The total number of six digit numbers formed using the digits 4, 5, 9 only and divisible b…

Question

The total number of six digit numbers formed using the digits 4, 5, 9 only and divisible by 6, is \_\_\_\_\_\_.

Answer: 81

Accepted Answer

For divisibility by 6, the number must be even (last digit 4) and digit sum divisible by 3.

All same digit (444444): sum works if divisible by 3; only 444444 qualifies among single-digit-repeats ending in 4: $1$ way.

Using two distinct digits:
- (4,5) as 444555 type arrangements ending in 4: $\dfrac{5!}{3!2!}=10$
- (4,9) type: $10$

Using three digits (4,5,9) mixed with repeats:
- 4,5,9,4,4,4 (four 4's,one 5,one 9): $\dfrac{5!}{3!}=20$
- 4,5,9,5,5,5 (four 5's --- but must end in 4, one 4 fixed, remaining four 5's,one 9 among 5 slots): $\dfrac{5!}{4!}=5$
- 4,5,9,9,9,9: $\dfrac{5!}{4!}=5$
- 4,5,9,4,5,9 (two each of 4,5,9 with a 4 fixed at end): $\dfrac{5!}{2!2!}=30$

$$\text{Total} = 1+10+10+20+5+5+30 = 81$$

Solutions · 1