Question

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

Answer 1

Answer:

x= 2, y= 2, and z= 6

Step-by-step explanation:

If this is a Diophantine equation, add 35 to both sides and factor the left:

(3x+1)(2y+7)(z+5) = 847 = 7 times 112

Each integer factorization of 847 into 3 factors leads to a different number/value of x, y, and z. If the first factor, (3x+1), is 1 more than a multiple of 3, and the second factor, (2y+7), is odd, then x, y, and z will be integers.

For example:

847 = 121 times -7 times -1 gives (x, y, z) = (40, -7, -6) because 121 times -7 times -1 is 847 as well, it checks out.

If x, y, and z need to be positive, then the three numbers/factors need to be greater than 1, 7, and 5. The only combination that works is 7 times 11 times 11, which gives (x, y, z) = (2, 2, 6).

:)