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Math challenge II-A Number theory Problem 2.22

 
 
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Math challenge II-A Number theory Problem 2.22
by Zeyin Wu - Saturday, November 6, 2021, 10:57 PM
 

The sample answer wrote 11 divides a-b+c-d. Isn’t it -a+b-c+d?

 
Picture of John Lensmire
Re: Math challenge II-A Number theory Problem 2.22
by John Lensmire - Friday, November 12, 2021, 10:33 AM
 

Thanks for your patience with my reply.

Short Answer:

Yes, it should be $d-c+b-a$ to be consistent and it is now updated.

Long Answer (+ preview of later chapters): 

Actually, as a divisibility rule it doesn't matter if you take the alternating sum "right to left" or "left to right". For example, with four digits it is okay to do $a-b+c-d$ or $d-c+b-a$. That is because the two alternating sums differ by $-1$, and thus,$$11 \mid (a-b+c-d) \text{ if and only if } 11 \mid d-c+b-a.$$So why do we want to stress doing the alternating sum "right to left"? What we'll show later is that actually this alternating sum also preserves the remainder when dividing by $11$.

Going back to our 4-digit example, this means that the remainder of $n=\overline{abcd}$ when dividing by $11$ is the same as the remainder when $d-c+b-a$ when dividing by $11$. Here, multiplying by $-1$ DOES change the remainder, so this order is important.

Hope this helps!