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M04 # 32 : Retired Discussions [Locked]

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Answer: For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or 12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

1st digit - can be filled in 4 ways (1,2,4,5). 2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5). 3rd digit - 3 ways 4th digit - 2 ways 5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

1st digit - can be filled in 4 ways (1,2,4,5). 2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5). 3rd digit - 3 ways 4th digit - 2 ways 5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

I believe the explanation above is wrong.

For an integer to be divisible by 3 the sum of its digits must be divisible by 3. Then you can have only 15 = 1+2+3+4+5 or 12 = 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second - 4*4! since 0 can not be the first digit. Therefore, the answer is 5!+4*4!=216 .

the questionn "...divisible by 3, without repeating the digits." <- does this mean that 3 shouldn't be repeated more than once for each digit? and why do multiply by 4

This question would've taken me way too long to figure out on the real thing; would've started looking at answer choices...

1) Need the digits to sum to a multiple of 3 (to satisfy "divisible by 3" rule):

12345 does the job 01425 does the job

2) Need to make a 5 digit number:

5! counts the number of ways of arranging 1, 2, 3, 4, 5 5!-4! counts the number of ways of arranging 0, 1, 2, 4, 5 taking out numbers that start with 0 (otherwise we would be counting 4 digit numbers as well!)

How many five-digit numbers can be formed using the digits 0, 1, 2, 3, 4 and 5 which are divisible by 3, without repeating the digits?

Answer: For an integer to be divisible by 3 sum of its digits must be divisible by 3. Then you can have only 15= 1+2+3+4+5 or 12= 0+1+2+4+5 as the sum. In the first case there are 5! permutations, and in the second there are 4*4! because 0 can not be the first digit. Therefore, the answer is 5! + (4*4!) - 216.

My only question here is: How do we come up with 4*4! for the second sum permutations? The 0 that can't be in first place has only four options so I think the logic is..how many times can we arrange one number into four spots? 4 differnt ways. Then 4! comes from how do we arrange 4 numbers into five spots? Help with the 4*4! explanation. I really wish these answers were more explanatory.

Thank you!

This question was posted in PS subforum. Below is my solution from there:

First step:

We should determine which 5 digits from given 6, would form the 5 digit number divisible by 3.

We have six digits: 0,1,2,3,4,5. Their sum=15.

For a number to be divisible by 3 the sum of the digits must be divisible by 3. As the sum of the six given numbers is 15 (divisible by 3) only 5 digits good to form our 5 digit number would be 15-0={1,2,3,4,5} and 15-3={0,1,2,4,5}. Meaning that no other 5 from given six will total the number divisible by 3.

Second step:

We have two set of numbers: 1,2,3,4,5 and 0,1,2,4,5. How many 5 digit numbers can be formed using this two sets:

1,2,3,4,5 --> 5! as any combination of these digits would give us 5 digit number divisible by 3. 5!=120.

0,1,2,4,5 --> here we can not use 0 as the first digit, otherwise number won't be any more 5 digit and become 4 digit. So, total combinations 5!, minus combinations with 0 as the first digit (combination of 4) 4! --> 5!-4!=4!(5-1)=4!*4=96

1st digit - can be filled in 4 ways (1,2,4,5). 2nd digit- can then be filled in 4 ways again - 0 , and 3 of the remaining digits from(1,2,4,5). 3rd digit - 3 ways 4th digit - 2 ways 5th digit - 1 way

So the total number of possibilities is 4*4*3*2*1, which is 4*4!

Kudos! I had the same question and that was a great explanation.