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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 = 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.