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# Integer Problem | AMC 10A, 2020 | Problem 17

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 $P(n) \geq 0$?

• $4900$
• $4950$
• $5000$
• $5050$
• $5100$

Number system

Probability

divisibility

## Check the Answer

Answer: $5100$

AMC-10A (2020) Problem 17

Pre College Mathematics

## Try with Hints

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........

$P(x)$ 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........

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

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

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