# How to find large modulus on Casio fx-991 MS

anes
Hi pals,
I am looking to find a modulus calculation using my casio fx-991 ms calculator

i know how to find modulus using small numbers

eg: 7 mod 3 = 1. This because 7 = 3(2) + 1, in which 1 is the remainder. To do this process on a simple calculator do the following: Take the dividend (7) and divide by the divisor (3), note the answer and discard all the decimals -> example 7/3 = 2.3333333, only worry about the 2. Now multiply this number by the divisor (3) and subtract the resulting number from the original dividend. so 2*3 = 6, and 7 - 6 = 1, thus 1 is 7mod3

but i need to find
5^36 mod 97

but when i do i don't get the full number in calculator it show 1.4500000x 10^34

but i don't know how to calculate mod use this type of result .

Thanks
Anes

Mentor
You'll have to do some calculations yourself, I think.

5^36 = (5^3)^12 = 125^12
125^12 mod 97 = (125 mod 97)^12 mod 97 (using computer notation, not mathematical notation).
125 mod 97 is easy to evaluate.
You can repeat those steps until the number is small enough for the calculator.

anes
Dear Mentor ,
I don't get your point fully . please solve (28)^12 mod 97 in next step.

Thanks
Anes

Mentor
(28)^12 = (28^2)^6 for example. You can just repeat that step.

anes
Thanks dear mentor i got the ultimate answer now

(28^2 mod 97)^6 mod 97 = (8)^6 mod 97 = 262,144 mod 97

which can find by

262,144/97 = 2702.515 take this 2702 as X

262,144 - 97*2702( we call it as X) = 50 (Ans)

Thanks alot i believe my answer is good for those who look in future

Anes

anes
For make this problem easeful , i gave 1 more example

5^58 mod 97

(5^2 mod 97)^29 mod 97

(25 mod 97)^29 mod 97
(25)^29 mod 97
25. (25)^8 mod 97 // because 29 cannot be factorized further
25.(25^4 mod 97)^7 mod 97

25.(6)^7 mod 97

25.91 mod 97

= 44(Ans)

Hope all understand

Thanks alot

Santhanahari
How can I get the solutions for solving (7^12) mod 71 ?? Correct answer is: 4, But I m getting a wrong answer...Anybody help me.

Mentor
Please open a new thread for new questions, this thread is from 2014. Anyway, the existing posts should help.