Question

How many five digit numbers that have a nine and a six as the first and third digit are not divisible by 3? Please explain HOW you get your solution.

Answer 1

Let n=9a6bc be the number (decimal representation:base10).
Let us find how many are divisible by 3 then we can subtract it from the total possible numbers to get the required answer.
For divisibility by 3, 3 must divide 9+a+6+b+c
or, 3 must divide a+b+c..

Answer 2

would that even work?

Answer 3

wat do u mean paulzord?

Answer 4

i was going to ask, howcome you only need a,b and c, but now that i read it again i think that should've been obvious

Answer 5

Isn't that brute forcing it? Would you have to try a lot of combinations to get the answer?

Answer 6

now possibilities (m is integer) {a,b,c}={3m,3m,3m},{3m+1,3m,3m+2}<--this one has 5 more combinations,{3m+1,3m+1,3m+1}.
You have to vary m from 0,1,2,3 and do it for the above 8 combinations.

Answer 7

this problem needs some bruteforcing...:( no matter whatever method u try!

Answer 8

So the answer is 666? Cool! Thanks. I thought there was a way where it was all algebra and no bruteforcing but I guess I was wrong.