Problems solving puzzle!

• May 10th 2008, 01:01 PM
Jonald
Problems 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, 01:19 PM
Moo
Quote:

Originally Posted by Jonald
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!

Hello,

Note that the sum of elements in the top-right square is 20.
The sum of elements in the bottom-right square is 30.
The sum of elements in the bottom-left square is 40.
So the sum of elements in the top-left square will be... ?
• May 10th 2008, 01:30 PM
Jonald
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, 01:41 PM
Moo
Quote:

Originally Posted by Jonald
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.

I'm sorry, I didn't know what level this was...
Following your idea (which I find great), (11+19+12)*a should be a perfect square.
11+19+12=42=2x3x7
The smallest a such that this is a perfect square is 2x3x7=42...
• May 10th 2008, 01:51 PM
Jonald
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, 01:54 PM
Moo
Quote:

Originally Posted by Jonald
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.

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 $p^{2x} \cdot q^{2y} \dots$

Since 11+19+12=42=2x3x7, then the least perfect square multiple of 42 will be 2²x3²x7².
Hence a=2x3x7=42
• May 10th 2008, 02:02 PM
Jonald
I get it now. Turns out it is the right answer as well. Thanks a lot for your help!