Absolutely modded

Algebra Level 2

Let $a,b,c$ be the positive integers such that $a<b<c$ . If the two curves :

$y$ $=$ $|x-a|$ $+$ $|x-b|$ $+$ $|x-c|$
$2x$ $+$ $y$ $=$ $2003$

have exactly one point in common then find the greatest possible value of $b$

The answer is 1001.

1 solution

A Former Brilliant Member
Apr 25, 2019

Since 0<a<b<c, the first equation has three points of singularity, at x=a, at x=b and at x=c. Since the two curves have only one common point, it must be at x=a. Therefore 2a+b+c-2a=2003. Since b+c>2b, therefore 2b<2003, so that the greatest magnitude of b is 1001.

Absolutely modded

The answer is 1001.

1 solution

0 pending reports