http://i177.photobucket.com/albums/w...nin/round9.png

I'm having difficulty trying to work out what number should replace the question mark. Any help would be greatly appreciated!

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- May 10th 2008, 02:01 PMJonaldProblems solving puzzle!
http://i177.photobucket.com/albums/w...nin/round9.png

I'm having difficulty trying to work out what number should replace the question mark. Any help would be greatly appreciated! - May 10th 2008, 02:19 PMMoo
- May 10th 2008, 02:30 PMJonald
Thanks for your help, however, i believe it is something far more complex. I will be told when i get the correct answer, however, in response to your suggestion, 8 is not the answer. I have noticed that if you multiply the numbers in the outer box by the corresponding number in the inner box and total the results you get 36 (six squared), 81 (nine squared) and 144 (twelve squared) eg. for the top right box: (2x5)+(2x6)+(2x7)=36.

I presumed that the last box should total 225 (fifteen squared) however, cannot find the number which would fit this as i got bored of trial and error after narrowing it down to between 5.3571427 and 5.3571429! I am not sure whether this is the correct approach, just an idea i had. Any more suggestions would be greatly appreciated. - May 10th 2008, 02:41 PMMoo
- May 10th 2008, 02:51 PMJonald
Thanks for the quick response. However, i'm not sure that i follow what you mean. I'm probably missing something fairly obvious but i don't see how what you said relates to the missing number.

- May 10th 2008, 02:54 PMMoo
Oh, let "a" be the missing number.

Assuming that the sum of the numbers in the outer square multiplied to the number in the inner square is the least perfect square possible, then :

(11+19+12)*a is a perfect square, and the least possible.

A perfect square is in the form

Since 11+19+12=42=2x3x7, then the least perfect square multiple of 42 will be 2²x3²x7².

Hence a=2x3x7=42 - May 10th 2008, 03:02 PMJonald
I get it now. Turns out it is the right answer as well. Thanks a lot for your help!