Question

Find the number of natural numbers which lie between 10${}^8$ and 10${}^9$ which have products of their digits as 6?

• A. 55
• B. 105
• C. 81
• D. 92
• E. 83

Answer 1

The correct option is C 81

OPTION C

10${}^8$<n<10${}^9$

n is a 9 digit number

Product of 6 is possible in the following cases

Exactly one digit is 6 and the remaining digits are 1

Number of possible numbers= 9

One digit is 2 and the other is 3, remaining are 1 = ${}^9$P${}_2$ = 72

Total possible numbers =72+9 = 81