5digit Numbers
Math brain teasers require computations to solve.
How many 5digit 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
Hide
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 
Back to Top
 

Follow Braingle!