2020 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 17 Hard

Define P(x)=\left(x-1^{2}\right)\left(x-2^{2}\right) \cdots\left(x-100^{2}\right). How many integers n are there such that P(n) \leqslant 0?

  • A.

    4900

  • B.

    4950

  • C.

    5000

  • D.

    5050

  • E.

    5100

Answer:E

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+\cdots+198 total possible values of n. Simplifying, there are 5000 possible numbers. Summing, there are (E) 5100 total possible values of \eta. \sim

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 \frac{50 \cdot 51}{2}=(\mathbf{E}) 5100

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