Let Z be the number of 8-digit numbers with 8 different digits, none of which is 0. How many 8 digit numbers exist that are divisible by 9, that have 8 different digits, none of which is 0. Answer in terms of Z.
I tried putting in 9, also tried putting in 18 at the end but soon realized that only some numbers with these endings will be divisible by 9, not all.
For a number to be divisible by 9, the sum of all its digits must be divisible by 9. So the sum of digits of a number made up of 8 digits can only be: $$(1 + 2 + \cdots + 9)-1 = 45 - 1 = 44$$ $$(1 + 2 + \cdots + 9)-2 = 45 - 2 = 43$$ $$(1 + 2 + \cdots + 9)-3 = 45 - 3 = 42$$ $$(1 + 2 + \cdots + 9)-4 = 45 - 4 = 41$$ $$(1 + 2 + \cdots + 9)-5 = 45 - 5 = 40$$ $$(1 + 2 + \cdots + 9)-6 = 45 - 6 = 39$$ $$(1 + 2 + \cdots + 9)-7 = 45 - 7 = 38$$ $$(1 + 2 + \cdots + 9)-8 = 45 - 8 = 37$$ $$(1 + 2 + \cdots + 9)-9 = 45 - 9 = 36$$ Of all the 9 sums only 36 is divisible by 9. As the possibility of a number having a particular sum of digits is equal for every sum means that $\dfrac{1}{9}$ of the numbers are divisible by 9. So the number of numbers divisible by 9 is $\dfrac{Z}{9}$
A number is divisible by 9 iff the sum of his digits equal 0 mod 9. So there are as many 8-digit numbers without the digit 0 that are divisible by 9 than 7-digit numbers without the digit 0, ie $\frac{Z}{9}$.
