How to find large modulus on Casio fx-991 MS

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  • #1
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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

it's answer is 50

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 .

please advise me

Thanks
Anes
 

Answers and Replies

  • #2
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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.
 
  • #3
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Dear Mentor ,
I don't get your point fully . please solve (28)^12 mod 97 in next step.

Thanks
Anes
 
  • #4
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(28)^12 = (28^2)^6 for example. You can just repeat that step.
 
  • #5
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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
 
  • #6
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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
 
  • #7
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.
 
  • #8
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Please open a new thread for new questions, this thread is from 2014. Anyway, the existing posts should help.
 

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