Question

How many numbers lie between 11 and 1111 which when divided by 9 leave a remainder of 6 and when divide by 21 leave a remainder of 12? ___

Answer 1

Let the possible number be N then it can be expressed as
N = 9k + 6
and N = $21l$ + 12
$\therefore$ 9k + 6 = $21l$ + 12
$\Rightarrow$ 9k - $21l$ = 6
or 3(3k - $7l$) = 6
or 3k = $7l$ + 2 or k$=\frac{7l+2}{3}$
So put the min. possible value of $l$ such that the value of k is an integer or in other words numerator (i. e., $7l$+ 2) will be divisible by 3.
Thus at $I$ =1, we get k = 3 (an integer). So the least possible number N = 9 $\times$ 3 + 6 = 21 $\times$ 1 + 12 = 33.
Now the higher possible values can be obtained by adding 33 in the multiples of LCM of 9 and 21. i.e., The general form of the number is 63m + 33. So the other number in the given range including 33 are 96, 159, 222, 285, 348, ..., 1104. Hence there are total 18 numbers which satisfy the given condition.