Asked by shelly
If x, y, and z are positive integers such that 6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z =481, find x + y + z.
Answers
Answered by
oobleck
what a load of rubbish.
Clearly, if x=y=z=1, then
6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 6+30+18+12+10+17+5 = 98, not 481
I do not think there is any solution with positive values.
Clearly, if x=y=z=1, then
6xyz + 30xy + 18xz + 12yz + 10x + 17y + 5z = 6+30+18+12+10+17+5 = 98, not 481
I do not think there is any solution with positive values.
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