A number theory problem by Srinivas Modem

Number Theory Level 3

If x , y , z x,y,z are positive integers, such that x y z + x y + x z + y z + x + y + z = 384 xyz + xy + xz + yz + x + y + z = 384 , find the value of x + y + z x+y+z .

20 35 18 23

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Srinivas Modem by Srinivas Modem , 51, India

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1 solution

Department 8
Nov 10, 2015

We have

x y z + x y + y z + z x + x + y + z = 384 x y z + x y + y z + z x + x + y + z + 1 = 385 ( 1 + x ) ( 1 + y ) ( 1 + z ) = 385 \Large{xyz+xy+yz+zx+x+y+z=384 \\ xyz+xy+yz+zx+x+y+z+1=385 \\ (1+x)(1+y)(1+z)=385}

Now as all variables are positive integers then they would be equal to the factors of 385 = 5 × 7 × 11 385=5 \times 7 \times 11 . So we see,

1 + x = 5 x = 4 1 + y = 7 y = 6 1 + z = 11 z = 10 \Large{1+x=5 \\ \boxed{x=4} \\ \\ 1+y=7\\ \boxed{y=6} \\ \\ 1+z=11\\ \boxed{z=10} }

So 3 + x + y + z = 23 3+x+y+z=23 .

Note :- For question creator please change the question to 3 + x + y + z 3+x+y+z .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

You can't say that 4,6,10 is the only possible solution triplet. There are 5 others too. So you infact don't need to show the individual values of x,y,z at all. We know that (1+x)(1+y)(1+z)=7×11×5 but we don't know about the order so if we simply add we get 3+x+y+z=23 which gives x+y+z=20

Kushagra Sahni - 5 years, 7 months ago

As Kushagra said, you can't assume the order of x, y, and z. So you need to either state WLOG that x<y<z or do what Kushagra said.

Kyle Coughlin - 5 years, 7 months ago

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