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March 4, 2015

March 4, 2015

Posted by **Ryan** on Wednesday, September 30, 2009 at 9:59pm.

- Pre-Calc (ugent!!!) -
**seth**, Wednesday, September 30, 2009 at 10:07pmhaha is this a last minute calendar problem by any chance? because it's october tomorrow and i have have this problem too! lets hope somebody answers soon...

- Pre-Calc (ugent!!!) -
**MathMate**, Wednesday, September 30, 2009 at 10:11pmThe digit combinations of a 5-digit number having a sum of 43 is quite limited.

The digits are 99997, 99988.

99997 can place the 7 in 5 positions, so the number of outcomes using 4-9's and a 7 is 5.

Using 3-9's and 2-8's has 5!/(3!2!)=10 ways of arranging the numbers.

Of these, only 98989, 97999 and 99979 are divisible by 11.

Can you continue?

Check my thinking.

- Pre-Calc (ugent!!!) -
**Ryan**, Thursday, October 1, 2009 at 6:36pmYes seth it is

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