Question

A number is divisible by $ 11 $ if the difference of the sums of even and odd places is or a multiple of $ 11 $
A. True
B. False

Answer 1

Hint: A number is divisible by $ 11 $ if the difference of sums of even and odd places is or a multiple of $ 11 $ .Divisibility Test by $ 11 $ - Subtract the last digit from the remaining leading truncated number. If the result is divisible by $ 11 $ , then so was the primary number. Apply this rule over and over again as necessary. For example ,
 \[19151 \to 1915 - 1 = 1914 \to 191 - 4 = 18 \to 18 - 7 = 11,\] so yes, \[19151\] is divisible by $ 11 $ .

Complete step by step solution:
This curious fact is easily shown using modular arithmetic. Since \[10n\] is congruent to \[\left( { - 1} \right)n{\text{ }}mod{\text{ }}11\] , we see that \[1,{\text{ }}100,{\text{ }}10000,{\text{ }}1000000,\] etc. have remainders one when divided by \[11\] , and \[10,{\text{ }}1000,{\text{ }}10000\] , etc. have remainders \[\left( { - 1} \right)\] when divided by \[11\] . Thus
 \[2728{\text{ }} = {\text{ }}2{\text{ }} \times {\text{ }}1000{\text{ }} + {\text{ }}7{\text{ }} \times {\text{ }}100{\text{ }} + {\text{ }}2{\text{ }} \times {\text{ }}10{\text{ }} + {\text{ }}8\] ,
so its remainder when divided by \[11\] is simply \[2\left( { - 1} \right){\text{ }} + {\text{ }}7\left( 1 \right){\text{ }} + {\text{ }}2\left( { - 1} \right){\text{ }} + {\text{ }}8\left( 1 \right),\] the alternating sum of the digits. (It’s sum is the negative of what we found above because the alternation here begins with a minus one .) But either way, if this alternating sum is divisible by eleven, then so is that the original number.
In fact, our observation shows more: that of course once we take the alternating sum of the digits read from right to left (so that the sign of the units digit is usually positive), then we obtain \[N{\text{ }}mod{\text{ }}11.\]
Therefore, the given statement is true.
A number is divisible by $ 11 $ if the difference of sums of even and odd places is or a multiple of $ 11 $ .
So, the correct answer is “Option A”.

Note: Questions similar in nature as that of above can be approached in a similar manner and we can solve it easily. Divisibility Test by $ 11 $ - Subtract the last digit from the remaining leading truncated number. If the result is divisible by $ 11 $ , then so was the primary number. Apply this rule over and over again as necessary.