1
$\begingroup$

Following line are quoted from the book Secrets of Mental Math by Arthur Benjamin and Michael Shermer.

To double-check your answer another way, you can use the method known as casting out elevens. It’s similar to casting out nines, except you reduce the numbers by alternately subtracting and adding the digits from right to left, ignoring any decimal point. If the result is negative, then add eleven to it. (It may be tempting to do the addition and subtraction from left to right, as you do with mod sums, but in this case you must do it from right to left for it to work.) For example: enter image description here

The same method works for subtraction problems: enter image description here

How can we explain why this method works?

$\endgroup$
3
$\begingroup$

The basis idea is $$ 10 \equiv -1 \pmod{11} $$ That is, for a natural number $a = \sum_{k=0}^n a_k10^k$, we have $$ a \equiv \sum_{k=0}^n a_k (-1)^k = a_0 - a_1 \pm \cdots + (-1)^k a_k \pmod{11} $$ That is, both $a$ and the alternating sum have the same remainder modulo eleven. As addition and subtraction works well modulo eleven, this method checks if the remainders modulo eleven are correct.

$\endgroup$
  • $\begingroup$ Is that $\sum_{k=0}^n a_k10^k = \sum_{k=0}^n a_k (-1)^k$ at your point of view; or something else? I couldn't understand why you have stated $\sum_{k=0}^n a_k (-1)^k$ here. $\endgroup$ – Tharindu Sathischandra Mar 17 '16 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.