Modular Arithmetic: Solve (21999+31998+51997) Divided by 7

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In summary, the conversation is about finding the smallest positive remainder of a sum divided by seven. The attempt at a solution involved using congruences to simplify the numbers, but the final answer was incorrect. It was found that the correct remainder is six, not two as originally thought. The reason for the error was due to not using whole numbers in one of the calculations.
  • #1
triac
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Homework Statement


Okay, so I'm going to find the smallest positive remainder of (21999+31998+51997) divided by seven.


Homework Equations





The Attempt at a Solution


Well, I did like this:
23 is congruent to 1 (mod 7). Therefore, 21999= (23)1999/3 is congruent to 1 (mod 7).
33 is congruent to (-1) (mod 7). Therefore, 31998=(33)666 is congruent to (-1)666=1 (mod 7).
53 is congruent to (-1) (mod 7). Therefore, 51997=(53)665*25 is congruent to (-1)665*4 = -4 (mod 7)

So, all in all the remainder should be two. However it says in the key that it is six, and I can't see where I'm wrong. Got any suggestions?
 
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  • #2
triac said:

Homework Statement


Okay, so I'm going to find the smallest positive remainder of (21999+31998+51997) divided by seven.


Homework Equations





The Attempt at a Solution


Well, I did like this:
23 is congruent to 1 (mod 7). Therefore, 21999= (23)1999/3 is congruent to 1 (mod 7).
33 is congruent to (-1) (mod 7). Therefore, 31998=(33)666 is congruent to (-1)666=1 (mod 7).
53 is congruent to (-1) (mod 7). Therefore, 51997=(53)665*25 is congruent to (-1)665*4 = -4 (mod 7)

So, all in all the remainder should be two. However it says in the key that it is six, and I can't see where I'm wrong. Got any suggestions?

Here's what I get:
23 is congruent to 1 (mod 7). Therefore, 21999= (23)666*2 is congruent to 2 (mod 7) (not 1 mod 7 as you had).
33 is congruent to (-1) (mod 7). Therefore, 31998=(33)666 is congruent to (-1)666=1 (mod 7).
53 is congruent to (-1) (mod 7). Therefore, 51997=(53)665*25 is congruent to (-1)665*4 = -4 (mod 7) = 3 mod 7.

Add 'em up and you get 6 mod 7.
 
  • #3
Ok, thanks a lot!
I just wonder, why is it wrong to do the way I did, why doesn't it work?
 
  • #4
Hi triac! :smile:
triac said:
I just wonder, why is it wrong to do the way I did, why doesn't it work?

Because 1999/3 isn't a whole number. :wink:
 

1. What is modular arithmetic?

Modular arithmetic is a branch of mathematics that deals with operations on integers or numbers that wrap around a fixed value, known as the modulus. It is often used to solve problems related to time, distance, and congruence.

2. How do you solve an equation using modular arithmetic?

To solve an equation using modular arithmetic, you first need to identify the modulus, which is the number that the operations will wrap around. Then, you perform the operations on the given numbers, keeping in mind that the result must be within the range of the modulus. If the result is larger than the modulus, you can repeatedly subtract the modulus from the result until it falls within the range.

3. What is the modulus in the equation (21999+31998+51997) divided by 7?

The modulus in this equation is 7, as indicated by the "divided by 7" part. This means that the operations will wrap around the number 7.

4. How do you solve (21999+31998+51997) divided by 7 using modular arithmetic?

To solve this equation using modular arithmetic, we can first calculate the sum of the three numbers, which is 105994. Then, we can repeatedly subtract 7 from this number until it falls within the range of 0 to 6. This gives us a final result of 4, which is the remainder when 105994 is divided by 7.

5. What is the final solution to (21999+31998+51997) divided by 7?

The final solution to this equation is 4, which is the remainder when 105994 is divided by 7 using modular arithmetic.

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