Numbers are to be built using only the digits 1,2,3,4, and 5 in such a way that each digit is only used once in each number. How many of these numbers will have the following property?
(a) The first digit is divisible by one
(b) The first 2 digits make a number that is divisible by 2
(c) The first 3 digits make a number that is divisible by 3
(d) The first 4 digits make a number that is divisible by 4
(e) All 5 digits make a number that is divisible by 5
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(a) Since every number is divisible by one, then the total possible permutations are 5, as you can't repeat the numbers and order matters.
(b) Since every number that is divisible by two is even, then it must end in 2 or 4 in this situation. Therefore(you could easily count them), the total possible permutations are 8.
(c) Since every number divisible by 3 is determined by adding the digits together(if the digits add up to a multiple of 3, then it is divisible), the only possible solutions are when the digits are (1,2,3),(1,3,5), and (3,4,5), Where there are 6 possible permutations of each, so the total number of permutations are 18(6+6+6=18).
(d) Since every number that is divisible by two is determined by looking at the last two digits of a number. I know that the multiples of 4 up to 100, but we could only use the multiples of 4 up to 52. The multiples you could possibly choose are 12,24,32, and 52. Every one of them has 50, so there are50+50+50+50=200 total possible permutations.
(e) Since all multiples of 5 ends in 5 or 0, we exclude zero right away. After some counting, you end up with 24 possible permutations.