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What is the sum of all the numbers formed by taking each of the digits 2,3,5,6 and 8 once?

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- Thread starter Greg Bernhardt
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- #1

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What is the sum of all the numbers formed by taking each of the digits 2,3,5,6 and 8 once?

- #2

BoulderHead

haha, maybe later.

- #3

jamesrc

Science Advisor

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6399936

- #4

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Originally posted by jamesrc

6399936

How'd ya get that answer?

- #5

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2 3 5 6 8

There are: (5)*(5-1)*(5-2)*(5-3)*(5-4) or 5*4*3*2*1 = 120 permutations for the above set of numbers.

One set of numbers add up to: 2+3+5+6+8 = 24.

So, the sum of all numbers formed from 120 permutations is:

120*24 = 2880.

- #6

jamesrc

Science Advisor

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If I answered the right question, for all of the 5-digit numbers you can make, each digit appears 24 times in each column, so you can write:

24*(22222+33333+55555+66666+88888)=6399936

Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.

24*(22222+33333+55555+66666+88888)=6399936

Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.

Last edited:

- #7

- 18,502

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Originally posted by jamesrc

If I answered the right question, for all of the 5-digit numbers you can make, each digit appears 24 times in each column, so you can write:

24*(22222+33333+55555+66666+88888)=6399936

Edit: Oh, I thought it was the sum of the numbers made by the combinations, not the sum of their digits.

James gets the point!

Here is an alternative way:

4!*(2+3+5+6+8)*(11111) = 6,399,936

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