What Is the Process to Solve for d in the Equation (d)(13) mod (2436) = 1?

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Homework Help Overview

The discussion revolves around finding the value of d in the equation (d)(13) mod (2436) = 1, which falls under the topic of modular arithmetic and number theory.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the process of finding d, with some suggesting the division of 2436 by 13 and discussing the implications of 2436 not being a multiple of 13. Others delve into the Euclidean division algorithm to express the relationship between the numbers involved.

Discussion Status

The discussion includes various approaches to the problem, with hints and insights shared among participants. There is an exploration of the modular inverse concept, and while some participants provide hints, there is no explicit consensus on a single method or solution.

Contextual Notes

Participants note that the numbers involved are large, which adds complexity to the problem. The distinction between 2436 and 2435 is highlighted as significant in the context of the problem.

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I need to figure out the value for d in this equation, but because the numbers are large, I don't know how.

(d)(13) mod (2436) = 1.

d times 13 mod 2436 should be equal to 1.

Is there a process I can use to find out the value of d.
 
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hi maxfails! smile:

hint: divide 2436 by 13 :wink:
 
tiny-tim said:
hi maxfails! smile:

hint: divide 2436 by 13 :wink:
2436 isn't a multiple of 13. Maybe I'm missing the point.
 
That "2435 isn't a multiple of 13" is the point! If it were, there would be no answer to this problem. "Modulus" problems are all about remainders. You are trying to find a number, n, such that some multiple of 2436, divided by 13n, has remainder 1, that is, such that 13n= 1+ 2436k for some integer k. That is the same as 13n- 2436k= 1.

Euclidean division algorithm:

13 divides into 2436 187 times with remainder 5- that is, 2436- (187)(13)= 5.

5 divides into 13 twice with remainder 3- that is, 13- (2)(5)= 3.

3 divides into 5 once with remainder 2- that is 5- 1(3)= 2.

2 divides into 3 once with remainder 1- that is, 3- 1(2)= 1.

Replace that "2" by 5- 1(3) from the previous equation: 3- 1(5- 1(3))= 2(3)- 5= 1. Replace the "3" in that with 13- 2(5): 2(13- 2(5))- 5= 2(13)- 5(5)= 1.
Replace the "5" in that by 2436- 187(13): 2(13)- 5(2436- 187(13))= 937(13)- 5(2436)= 1.

Now that I used 2436 rather than the mistaken 2435, I get the same result as Mensenator.
 
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maxfails said:
I need to figure out the value for d in this equation, but because the numbers are large, I don't know how.

(d)(13) mod (2436) = 1.

d times 13 mod 2436 should be equal to 1.

Is there a process I can use to find out the value of d.

Yes, find the modular imverse of 13 & 2436.

the answer is then invert(13,2436) * 1 % 2436.

Should be 937.
 
lol, you are in my security class.
prolly Ian.
 

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