Do Not Use a Calculator: Divisibility By 11

True story. Wait to think logically.

An $n+1$ digit number can be written as $10^n a_n+10^{n-1} a_{n-1}+10^{n-2} a_{n-2}+\cdots +10^1 a_1+10^0 a_0$ .Let us name it $N$ .In order for it to be divisible by 11,we must have: $N\equiv 0 \mod 11\\ 10^n a_n+10^{n-1} a_{n-1}+\cdots + 10^0 a_0 \equiv 0 \mod 11$ Now,we have 2 cases:

Case 1

$\text{n is even}$ Then $n=2k$ for $k\geq 0$ .Note that: $10^n\equiv (10^k)^2 \equiv (-1)^2\equiv \boxed{1 \mod 11}$

Case 2

$\text{n is odd}$ Then $n=2m+1$ for some $m$ .Note that: $10^n \equiv (10^n)^2 \times 10 \equiv 1\times -1 \equiv \boxed{-1 \mod 11}$

Assume without loss of generality that $n$ is even (If $n$ was odd the following proof would still hold,with some minor changes;if you want,you can prove it for practice :) ).Then: $10^n a_n+10^{n-1} a_{n-1}+\cdots +10^0 a_0\\ 1(a_n)+(-1)(a_{n-1})+(1)(a_{n-2})+\cdots + (1)(a_0)\equiv 0 \mod 11\\ \boxed{a_n-a_{n-1}+a_{n-2}-a_{n-3}+\cdots + a_0 \equiv 0\mod 11}$ Hence $N$ is divisible by $n$ iff $a_n-a_{n-1}+a_{n-2}+\cdots + a_0$ is divisible by $11$

U r wrong.

that's the formula: the sum of the odd number digits-the sum of the even number digits if the answer is divisible by 11 then yes if the result is much too big try the trick again

Thank you, I agree. There will be a remainder or extended decimal equivalent. There is no whole number resolution. Therefore, the question is misleading at best. Technically, all numbers outside of 0 are divisible if we are going to assume the question is not speaking of whole numbers.

I don't know how it's not divisible by 11. If dividing by 11, the answer is 5,162,039,263,187. Double checked on a calculator. No remainder or decimal. I think you may have a typo when inputting. Skip the calculator.