# Thread: Algebra solution to (unusual) digit reversal problem

1. ## Algebra solution to (unusual) digit reversal problem

The problem is as follows:

"By a mistake a number was doubled instead of raising it to the power of two. The result was a two digit number which was the reverse of what it should have been. What was the original number?"

I realized rapidly that the solution must be a number between 5 and 9. Any less than 5 and doubling does not result in a two digit number, any more then 9 and squaring does not result in a two digit number.

Clearly the answer is 9 since $2·9 = 18$ and $9^2 = 81$. [Simple, well below my pay grade.]

But is there an algebra solution to this problem?

I got as far as this simple system...

$\displaystyle 2n = 10a + b \Longrightarrow n = 5a + \frac{b}{2}$
$\displaystyle n^2 = 10b + a$

Substituting $n$ from the top equation into the bottom one leads to nothing useful.

$\displaystyle (5a + \frac{b}{2})^2 = 10b + a \Longrightarrow 100a^2 + 20ab + b^2 = 40b + 4a$

Am I missing some clever way of solving the problem using algebra?

Thanks.

2. ## Re: Algebra solution to (unusual) digit reversal problem

$2n=10a+b$ and $n^2=10b+a$ we get

$n(n+2)=11(a+b)$

so $11$ must divide $n+2$ since it cannot divide $n$

using the condition $5\leq n\leq 9$

3. ## Re: Algebra solution to (unusual) digit reversal problem

I'm sorry but I do not follow the logic. Could you spell it out for me please?

Note: I understand the equation addition, just not what follows.

Thanks.

4. ## Re: Algebra solution to (unusual) digit reversal problem

if a prime number p divides the product xy then
p divides x or p divides y

we have that 11 divides n(n+2) so
11 divides n or 11 divides n+2

but 11 cannot divide n since n is smaller than 11
so 11 must divide n+2 which means
n=9

5. ## Re: Algebra solution to (unusual) digit reversal problem

Okay, got it. Thanks for your help.