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Find the number of 10-digit numbers where the sum of the digits is divisible by 5.

 Nov 8, 2019
 #2
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The answer is 5*10^6 = 5,000,000.

 Nov 8, 2019
 #3
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xxxxxxxxx

 Nov 8, 2019
edited by Guest  Nov 8, 2019
 #4
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*****

 

 

cool cool cool

CPhill  Nov 8, 2019
edited by CPhill  Nov 8, 2019
 #5
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Your question is not clear enough ! For example, can the digits be repeated such as:

 

6 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
5 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
4 + 3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
4 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
3 + 3 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 15
3 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 15
2 + 2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 15........etc.

If they can be repeated, then it is a matter of adding partitions of: 5(10 parts) + 10(10 parts) + 15(10 parts).........+90(10 parts), which would be a very involved undertaking.

 

If they cannot be repeated, that is, you can only use each digit once such as:1234567890, then that might be very easy to calculate. Since the sum of the 10 digits [0 to 9 =45, which is divisible by 5], then the arrangement of the 10 digits wouldn't matter at all, since they will always sum up to 45.

Therefore, if you started with: 1234567890, then you will have:9 * 9! (on the assumption that all 10 digits be different from each other). So, 9 * 9! =3,265,920.

And that is my attempt. Others should take a crack at it!

 Nov 8, 2019
 #6
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 Nov 12, 2019
edited by Melody  Nov 12, 2019

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