### Brain Teasers

# 5-digit Numbers

How many 5-digit numbers are there that do not contain the numbers 3 and 5, and are multiples of 4?

(It cannot start with 0, ie. 01234 is not a 5 digit number)

(It cannot start with 0, ie. 01234 is not a 5 digit number)

### Hint

Permutations### Answer

9408!The rule of divisibility says that a number is a multiple of 4 if its last 2 numbers are multiples of 4. So the last 2 numbers can be 21 different options:00, 04, 08, 12, 16, 20, 24, 28, 40, 44, 48, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 because they are multiples of 4 and do not have 3's or 5's.

We have the last 2 digits we need to know how many can be in the first digit: 1, 2, 4, 6, 7, 8, 9 and the second and third numbers can have these 7 and number 0 too.

By using permutations we give the possible numbers in each digit and the last 2 digits together.

1st*2nd*3rd*(4th and 5th)=?

7x8x8x21=9408

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## Comments

yeah pretty good work

Thnks! Please give me more suggestions!!!!

Nicely explained. I rushed in and got 7*8*8*20 by halving the number of numbers dividible by 2. I had a suspicion that it was wrong so you've set me straight.

In fact, if you prohibit any two odd digits (instead of 3 and 5), you still get 9408. But, which two digits would you prohibit to have the least number of solutions -- 4032? Thanks for the fun puzzle.

hi

Dude nice one i couldnt figure that out if my life depende on it but it was sorda fun to also see a difrent teaser for once.

I don't quite realise how to work that out yet.

Not the kind of thing you learn in year four

Not the kind of thing you learn in year four

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