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.
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.
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.
