Question

Which is the smallest six digit number divisible by 111?
$
  (a){\text{ 111111}} \\
  (b){\text{ 110011}} \\
  (c){\text{ 100011}} \\
  (d){\text{ 100111}} \\
 $

Answer 1

Hint: In this question form the smallest possible 6 digit number first and check whether the number formed is divisible by 111 or not, if not then see how much remainder is left behind and use it to find the number that we need to add to this least 6 digit number in order to make it divisible by 111.

Complete step-by-step answer:
The possible digits are (0, 1, 2, 3, 4, 5, 6, 7, 8 and 9)
Now as we know that the number cannot start from zero otherwise the number converts into a five digit number rather than six digit number for example (012345 = 12345).
So the smallest six digit number starts from 1 and the rest of the terms are filled by the least digit which is zero.
So the smallest six digit number is = 100000.
Now we have to find out the smallest six digit number which is exactly divisible by 111.
So first divide 100000 by 111 and calculate quotient and remainder.
$ \Rightarrow 100000 = 900\dfrac{{100}}{{111}}$
So the quotient is 900 and the remainder is 100.
So as we know 111 = (100 + 11), if we add 11 in the remainder then the number which is formed is the least six digit number which is exactly divisible by 111.
Or we can say add 11 in the least six digit number so that the number which is formed is the least six digit number which is exactly divisible by 111.
So the least number is (100000 + 11) = 100011.
So this is the required least six digit number which is exactly divisible by 111.
Hence option (C) is correct.

Note: Whenever we come across a mixed fraction of the form $q\dfrac{r}{p}$, q is the quotient, r is the remainder and p is the divisor. If the divisor p can be broken into reminder + s, then this s when added to the remainder gives the next possible number which is divisible by that divisor. This concept is very handy while solving such problem statements.