Find the smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case.

Given: 12, 15 and 18.

To find: Here we have to find the smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case.

Solution:

The smallest number that when divided by 12, 15 and 18 leaves a remainder of 0 in each case is the LCM of 12, 15 and 18.

So,

The smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case is = LCM of 12, 15 and 18 $+$ 3

Now,

Finding LCM of 12, 15 and 18:

Writing all the numbers as a product of their prime factors:

Prime factorization of 12:

• 2 $\times$ 2 $\times$ 3 = 2² $\times$ 3¹

Prime factorization of 15:

• 3 $\times$ 5 = 3¹ $\times$ 5¹

Prime factorization of 18:

• 2 $\times$ 3 $\times$ 3 = 2¹ $\times$ 3²

Multiplying highest power of each prime number:

• 2² $\times$ 3² $\times$ 5¹ = 180

Thus,

LCM(12, 15, 18) = 180

The smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case is = LCM of 12, 15 and 18 $+$ 3

The smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case is = 180 $+$ 3

The smallest number that when divided by 12, 15 and 18 leaves a remainder of 3 in each case is = 183

So, the required number is 183.

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Updated on: 10-Oct-2022

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