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

derpiiXD Apr 5, 2020

#1**+3 **

Here's the answer: 10

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

We can group them and factor:

6xy(z+5)+21x(z+5)+2y(z+5)+7z=812

(z+5)(6xy+21x+2y)+7z=812

Now if we add 35 to both sides, we can put 7z into parenthesis too,

(z+5)(6xy+21x+2y)+7z+35 --> 7(z+5)=812+35

(z+5)(6xy+21x+2y+7)=847

This simplifies to

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

Now we can find the factors of 847, which are:

1, 7, 11, 77, 121, 847.

Now, we must find three positive integer values that multiply to 847. There are three of them, which are

and 7x11x11(they don't have to be distinct)

(z+5)(3x+1)(2y+7) take the place of the three integers.

Now we must test all three possibilities.

z+5=1

z=-4

Since it is negative, 1 does not work

3x+1=1

3x=0

x=0

But zero is neither positive or negative, so 1 does not work for this one either

2y+7=1

2y=-6

y=-3

Since this is also negative, it does not work

This means that

1x7x121

1x11x77

do not work.

That means that (z+5)(3x+1)(2y+7)=7x11x11

BUT...

Which is which?

We must test again...

z+5=7

z=2 This works...

3x+1=11

3x=10

x=10/3

This does NOT work because z must be an integer.

So

z+5 must equal 11

z=6

Now we need to find the other 11 and the 7

This is easy, because we already know that 3x+1 cannot equal 11, so it equals 7.

3x+1=7

3x=6

x=2

This means that

2y+7=11

2y=4

y=2

NOW WE JUST ADD THEM UP

2+2+6=10

The answer is 10

XD

derpiiXD Apr 5, 2020