SOLUTION: Find the sum of three positive integers whose product is 144 if the sum of their squares is 149.

Question 523925: Find the sum of three positive integers whose product is 144 if the sum of their squares is 149.

Found 2 solutions by Alan3354, lmeeks54:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
Find the sum of three positive integers whose product is 144 if the sum of their squares is 149.
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One of the numbers is odd to give the 149.
The odd number has to be 1, 3 or 9.
--> 9, 2 & 8

Answer by lmeeks54(111) (Show Source): You can put this solution on YOUR website!
Algebraically speaking, this is tricky. But not too much. Normally, to reduce a system of equations with n unknowns, you need at least n # equations. We have two:
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x * y * z = 144
x^2 + y^2 + z^2 = 149
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We also have the condition that x, y, and z are all positive integers. This helps.
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The problem wants us to find the sum of x + y + z
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This can be solved with logic and reasoning, and some trial and error. I used an MS Excel spreadsheet to speed up the calculations (so I could focus on the logic), but you could do it just as easily with pencil and paper.
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The easiest place to start is with the products of x * y * z = 144
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144 = 12^2
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If we only had two integers to multiply, starting with 12 * 12 is a great start. But we have three. So, factoring 12 and something like 3 * 4 * 12 makes a good guess.
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This gets us x * y * z = 144 okay, but not the sum of the squares:
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3^2 + 4^2 + 12^2 = 169, which is greater than the 149 we are looking for. But that gives us another clue:
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Keeping under the sum of squares constraint with an integer as large as 12 is probably impossible. So, we are narrowing down the possible solutions.
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Next tactic, factor 144:
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1, 144
2, 72
3, 48
4, 36
5, 28.8 (doesn't work, not an integer)
6, 24
7, 20.5 (doesn't work, not an integer)
8, 18
9, 16
10, 10.4 (doesn't work, not an integer)
11, 13.1 (doesn't work, not an integer)
12, 12 (already know we can't use a 12, teh sum of squares gets too big)
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No point in going any larger because we know 12 and two other factors won't work.
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Somewhere in that top range, we can use the first value and then factor the second value into two more values. After narrowing it down like this, with some quick trial and error, we find that:
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2 * 8 * 9 = 144
2^2 + 8^2 + 9^2 = 149
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We have satisfied both conditions (equations) and know that our x, y, and z are:
2, 8, 9
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The original problem asked for the sum of these values. That is:
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2 + 8 + 9 = 19
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Cheers, Lee
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