Question

Which is the largest five digit number exactly divisible by 91?
$
  (a){\text{ 99921}} \\
  (b){\text{ 99918}} \\
  (c){\text{ 99981}} \\
  (d){\text{ 99971}} \\
  {\text{(e) None of these}} \\
 $

Answer 1

Hint: In this question form the largest possible 5 digit number first and check whether the number formed is divisible by 91 or not, if not then see how much remainder is left behind and use it to find the number that needs to be subtracted from this greatest 5 digit number in order to make it divisible by 91.

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 four digit number rather than five digit number for example (01234= 1234).
So the smallest five digit number starts from 1 and the rest of the terms are filled by the least digit which is zero.
So the smallest five digit number is = 10000.
And in the largest five digit number all the digits are filled by the largest digit which is 9.
Therefore the largest 5 digit number is 99999.
Now we have to find out the largest five digit number which is exactly divisible by 91.
So first divide 99999 by 91 and calculate quotient and remainder.
$ \Rightarrow 99999 = 1098\dfrac{{81}}{{91}}$
So the quotient is 1098 and the remainder is 81.
So if we subtract the remainder from the largest five digit number the number formed is the required largest 5 digit number which is exactly divisible by 91.
So the largest 5 digit number is (99999 – 81) = 99918.
So this is the required largest five digit number which is exactly divisible by 91.
Hence option (B) 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 we can subtract the remainder from the largest five digit number then the number formed is the required largest number which is exactly divisible by p. This concept is very handy while solving such problem statements.