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I came across the following puzzling problem in an elementary algebra textbook:

Problem. Prove that the sum of a two-figure number and the number written with the same digits in the reverse order is divisible by 11.

Assuming that the numbers are written in the decimal base, I understood that I need to prove that:

Goal. $\forall x \in \mathbb N : (0 \leq x \leq 99) \rightarrow \exists k\in \mathbb N: x + (10 \cdot rem(x, 10) + 1\cdot div(x, 10)] = 11k)$,

where $rem$ and $div$ satisfy: $div(x, y) + rem(x,y) = x$, for all $x \in \mathbb N$. Informally, the claim is that any number between $0$ and $99$ is s.t. it and its 'mirror' number, so to speak, add up to some number divisible by $11$. The proof of this, needless to say, got a bit out of hands. I looked up the solution:

Solution. We have $\overline{ab} = 10a + b$ and $\overline{ba} = 10b + a$. Therefore, $\overline{ab} + \overline{ba} =$ $= 10a + b + 10b + a =$ $11a + 11b = 11(a + b)$, which is obviously divisible by $11$.

This solution puzzles me because it doesn't start by breaking a two-figure number into pieces and by manipulating those pieces to obtain the result, but assumes that we have a function $\overline{x}$ on numerals s.t. $\overline{x}$ returns $10a + b$, where $a,b \in \mathbb N \leq9$. Is that a legitimate thing to do? I feel like I am overcomplicating things because of my inability to see that it's an acceptable move.

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2 Answers 2

up vote 3 down vote accepted

No, I think you are misunderstanding the solution. The overbar notation in context seems to be this author's way of denoting a multidigit number. So $ab$ means $a$ times $b$, but $\overline{ab}$ means the two digit number with an $a$ in the tens place and a $b$ in the ones place. Therefore $\overline{ab}=10a+b$ is just rewriting the two-digit number in expanded form.

The thrust of the argument is: Rather than think about a number $x$ in the range $0 \leq x \leq 99$, instead think about a number of the form $10a + b$ where $a,b$ are single-digits (i.e. in the set 0 through 9). Then its reverse is $10b + a$, and the rest of the proof is trivial.

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I do understand that $\overline{ab}$ means the name of digit $a$ catenated with the name of digit $b$, and not $a \cdot b$. But yes, thanks for your contribution; also helpful. –  Readingtao May 2 at 0:02

It is a legitimate move, but don't get too confused by it: They simply made up some notation to stand for "the number with digits $ab$ in that order."
That is, they intend to distinguish $\overline{ab}$ from $a\cdot b$ by using an overline. This proof could have been written omitting the overline, but there would have been more words.

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Thank you! That's helpful. I somehow was feeling that instead proving things about numbers they were proving things about names of numbers. I understand that's not right. –  Readingtao May 2 at 0:00

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