Finding remainder: looking for a formula

[defeated_soldier]

I want a formula to find a remainder when its divided by some other number.

For example ,

Find the remainder when 5 times the units digit of (37)^735 is divided by 4

PLease note, i am not asking for answer . I would like to solve myself....so please dont provide me the answer or solve it here .

you could tell me some step by steps to attak this problem.

I just looking for the way people solve this kind of problem .

I have heard there are Fermat/Euler remainder theorem kind of stuff , which i really dont understand and surfed net to master this technique but failed.


I'll highly appreciate , if you guys kindly help me to master the technique .
I would like to apply the same formula in similar kind of problem , so i am not looking for a problem specific answer.

What i am expecting is , the procedure to find remainder by fermat/eulaer (whichever is appropriated) theorem.

can you guys help me out .


Thank you very much for reading my post

 

[Denis]

Re: Finding remainder

I want a formula to find a remainder when its divided by some other number.

Keep it simple, d_s:

r = remainder
n = number
d = divisor

r = n - d * |n / d|

|| means integral part; like |37 / 5| = 7

 

[soroban]

Re: Finding remainder

Hello, defeated_soldier!

I don't have a formula for you, but I have an approach . . .

Find the remainder when 5 times the units-digit of \(\displaystyle (37)^{735}\) is divided by 4

We are raising \(\displaystyle 37\) to consecutive powers and examiing its units-digit.

Consider the units-digit of consecutive powers of \(\displaystyle 7\).

\(\displaystyle \begin{array}{ccc}7^1 &\:\rightarrow\: &7 \\ 7^2 & \:\rightarrow\: & 9 \\ 7^3 & \:\rightarrow\: & 3 \\ 7^4 & \:\rightarrow\: & 1 \\ 7^5 & \:\rightarrow\: & 7 \\ & \vdots \\ \end{array}\)

The units-digits occur in a four-digit cycle: \(\displaystyle 7\)-\(\displaystyle 9\)-\(\displaystyle 3\)-\(\displaystyle 1\)

Can you determine the units-digit of the \(\displaystyle 735^{th}\) power?

 

[defeated_soldier]

Thanks sorobon,

nice tips .

cream thing , that i learnt from your advice is ,

try writing down on the paper couple of terms and then look for a cycle ...........there must be some cycle . then use that cycle to find the last digit finally.

in this case , last digit is 3

3x5=15

15/4 = 3 is the remainder