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January 31st, 2018, 04:44 PM   #1
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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.

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January 31st, 2018, 05:21 PM   #2
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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
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February 1st, 2018, 03:31 AM   #3
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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.
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Last edited by Country Boy; February 1st, 2018 at 03:37 AM.
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