Try this beautiful problem from Number theory based on Integer.

## Integer Problem – AMC-10A, 2020- Problem 17

Define \(P(x)=(x-1^2)(x-2^2)……(x-{100}^2)\)

How many integers \(n\) are there such that \(P(n) \geq 0\)?

- \(4900\)
- \(4950\)
- \(5000\)
- \(5050\)
- \(5100\)

**Key Concepts**

Number system

Probability

divisibility

## Check the Answer

But try the problem first…

Answer: \(5100\)

AMC-10A (2020) Problem 17

Pre College Mathematics

## Try with Hints

First hint

Given \(P(x)=(x-1^2)(x-2^2)……(x-{100}^2)\). at first we notice that \(P(x)\) is a product of of \(100\) terms…..now clearly \(P(x)\) will be negetive ,for there to be an odd number of negetive factors ,n must be lie between an odd number squared and even number squared.

can you finish the problem……..

Second Hint

\(P(x)\) is nonpositive when \(x\) is between \(100^2\) and \(99^2\), \(98^2\) and \(97^2 \ldots\) , \(2^2\) and \(1^2\)

can you finish the problem……..

Final Step

Therefore, \((100+99)(100-99)+((98+97)(98-97)+1)\)+….

.+\(((2+1)(2-1)+1)\)=\((200+196+192+…..+4)\) =\(4(1+2+…..+50)=4 \frac{50 \times 51}{2}=5100\)

## Other useful links

- https://www.cheenta.com/divisibility-problem-from-amc-10a-2003-problem-25/
- https://www.youtube.com/watch?v=XOrePzJWFiE

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