(I am aware there are similar questions on the forum)
What is the Question?
A positive integer (in decimal notation) is divisible by $11$ if and only if the
difference of the sum of the digits in even-numbered positions and the sum of digits in odd-numbered positions is divisible by $11$.
For example consider the integer 7096276.
The sum of the even positioned digits is $0+7+6=13.$ The sum of the odd positioned
digits is $7+9+2+6=24.$ The difference is $24-13=11$, which is divisible by 11.Hence 7096276 is divisible by 11.
(a)
Check that the numbers 77, 121, 10857 are divisible using this fact, and that 24 and 256 are not divisible by 11.
(b)
Show that divisibility statement is true for three-digit integers $abc$.
Hint: $100 = 99+1$.
What I've Done?
I've done some research and have found some good explanations of divisibility proofs. Whether that be 3,9 or even 11. But…the question lets me take it as fact so I don't need to prove the odd/even divisible by 11 thing.
I need some help on the modular arithmetic on these.
For example… Is 77 divisible by 11? $$7+7 = 14 \equiv …$$
I don't know what to do next.
Thanks very much, and I need help on both (a) and (b).
Best Answer
In order to apply the divisibility rule, you have to distinguish between odd and even positioned digits. (It doesn't matter how you count.)
Example: In 77 the first position is odd, the second even, so you would have to calculate $7-7=0$, which is divisible by 11.
Now it should be easy for you to understand what you try to prove in (b): If a,b,c are three digits abc is the number $100a+10b+c$. You know what it means to say that this number is divisible by 11. You have to prove that $$11\vert (a+c)-b \Leftrightarrow 11\vert 100a+10b+c$$ or with modular arithmetic $$ (a+c)-b \equiv 0 \pmod{11}\Leftrightarrow 100a+10b+c\equiv 0 \pmod {11}\; .$$ I don't want to spoil the fun so I leave it there.
P.S. Sorry, I hadn't noticed the answer posted in the meantime.