
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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November 28th, 2016, 10:09 PM  #1 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3  a simple but timeconsuming problem
The problem I am going to write down is simple but very tedious. (ex) Determine the sum of all the 3digit numbers that can be made by choosing 3 different numbers from the list 1, 2, 3, 4, 5, 6, 7. Forming 3digit 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. 
November 28th, 2016, 10:23 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,052 Thanks: 1395 
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.

November 28th, 2016, 11:32 PM  #3  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,495 Thanks: 753  Quote:
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 $T = 49(1+2+3+4+5+6+7) + 490(1+2+3+4+5+6+7)+4900(1+2+3+4+5+6+7) = (49+490+4900)(28 ) = 152292$  
November 29th, 2016, 04:58 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,593 Thanks: 938 Math Focus: Elementary mathematics and beyond 
As the digits are required to be different there are 210 threedigit 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.

November 29th, 2016, 07:15 AM  #5  
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,495 Thanks: 753  Quote:
shouldn't be answering these post midnight.  
November 29th, 2016, 12:48 PM  #6 
Senior Member Joined: Apr 2008 Posts: 176 Thanks: 3 
Thanks everyone for all your help.


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