Reasoning about Integers
Date: 10/06/2004 at 08:40:38 From: Kim Subject: Algebra When positive integers p and q are divided by an even positive integer t, they have remainders 2 and t/2, respectively. What is the remainder when the product pq is divided by t? I'm not sure where to start. 2 * t/2 = t. Therefore, you are left with t. So t/t = 1, so the remainder is 0?
Date: 10/06/2004 at 15:25:20 From: Doctor Achilles Subject: Re: Algebra Hi Kim, Thanks for writing to Dr. Math. Your reasoning is correct: you will end up with remainder 0. Here is how I proved it: p/t gives you remainder 2. That means that there is some other integer "n" that you can multiply by t and then add 2 to get p. n*t + 2 = p q/t gives you remainder t/2. Using the same reasoning, there is some integer m such that: m*t + t/2 = q Also, a remainder is always an integer. Therefore t/2 is an integer. Therefore, t is even. Now, we need to multiply p and q together: p*q = (nt+2)(mt+2) If we do this, we find that: p*q = nmt^2 + 2mt + nt^2/2 + t We can then factor "t" out of the answer: p*q = t(nmt + 2m + nt/2 + 1) If we divide both sides by t, we get: pq/t = nmt + 2m + nt/2 + 1 We know from the problem that t, n, and m are integers. That means that nmt is an integer and 2m is an integer. We also know that t is even. That means that nt is even. So nt/2 is an integer. So, pq/t gives us a sum of integers. Therefore pq/t is an integer. Therefore pq/t has no remainder. Hope this helps. If you have other questions or you'd like to talk about this some more, please write back. - Doctor Achilles, The Math Forum http://mathforum.org/dr.math/
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