Q: How many 5-digit numbers 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)
A: 9408
Multiples of 4 without 3s and 5s have the following last two digits: 00, 04, 08, 12, 16, 20, 24, 28, 40, 44, 48, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 this makes 21 possibilities for the last two digits together
A: 9408
Multiples of 4 without 3s and 5s have the following last two digits: 00, 04, 08, 12, 16, 20, 24, 28, 40, 44, 48, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96 this makes 21 possibilities for the last two digits together
The second and third digits have 8 possibilities: 0, 1, 2, 4, 6, 7, 8, 9
By using permutations, we multiply the different possibilities:
1st*2nd*3rd*(4th and 5th)=
7 x 8 x 8 x 21=9408
