INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

May 5, 2020

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 \(P(n) \geq 0\)?

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

Key Concepts

Number system



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

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

Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.