My favorite number is 1

Number Theory Level 3

x y z + x y + x z + y z + x + y + z = 2000 xyz + xy + xz + yz + x + y + z = 2000

Let x , y , z x,y,z be positive integers greater than 1 such that the equation above is fulfilled. Find the value of x + y + z x+y+z .


The answer is 52.

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Francehelder Santos by Francehelder Santos , 27, Brazil

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1 solution

Otto Bretscher
Oct 21, 2015

Nice problem! I was really curious to find out how there could be a unique solution...

We can write the given equation as ( x + 1 ) ( y + 1 ) ( z + 1 ) = 2001 (x+1)(y+1)(z+1)=2001 . Now x + 1 , y + 1 , x+1,y+1, and z + 1 z+1 must be the three prime factors of 2001 = 3 × 23 × 29 2001=3\times23\times29 , so that x + y + z = 3 + 23 + 29 − 3 = 52 x+y+z=3+23+29-3=\boxed{52}

PS: The condition "greater than 1" does not seem to be necessary, making the problem even more elegant.

9 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good usage of Simon's Favorite Factoring Trick.

Yes, the condition of greater than 1 is not needed. What we mostly need was for x , y , z > 0 x, y, z > 0 so that x + 1 > 1 x + 1 > 1 and so we can use the uniqueness of prime factorization.

Exactly! Symmetric sums relate so fast to Viete's and thus, Weirestrass.

Kartik Sharma - 5 years, 7 months ago

Even I was beginning to start making the natural numbers which will give 2001 as a product but my observation saved me from that work as the 3 were primes which gave the same sum.

Satyajit Ghosh - 5 years, 7 months ago

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