Try this beautiful problem from Number theory based on Integer.
Define \(P(x)=(x-1^2)(x-2^2)......(x-{100}^2)\)
How many integers \(n\) are there such that \(P(n) \geq 0\)?
Number system
Probability
divisibility
Answer: \(5100\)
AMC-10A (2020) Problem 17
Pre College Mathematics
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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