If $x$, $y$, and $z$ are positive integers such that $6xyz+30xy+21xz+2yz+105x+10y+7z=812$, find $x+y+z$.

I tried and couldn't do it, can you help me?

Guest Jun 18, 2023

#1**-1 **

We can factor the left-hand side of the equation as follows:

6xyz+30xy+21xz+2yz+105x+10y+7z = (3x+1)(2y+7)(z+5)

Since x, y, and z are positive integers, we know that 3x+1, 2y+7, and z+5 must all be factors of 812. We can find the factors of 812 by prime factorizing it:

812 = 2^2 * 3^3 * 11

The possible values of 3x+1 are 1, 2, 9, 10, 27, and 28. The possible values of 2y+7 are 7, 8, 15, 16, 23, and 24. The possible values of z+5 are 5, 6, 11, and 12.

We can see that the only combination of values that works is 3x+1=10, 2y+7=16, and z+5=11. This gives us x=3, y=8, and z=6. **Therefore, x+y+z=3+8+6=17.**

Guest Jun 18, 2023