My Math Forum Divison - Remainder of Positive Integer

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 January 31st, 2018, 04:44 PM #1 Newbie   Joined: Jan 2018 From: Toronto Posts: 12 Thanks: 0 Divison - Remainder of Positive Integer The remainder of any positive integer when divided by 100 is the integer made of the two rightmost digits - True or False and why? - I am unsure how to answer this question. I say true, but don't know how to answer the why. Any suggestions on how two answer that question is greatly appreciated. Thanks.
 January 31st, 2018, 05:21 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,549 Thanks: 1399 any positive integer can be written as $i = 100k + r,~0 \leq r < 100$ so clearly $\dfrac{i}{100} = r$ $r$ is the rightmost two digits Thanks from Tricia
 February 1st, 2018, 03:31 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 Ooh, I just cannot look at "$\displaystyle \frac{i}{100}= r$" without cringing! What you mean, of course, is that is i= 100k+ r then $\displaystyle \frac{i}{r}= \frac{100k+ r}{100}= k+ \frac{r}{100}$ so that the remainder is r. Tricia, do you understand that our numeration system is "base 10"? That is that the number "3215" means $\displaystyle 3\times 10^3+ 2\times 10^2+ 1\times 10+ 5= 3 \times 1000+ 2\times 100+ 1 \times 10+ 5= (3\times 10+ 2)\times 100+ 1\times 10+ 5$. If we divide by 100, that "100" cancels the "100" I factored out of the first two digits leaving a quotient of $\displaystyle 3\times 10+ 2= 32$ and a remainder of $\displaystyle 1\times 10+ 5= 15$, the last two digits. Thanks from Tricia Last edited by Country Boy; February 1st, 2018 at 03:37 AM.

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