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I couldn't solve the problem in my initial attempt, so had to refer to the official solution provided where it was given as follows:

$P^5+9^5$ is divisble by $P+9$ for all P, where P is a natural number.

And now $P^5+9^5$ can be written as $(P^5+99)+58950$ where we know that complete term is a multiple of $P+9$ and also as per the question requirement $P^5+99$ is a multiple of $P+9$ which implies that 58950 also has to be a multiple of $P+9$.

Hence, the max value of $P+99=58950$ or $P=58941$

My question is why are we restricting to $9^5$, why cant we take a bigger number than this and then express it with the help of $99+ something$ as done in the solution.

Also is there any other method to solve this problem ?

Vasu Gupta
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1 Answers1

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The point is that $x^5 + 9^5$ is divisible by $x + 9$ (as a polynomial over the integers) so that $P^5 + 9^5$ is divisible by $P + 9$ for any natural number $P$. Since no constant other than $0$ is divisible by $x + 9$, there is no "bigger number than this" that will work.

Robert Israel
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  • I am not able to follow your reasoning. My question is why do we need to take only $9^5$ , surely we know that $x^5+9^5$ is divisible by $x+9$, but what if there is some other number, lets say $'a'$ such that it is more than $9^5$ and also follows the condition of $x^5+a$ being divisible by $x+9$ ? – Vasu Gupta May 19 '24 at 05:10
  • Please strive not to post more (dupe) answers to dupes of FAQs. This is enforced site policy, see here. It's best for site health to delete this answer (which also minimizes community time wasted on dupe processing.) – Bill Dubuque May 19 '24 at 05:28
  • @Vasu We're using $,a\mid b = qa+r\iff a\mid r.,$ That remains true no matter which multiple $qa$ of $,a,$ we choose, but generally to simplify matters we choose one that minimizes the magnitude of the "residue" $,r,$ (so $,r,$ is the remainder and $,q,$ is the quotient), i.e. in divisibility mod reduction we choose our "simpler" residue $,r\equiv b\pmod{!a},$ to be the remainder $,r = (b\bmod a).\ \ $ – Bill Dubuque May 19 '24 at 06:00