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# Question

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Here is the question that I'm stuck on:

"When four positive integers are divided by $$11$$, the remainders are  $$2,4,6,$$ and $$8$$ respectively. When the sum of the four integers is divided by $$11,$$ what is the remainder?"

I'm confused on how to solve this.

May 23, 2023

#1
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Let the four integers be a,b,c,d. We know that a≡2(mod11), b≡4(mod11), c≡6(mod11), and d≡8(mod11). Adding all of these congruences, we get [a+b+c+d\equiv 2+4+6+8\equiv 20\pmod{11}.]Therefore, the remainder when a+b+c+d is divided by 11 is 0​.

May 23, 2023
#3
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When you have a question like this, you have to specify something like" what are the SMALLEST 4 positive integers.....etc.", because the LARGEST can go to infinity.

So, the smallest 4 positive integes are stated in your question! They are: 2, 4, 6, 8 because:

2 mod 11 ==2

4 mod 11 ==4

6 mod 11 ==6

8 mod 11 ==8

Their sum{2 + 4 + 6 + 8} ==20 mod 11 ==9 - the remainder.

P.S. If you had specified that the 4 numbers were all 2-digit numbers and you wanted the SMALLEST 2-digit numbers, then you could have:13, 15, 17, 19. If you divide these 4 by 11, you will have the same remainders of: 2, 4, 6, 8......and so on.

May 23, 2023