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Is there an easy to solve this problem? I can find the answer by using a complicated rule that I don't understand. Even if I try to remember this rule, I probably will forget about it a year later. The rule is "To find out if a number is divisible by 11, add every other digit and call that sum $x$. Add together the remaining digits, and call that sum $y$. Take the positive difference of $x$ and $y$. If the difference is zero or a multiple of eleven, then the original number is a multiple of eleven."

A and B are non-zero digits for which A468B05 is divisible by 11. What is A+B?

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  • $\begingroup$ Can anyone tell me why divisibility is so important in math? $\endgroup$ – learning May 26 '16 at 3:11
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    $\begingroup$ that's a very loaded question and if you're serious about it I suggest posting it as a separate question or at least searching this site for similar questions. I think a number theorist can better elaborate but one reason is that divisibility and primes are central ideas in cryptography. $\endgroup$ – tilper May 26 '16 at 3:37
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    $\begingroup$ Divisibility is quite important in Diophantine equations. You can often restrict the possible solutions to particular residue classes. The classic divisibility tests that depend on the base $10$ representation of numbers, not so much. $73$ is a prime in any base. The fact that you can test divisibility by $9$ with a trick involving the digits is particular to base $10$ (and some others) so has little mathematical content. $\endgroup$ – Ross Millikan May 26 '16 at 3:51
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$$x=4+8+0$$

$$y=A+6+B+5$$

Case 1: $x-y=0$

$$12-11-(A+B)=0$$

$$A+B=1$$

This means either $A$ or $B$ is $0$, so it doesn't work.

Case 2: $x-y=-11$

$$12-11-(A+B)=-11$$

$$A+B=12$$

$A$ and $B$ can be several combinations of positive nonzero integers that satisfies $A+B=12$, so this is the correct answer.

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