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April 20, 2014

April 20, 2014

Posted by **Emma** on Saturday, July 7, 2007 at 8:19am.

I got the answer of 31 students. am i right? if i am not,could someone please show me how to figure this out?

Look at the sums:

ways to get sum of 1...10

ways to get sum of 2....20,11

How can three students get 3> 2,0;1,1, disallow 0,2

How can three students get 4> 4,0;22, 13, 31

get 5? 32, 23, 14, 41, 50

get 6 60, 51,15,24,42, 33

get 7 25, 16, 52, 61, 70, 43,34

get 8 17,71, 80, 62, 26, 35, 53,44

get 9 45, 54, 63,36, 27,72,`18,81,90

get 10 19,91,28,82,37,73,46,64, 55

get 11 29,92,38,83,74,47,56,65

notice the pattern...the ways to get a sum is always Sum until 9, then it is receeding in order.

get 12 ...7 ways (39,93,48,84,57,75, 66

get 13 six ways

get 14 five ways

get 15 four ways

get 16 three ways

get 17 two ways

get 18 one way.

Ok, then to have three in one sum, you will have two single sums, 15 of two sums, and one of three sums.

Number kids: 2 + 15*2 + 1*3

check my thinking.

- maths -
**ugen**, Sunday, July 29, 2012 at 8:55pm31 students

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