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# Help :)

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

Mar 3, 2020

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$$\displaystyle 6xyz +30xy+21xz+2yz+105x+10y+7z=812.$$

Factorise.

$$\displaystyle (3x+1)(2y+7)(z+5)=812 = 2\times 2\times7\times29.$$

The only x, y, z that fit are x =1, y = 11 and z = 2, so x + y + z = 14.

Mar 4, 2020