(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. 

Answer by greenestamps(13198)   (Show Source): You can put this solution on YOUR website!
 


Most of the answers from the other tutor are not right; and his post doesn't really tell you much about HOW to find the answers.


(a) First digit divisible by 1

Obviously each of the given 5 digits is divisible by 1, so the 5 digits can be arranged in 5! = 120 ways.

ANSWER: 120


(b) First 2 digits form a number divisible by 2

The second digit must be even, so it must be 2 or 4 (2 choices); the other 4 digits can be arranged in 4!=24 ways.  Total: 2*24 = 48

ANSWER: 48


(c) First 3 digits form a number divisible by 3

The sum of the 5 digits is 15, which is divisible by 3.  If the first 3 digits are divisible by 3, then the sum of the last 2 digits is divisible by 3.  There are 4 such combinations -- (1,2), (1,5), (2,4), and (4,5).  So there are 4 ways to group the 5 digits into a group of 2 and a group of 3, in which the sum of the digits in each group is divisible by 3. The group of 2 digits can be arranged in any of 2! = 2 ways; the group of 3 digits can be arranged in any of 3! = 6 ways.  So the total number of 5-digit numbers using those digits in which the first 3 digits make a number divisible by 3 is 4*2*6 = 48.

ANSWER: 48:


(d) First 4 digits form a number divisible by 4

A 4-digit number is divisible by 4 if the last 2 digits form a number divisible by 4.  The 2-digit numbers using these digits that are divisible by 4 are 12, 24, 32, and 52.  That's 4 possibilities for the 3rd and 4th digits; the remaining 3 digits can be arranged in any of 3! = 6 ways.  So the total number of 5-digit numbers using these digits in which the first 4 digits form a number divisible by 4 is 4*6 = 24.

ANSWER: 24


(e) The 5-digit number is divisible by 5

With these digits, the final digit must be 5; the remaining 4 digits can be arranged in any of 4! = 24 ways.

ANSWER: 24


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Added in response to the student's message regarding this response....


Your comment doesn't make sense, and/or it is poorly written.


The sum of the 5 digits 1, 2, 3, 4, and 5 is 15, which is divisible by 3.  So if the sum of the first three of those digits is divisible by 3, then the sum of the last two of those digits is also divisible by 3.


 

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