Help!

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Help!

Postby Blessedmum » Mon Mar 01, 2010 12:53 am

Hi, any method to solve the problem below rather than trial n error?

Write 1 to 10 in the circles so that the sum of the numbers on each side of the pentagon is equal to 16. (There are 3 circles in each side of the pengaon). I'm still working on the trial n error method!!!! Help!

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Postby CoffeeCat » Mon Mar 01, 2010 4:25 pm

hihi, here's my method

first i sum 1+2+...+10 = 55
then i find 16 *5=80.
(this is a common step in such puzzles to find clues about the repeated values).
80-55=25. Notice when i add 16 5 times (the 5 sides) i added each corner twice. therefore the sum of the numbers at the corners are 25.

I choose the odd numbers 1, 3, 5, 7, 9 as my corner numbers (just a hunch because 25/5=5).
the even numbers, 2, 4, 6, 8, 10 are going into the middle.
so i find 16-2 =14
16-4=12
16-6 = 10
16-8 =8
16-10=6
so now i know i must find unique pairings of the corner numbers to add up to the right hand side numbers .
there is only 1 pairing for 14(the biggest of the 5). 9+5
so the first side i have the 3 numbers, 9+2+5
now i look at 6 (the smallest of the 5). 5+1 (only 1 pairing)
so the second side is 5+10+1
only 2 corner numbers left, 3 and 7.
only way to get 12 is 3+9. so another side is 3+4+9
only way to get 8 is 7+1. so another side is 7+8+1
leaving the last side 3+6+7.

if i didn't explain clearly enough, this is my set of solution
9 + 2 +5
5 + 10 +1
1+ 8+7
7+ 6+3
3 +4 +9

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Postby Blessedmum » Mon Mar 01, 2010 8:07 pm

Hi Coffeecat!!

wow...amazing.... but I need time to digest these.....

Thanks!!!

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Postby CoffeeCat » Mon Mar 01, 2010 8:25 pm

hmmm may i ask where you get this interesting question from?

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Postby Blessedmum » Mon Mar 01, 2010 9:04 pm

From Catholic High's Worksheet.

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