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November 28th, 2016, 11:09 PM   #1
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a simple but time-consuming problem

The problem I am going to write down is simple but very tedious.

(ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from the list 1, 2, 3, 4, 5, 6, 7.

Forming 3-digit numbers from 1 to 3 is easy.

123
132
213
231
312
321

Adding them up gives 1332.

Things can get very ugly with 7 numbers.

Is there an easier way to approach the problem? Thanks.
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November 28th, 2016, 11:23 PM   #2
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In your example, 1332 = 222(1 + 2 + 3). A similar calculation gives the total for each of the other 34 choices of the three digits. For example, 222(1 + 2 + 4) = 1554.
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November 29th, 2016, 12:32 AM   #3
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Quote:
Originally Posted by davedave View Post
The problem I am going to write down is simple but very tedious.

(ex) Determine the sum of all the 3-digit numbers that can be made by choosing 3 different numbers from the list 1, 2, 3, 4, 5, 6, 7.

Forming 3-digit numbers from 1 to 3 is easy.

123
132
213
231
312
321

Adding them up gives 1332.

Things can get very ugly with 7 numbers.

Is there an easier way to approach the problem? Thanks.
there are $7^3=343$ total 3 digit numbers using digits 1-7.

Each digit appears the same number of times as the others.

Thus each digit appears $\dfrac {343}{7} = 49$ times in the 1's, 10's, and 100's column.

Summing this we get

$T = 49(1+2+3+4+5+6+7) + 490(1+2+3+4+5+6+7)+4900(1+2+3+4+5+6+7) =
(49+490+4900)(28 ) = 152292$
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November 29th, 2016, 05:58 AM   #4
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As the digits are required to be different there are 210 three-digit numbers that can be formed. The required total is 93240, obtained by adapting the sum romsek gave, i.e., by using coefficients of 30, 300 and 3000.
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November 29th, 2016, 08:15 AM   #5
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Quote:
Originally Posted by greg1313 View Post
As the digits are required to be different there are 210 three-digit numbers that can be formed. The required total is 93240, obtained by adapting the sum romsek gave, i.e., by using coefficients of 30, 300 and 3000.
bah, missed the word different.

shouldn't be answering these post midnight.
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November 29th, 2016, 01:48 PM   #6
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Thanks everyone for all your help.
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