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What is the sum of the 3 digit numbers that can be formed

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What is the sum of the 3 digit numbers that can be formed [#permalink] New post 02 Jun 2005, 02:52
What is the sum of the 3 digit numbers that can be formed from the digits 6,7,8 and 9?

Thanks
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Re: Is there a quick way to solve a problem of tis kind? [#permalink] New post 02 Jun 2005, 06:26
krishrads wrote:
What is the sum of the 3 digit numbers that can be formed from the digits 6,7,8 and 9?

Thanks


4*4*4 = 64 numbers

6+7+8+9 = 30

16*30 = 480 = sum of units digit of the 64 numbers (same for tens and hundreds digits)

total sum = 480*(1+10+100) = 53280.
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 [#permalink] New post 02 Jun 2005, 07:47
Thanks Dan.Unfortunately I didn't quite understand how you did it.Culd you pls explain the answer.

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Re: Is there a quick way to solve a problem of tis kind? [#permalink] New post 02 Jun 2005, 11:18
krishrads wrote:
What is the sum of the 3 digit numbers that can be formed from the digits 6,7,8 and 9?

Thanks


(6+7+8+9)*3! = 30*6=180 for sum for 1 line
so sum is 19980
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 [#permalink] New post 02 Jun 2005, 11:28
Thanks Sparky what is the logic you followed?Unfortunately I don't know what the right answer is.
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 [#permalink] New post 02 Jun 2005, 11:35
krishrads wrote:
Thanks Sparky what is the logic you followed?Unfortunately I don't know what the right answer is.


here we go,
The last digits can be 6,7,8, or 9
so if it is 9, for example, there are 2 more digits before it that need to be chosen from 6,7, and 8 (order matters here) so 3!/1! permutations = 3!=6

so there will be 6 numbers with 9 in the end, 6 numbers with 8, etc
total 6x4 = 24 numbers

so sum of last digits will be (6+7+8+9)*3!

do the same the middle and the first digits

and you get the number

Dan did the same thing basically but he accounted for repeats, so in his case 999 is possible but it isn't possible in my solution
  [#permalink] 02 Jun 2005, 11:35
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