3-digit number permutation problem

[zipup]

1. 3-digit numbers are constructed from the digits 0,1,2,3,4,5,6,7,8,9 using each digit at most once. How many such number are divisible by 5?



2. Just simply permutation, the answer is 136.



3. 8 x 7 x 2 = 128

 

[Sisplat]

Firstly, 8*7*2 is not equal to 128.

My calculation is:

Firstly for numbers ending in 0,

9*8 possible ways of choosing 2 digits.

For numbers ending in 5,

8*7, since we want to exclude 0

9*8 + 8*7 = 128.

 

[daschaich]

Firstly, 8*7*2 is not equal to 128.

My calculation is:

Firstly for numbers ending in 0,

9*8 possible ways of choosing 2 digits.

For numbers ending in 5,

8*7, since we want to exclude 0

9*8 + 8*7 = 128.

I think this undercounts by excluding the eight numbers 105, 205, etc. (Eight since 505 is still out.) We just want to exclude 0 from the first digit, not from the second.

The counting then becomes 9*8 + 8*8 = 136, which seems to be desired answer.

PS. Why is this filed under "Advanced Physics" instead of "Precalculus Mathematics"?