Welcome to our community

Be a part of something great, join today!

Problem Of The Week #376 Jul 24th, 2019

Status
Not open for further replies.
  • Thread starter
  • Admin
  • #1

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Here is this week's POTW:

-----

Let $P(x)$ be any polynomial with integer coefficients such that $P(21)=17,\,P(32)=-247$ and $P(37)=33$.

Prove that if $P(N)=N+51$ for some integer $N$, then $N=26$.

-----

Remember to read the POTW submission guidelines to find out how to submit your answers!
 
  • Thread starter
  • Admin
  • #2

anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,894
Congratulations to kaliprasad for his correct solution(Cool)!

Model solution from other:
If $P(N)=N+51$ for some integer $N$, then $P(x)-x-51=(x-N)Q(x)$ for some polynomial $Q(x)$ by the factor theorem.

Note $Q(x)$ has integer coefficients because $P(x)-x-51=P(x)-P(N)-(x-N)$ is a sum of $a_i(x_i-N_i)$ terms with $a_i's$ integer.

Since $Q(21)$ and $Q(37)$ are integers, $P(21)-21-51=-55$ is divisible by $21-N$ and $P(37)-37-51=-55$ is divisible by $37-N$ is 16, we must have $N=26$ or 32.

However, if $N=32$, then we get $-247=P(32)=32+51$, a contradiction. Therefore, $N=26$.


 
Status
Not open for further replies.