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AMC 8 Math Olympiad USA Math Olympiad

Integer Problem | AMC 10A, 2020 | Problem 17

Try this beautiful problem from Number theory based on Integer from AMC-10A, 2020. You may use sequential hints to solve the problem.

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


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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