The number of 5-digit natural numbers, such that the product of their digits is 36 , is __________.

Solution

Maths sir · 3 steps

Break 36 into digit groups first, then count the arrangements of each group and add them up.

1. Allowed digits. 36 = 2² × 3², so no digit can be 0, 5, 7 or 8. Only 1, 2, 3, 4, 6 and 9 can appear, and the 1s fill up whatever places are left.
2. List the groups. Write 36 as a product of digits greater than 1, using at most five of them: 9 × 4, 9 × 2 × 2, 6 × 6, 6 × 3 × 2, 4 × 3 × 3 and 3 × 3 × 2 × 2. Pad each one with 1s to make five digits.

   Five digits	Arrangements	Count
   9, 4, 1, 1, 1	5!/3!	20
   9, 2, 2, 1, 1	5!/(2! 2!)	30
   6, 6, 1, 1, 1	5!/(2! 3!)	10
   6, 3, 2, 1, 1	5!/2!	60
   4, 3, 3, 1, 1	5!/(2! 2!)	30
   3, 3, 2, 2, 1	5!/(2! 2!)	30

3. So. No digit is 0, so every arrangement is a genuine 5-digit number. The total is 20 + 30 + 10 + 60 + 30 + 30 = 180.

The common slip: 150. Missing the group 3, 3, 2, 2, 1, where 36 uses four digits greater than 1, drops 30 numbers and gives 150.

The idea to keep. For a digit-product question, factorise the number into primes, list every way to group them into single digits, pad with 1s, then count arrangements with repeated digits.

Answer from the NTA final answer key. Solution by PrepWiser.