Solving a simple addition / multiplication puzzle.

[Albert Chan]

CMS Vol 32 #5, Sep06, page 12: M203 Mayhem solution
The solution skipped trial and errors, and started with a = 79*4 = 316

Here is another solution, using simple parity check.
To get sum of 711, even a, (b+c+d) is odd, we need an odd number.

Odd variable possibilities, b = [75, 125, 225]
Based on AM/GM inequality, variables within 114 .. 150 --> b = 125 (see post #11)

Assuming we don't know that, it is still simple to check them all

a = 316:
b =   75 --> (c+d)=320, (c*d)=30000 --> (c-d)² = -17600 < 0
b = 125 --> (c+d)=270, (c*d)=18000 --> (c-d)² = 900 = 30² --> (c,d)=(270±30)/2
b = 225 --> (c+d)=170, (c*d)=10000 --> (c-d)² = -11100 < 0

--> (a, b, c, d) = (316, 125, 150, 120)

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Another way is to figure out how 5^6 factor get splitted among (b,c,d).
5^4 = 625 > 319, outside AM≥GM criteria, we are down to 2 choices.

5^6 = 5^(3+2+1)
5^6 = 5^(2+2+2)

(b+c+d) = 711 - 79k = 79*(9-k)

If k=4, RHS divisible by 5, but still not divisible by 25
--> a = 79*4 = 316, b = (125 or 250) are the only cases to check.

Again, assume we don't know (b,c,d) valid range = 114 .. 150 --> b = 125

If b = 250, AM(c,d) = (711-a-b)/2 = 72.5, GM(c,d) = √(711e6/a/b) = √9000 ≈ 94.9
AM < GM, impossible.

If b = 125, AM(c,d) = (711-a-b)/2 = 135, GM(c,d) = √(711e6/a/b) = √18000 ≈ 134.2
AM > GM: (c,d) = AM ± √(AM^2 - GM^2) = 135 ± √225 = 135 ± 15

--> (a, b, c, d) = (316, 125, 150, 120)