New Year's Countdown Day 31: One Last Day
Consider the equation
Let be integers such that is a solution to the equation, and is minimized. Find the value of
This problem is part of the set New Year's Countdown 2017 .
The answer is 33.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
First, observe that ( x , y ) = ( − 1 , 3 ) is a solution to the equation, since 2 0 ( − 1 ) + 1 7 ( 3 ) = − 2 0 + 5 1 = 3 1 . Then, any Diophantine solution to the equation can be expressed as
( x , y ) = ( − 1 + 1 7 n , 3 − 2 0 n )
for some integer n , because 2 0 ( − 1 + 1 7 n ) + 1 7 ( 3 − 2 0 n ) = − 2 0 + 2 0 ( 1 7 ) n + 5 1 − 1 7 ( 2 0 ) n = 3 1 .
The sum of these two expressions is
− 1 + 1 7 n + 3 − 2 0 n = 2 − 3 n ,
and our task is to minimize its absolute value. This is achieved when n = 1 to give ∣ 2 − 3 ( 1 ) ∣ = 1 as the minimum possible value of ∣ x + y ∣ .
Therefore, a = − 1 + 1 7 ( 1 ) = 1 6 and b = 3 − 2 0 ( 1 ) = − 1 7 , and a − b = 1 6 − ( − 1 7 ) = 3 3 .
This concludes the New Year's Countdown 2017 series. Thanks to everyone on Brilliant for your support this year. I hope that 2017 has treated you well, and may 2018 go even better for everybody. Happy New Year!