2020 AMC 10A Problems/Problem 17

Revision as of 10:47, 19 October 2020 by Icematrix2 (talk | contribs)

Define \[P(x) =(x-1^2)(x-2^2)\cdots(x-100^2).\] How many integers $n$ are there such that $P(n)\leq 0$?

$\textbf{(A) } 4900 \qquad \textbf{(B) } 4950\qquad \textbf{(C) } 5000\qquad \textbf{(D) } 5050 \qquad \textbf{(E) } 5100$

Solutions

Solution 1

Notice that $P(x)$ is a product of many integers. We either need one factor to be 0 or an odd number of negative factors.

Case 1: There are 100 integers $n$ for which $P(x)=0$

Case 2: For there to be an odd number of negative factors, $n$ must be between an odd number squared and an even number squared. This means that there are $2+6+\dots+198$ total possible values of $n$. Simplifying, there are $5000$ possible numbers.

Summing, there are $\boxed{\textbf{(E) } 5100}$ total possible values of $n$. ~PCChess

Solution 2

Notice that $P(x)$ is nonpositive when $x$ is between $100^2$ and $99^2$, $98^2$ and $97^2 \ldots$ , $2^2$ and $1^2$ (inclusive), which means that the amount of values equals $((100+99)(100-99) + 1) + ((98+97)(98-97)+1) + \ldots + ((2+1)(2-1)+1)$.

This reduces to $200 + 196 + 192 + \ldots + 4 = 4(1+2+\ldots + 50) = 4 \cdot\frac{50 \cdot 51}{2} = \boxed{\textbf{(E) } 5100}$

~Zeric

Solution 3 (end behavior)

We know that $P(x)$ is a $100$-degree function with a positive leading coefficient. That is, $P(x)=x^{100}+ax^{99}+bx^{98}+...+\text{(constant)}$.

Since the degree of $P(x)$ is even, its end behaviors match. And since the leading coefficient is positive, we know that both ends approach $\infty$ as $x$ goes in either direction.

\[\lim_{x\to-\infty} P(x)=\lim_{x\to\infty} P(x)=\infty\]

So the first time $P(x)$ is going to be negative is when it intersects the $x$-axis at an $x$-intercept and it's going to dip below. This happens at $1^2$, which is the smallest intercept.

However, when it hits the next intercept, it's going to go back up again into positive territory, we know this happens at $2^2$. And when it hits $3^2$, it's going to dip back into negative territory. Clearly, this is going to continue to snake around the intercepts until $100^2$.

To get the amount of integers below and/or on the $x$-axis, we simply need to count the integers. For example, the amount of integers in between the $[1^2,2^2]$ interval we got earlier, we subtract and add one. $(2^2-1^2+1)=4$ integers, so there are four integers in this interval that produce a negative result.

Doing this with all of the other intervals, we have

$(2^2-1^2+1)+(4^2-3^2+1)+...+(100^2-99^2+1)$. Proceed with Solution 2. ~quacker88

Video Solution

https://youtu.be/RKlG6oZq9so

~IceMatrix

See Also

2020 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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All AMC 10 Problems and Solutions

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