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# 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 = 2
^{2}$\times$ 3^{1}

Prime factorization of 15:

- 3 $\times$ 5 = 3
^{1}$\times$ 5^{1}

Prime factorization of 18:

- 2 $\times$ 3 $\times$ 3 = 2
^{1}$\times$ 3^{2}

Multiplying highest power of each prime number:

- 2
^{2}$\times$ 3^{2}$\times$ 5^{1}= 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.