Why this is happening?

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stevecowall

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Nov 9, 2011
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Well as I across online, I found this problem:
Determine the remainder when
number.png
is divided by 37

This seems to be easy but I have been trying to figure out for so long that I need freemathhelp.com. I type on both my CASIO fx-115ES and TI-83 Plus but the result or what it shows was "Error". Can you show me some how to find the solution of this?
 
How about you start first by showing us something that you tried. :cool:
 
stevecowall said:
As i told you in the head, i tried with the calculator but it keeps putting up error.
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Of course, the calculator cannot handle that number. It will output an overflow error, its just too big.

This will have to be done analytically (by hand)... What course are you in? Have you covered similar problems like this?
 
Hello, stevecowall!

\(\displaystyle \text{Determine the remainder when }36^{553}\text{ is divided by 37.}\)
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Obviously, if we could get this answer on a calculator, it is not much of a math problem, is it?


Did you try to crank out a few remainders when dividing by 37?


. . \(\displaystyle \begin{array}{|ccc|c|}\hline & 36^n && \text{Remainder} \\ \hline 36^1 &=& 36 & 36 \\ 36^2 &=& 1,\!296 & 1 \\ 36^3 &=& 46,\!656 & 36 \\ 36^4 &=& 1,\!679,\!616 & 1 \\ \vdots && \vdots & \vdots \\ \hline\end{array}\)


Got it?
 
stevecowall said:
Well as I across online, I found this problem:
Determine the remainder when
View attachment 1502
is divided by 37


This seems to be easy but I have been trying to figure out for so long that I need freemathhelp.com. I type on both my CASIO fx-115ES and This seems to be easy but I have been trying to figure out for so long that I need freemathhelp.com. I type on both my CASIO fx-115ES and TI-83 Plus but the result or what it shows was "Error". Can you show me some how to find the solution of this?
Click to expand...


As you've found out, your calculator isn't going to be much help on this one, because 36553 is just TOO BIG for your calculator to handle.

What to do in such a situation??

Well, sometimes it helps to start with a much simpler problem, and HOPE you can see a pattern emerging that will allow you to solve the more difficult problem.

So....here's what you might want to do.

Find the remainder for each of THESE divisions:

361 / 37

362 / 37

363 / 37

364 / 37

Look at your results. Do you see a pattern? Will that pattern help you to find the remainder when you divide 36553 by 37?
 
WOW!! This problem seems to be easy. I just don't understand why i didn't get it earlier. Thanks you guys JeffM, Mrspi, and soroban.
 
I could swear there was a way to do problems like this using Taylor series expansion... am i mistaken?
 
When you are dividing by a prime, you may always use fermat's little theorem (restated):

\(\displaystyle \text{ If } p \text{ is prime then for any integer } a \text{ it is true that}\)
\(\displaystyle a^{p-1} \text{ when divided by } p \text { has a remainder of 1}.\)

In this case, \(\displaystyle \frac{36^{36}}{37} = k + \frac{1}{37}\) for some integer \(\displaystyle k\).

You may rewrite \(\displaystyle 36^{553} = (36^{36})^{15} \cdot 36^{13}\).

So your problem reduces to calculating the remainder of "only" \(\displaystyle 36^{13}\) when divided by \(\displaystyle 37\).

For small primes, it could make the difference of being able to use a calculator.

If you want to learn more tricks, you can look at sources for "modular arithmetic". For example, this problem becomes trivial if you know that 36 and -1 give the same remainders when being divided by 37. So you only need to calculate \(\displaystyle (-1)^{553} = -1\), then add \(\displaystyle 37\) to obtain the "correct" (positive) answer of \(\displaystyle 36\).
 
stevecowall said:
As i told you in the head, i tried with the calculator
Click to expand...

You told us only that you used a calculator, but you did not tell us what you actually did with the calculator.

It never occurred to me that you were literally trying to do the posted division.

You told us that you had worked "so long" at it; hence, I naturally assumed that you had worked on something else.

Some of those efforts is what I wanted to see. :cool:
 
renegade05 said:
I could swear there was a way to do problems like this using Taylor series expansion... am i mistaken?
Click to expand...

These are integer problems with integer answers. Taylor's series most probably won't work right....
 
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