5-digit Numbers
Math brain teasers require computations to solve.
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)
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
bumbles2003
Apr 24, 2003
| yeah pretty good work |
Mhalun
Apr 24, 2003
| Thnks! Please give me more suggestions!!!! |
jimbo  
Apr 25, 2003
| 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. |
jimrcook
Apr 28, 2003
| 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. |
smart432890
Apr 29, 2003
| hi |
EverquestBoy
May 21, 2003
| 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. |
jacintan
Jun 13, 2003
| I don't quite realise how to work that out yet.
Not the kind of thing you learn in year four |
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