Christmas Streak 48/88: Diophantine #2

Number Theory Level 2

$xy+5x+3y=290$

$x$ and $y$ are positive integers satisfying the equation above. Find the value of $xy$ .

The answer is 112.

2 solutions

Boi (보이)
Nov 18, 2017

$xy+5x+3y+15=305 \\ (x+3)(y+5)=5\times 61 \\ x+3 = 5,~ y + 5 = 61 ~(\because x+3>3,~y+5>5) \\ x=2,~y=56\\ xy=\boxed{112}.$

what led you to add 15?

Dayo Fayanju - 3 years, 6 months ago

The problem title suggests factoring. You can only factor the original left-hand side after you add a constant. Adding 15 here gives you a nice factoring, and it just so happens to give you two prime factors on the right-hand side, simplifying the problem. It's a common trick with algebra problems to add a constant to both sides of the equation for the sake of a factorization.

Oli Hohman - 3 years, 6 months ago

x = 12 and y = 22 also works to give xy = 264.

Andrew Bourke - 3 years, 6 months ago

Check your math. x=12 and y=22 are far too large to satisfy this equation.

Oli Hohman - 3 years, 6 months ago

Thinking it was 390 for some silly reason!

Andrew Bourke - 3 years, 6 months ago

Amed Lolo
Dec 3, 2017

Y(3+x)+5x=290 so y=290-5x/(3+x) so by trying and error put x=2 so y=56. X&y satisfy the above conditions to be postive & integer numbers so xy=112#

x=1 y=285/4

Genti Haxhillari - 3 years, 6 months ago

x and y are positive integers

Fake Account - 8 months ago

Christmas Streak 48/88: Diophantine #2

The answer is 112.

2 solutions

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