs most often as a factor and write it that many times in a new list.

2, 2, 2, 2, 2, 3, 3

Multiply these factors together to find the LCM.

LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

Therefore,

smallest number = (LCM of 12, 18, 32 ) + 5

smallest number = 288 + 5 = 293

Thus smallest number which when divided by 12 18 and 32 leaves remainder 5 in each case​ is 293

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