The total number of four digit numbers such that each of the first three digits is divisib…

Question

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

Answer: 1086

Accepted Answer

Let the number be $\overline{abcd}$ where $a,b,c$ are each divisible by $d$ (and $a\ne0$).

- $d=1$: $a,b,c$ any digit ($a\ne0$): $9\times10\times10=900$
- $d=2$: $a,b,c\in\{2,4,6,8\}$ (also 0 for $b,c$; but must be divisible by 2, and $a\ne0$): $4\times5\times5=100$
- $d=3$: $a,b,c\in\{3,6,9\}$ (plus 0 for $b,c$): $3\times4\times4=48$
- $d=4$: $\{4,8\}$ plus 0: $2\times3\times3=18$
- $d=5$: $\{5\}$ plus 0: $1\times2\times2=4$
- $d=6,7,8,9$: only digit itself qualifies (plus 0 for $b,c$): $1\times2\times2=4$ each, total $4\times4=16$

$$\text{Total} = 900+100+48+18+4+16 = 1086$$

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