Find the least number which when divided by 6,15 and 18 leave remainder 5 in each case.

Given: 

6, 15 and 18.

To do: 

We have to find the smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case.

Solution:

The smallest number that when divided by 6, 15 and 18 leaves a remainder of 0 in each case is the LCM of 6, 15 and 18.

So,

The smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=$ LCM of 6, 15 and 18 $+$ 5

Now,

Finding LCM of 6, 15 and 18:

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

Prime factorization of 6:

$6=2\times3$

Prime factorization of 15:

$15=3\times5$

Prime factorization of 18:

$18=2\times3\times3$

LCM of 6, 15 and 18 $=2\times3\times3\times5$

$=90$

Thus,

LCM$(6, 15, 18) = 90$

The smallest number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=$ LCM of 6, 15 and 18 $+$ 5

The least number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is $=90+5$

$=95$

The least number that when divided by 6, 15 and 18 leaves a remainder of 5 in each case is 95.