The total number of four digit numbers such that each of first three digits is divisible by the last digit, is equal to ____________.

Solution

Maths sir · 3 steps

Fix the last digit first, then count the choices for each of the other three digits.

1. Last digit. The last digit d must divide the others, so d ≠ 0. That leaves d = 1, 2, …, 9.
2. Other three digits. The first digit must be a non-zero multiple of d. The second and third digits can be any multiple of d from 0 to 9. 0 counts, because 0 is divisible by every d.

   Last digit d	First digit choices	Second, third digit choices	Numbers
   1	1 to 9: 9	0 to 9: 10 each	9 × 10 × 10 = 900
   2	2, 4, 6, 8: 4	0, 2, 4, 6, 8: 5 each	4 × 5 × 5 = 100
   3	3, 6, 9: 3	0, 3, 6, 9: 4 each	3 × 4 × 4 = 48
   4	4, 8: 2	0, 4, 8: 3 each	2 × 3 × 3 = 18
   5, 6, 7, 8, 9	d itself: 1	0 or d: 2 each	1 × 2 × 2 = 4 each, so 20

   900 + 100 + 48 + 18 + 20 = 1086

3. So. There are 1086 such four digit numbers.

The common slip: 833. Leaving 0 out of the second and third digits gives 729 + 64 + 27 + 8 + 5 = 833. 0 is a multiple of every d, so it is allowed in those places.

The idea to keep. When one digit controls the others, fix that digit first and multiply the choices for the rest. Remember that 0 is divisible by every non-zero digit, but it cannot be the leading digit.

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