# Which two digit number when you find the product of the digits yields a number that is half the original?

Which two digit number when you find the product of the digits yields a number that is half the original?

Let x=$ab$ be the $2$-digit number. So $x=10a+b$.

Then $ab=\frac{x}{2} \implies ab=\frac{10a+b}{2} \implies 2ab=10a+b \implies b=\frac{10a}{2a-1}$. I guess $a=3$ and get $b=6$. So the answer is $36$.

But how can this be done without guessing.

• When there are only ten options to try, the quickest way will be to just try them all. Anything else is overkill. – vadim123 Mar 18 '16 at 21:47
• Maybe rewrite as $5+\frac{5}{2a-1}$. The number $5$ has few divisors. – André Nicolas Mar 18 '16 at 21:51

Given $b=\frac {10a}{2a-1}$, and $b$ is a whole number, you need $2a-1$ to divide evenly into $10$ because it can't share any factors with $a$. There are only two odd factors of $10, 1$ and $5$, so $2a-1$ must be one of these. $1$ doesn't work because you get $b=10$ which is not acceptable.

Starting with $$ab = \frac{1}{2}(10 a + b) = 5 a + (1/2) b$$ This is only one equation with two unknowns, so there is one degree of freedom.

I would note the additional constraints $$b \bmod 2 = 0 \\ a, b \in \Sigma_d = \{ 0, \dotsc, 9 \}$$

So we take the more constrained variable, $b$, as free variable (to minimize trials) and have: $$(b - 5) a = (1/2) b \iff \\ a = \frac{b}{2b - 10}$$ Now you could simply try $0,2,4,6,8$ for $b$ and get $0, -1/3, -2, 3, 4/3$ for $a$, where $$(a, b) \in \{ (0,0), (3, 6) \}$$ are the feasible solutions.

However due to the normalization to remove heading $0$ digits $00$ is no valid decimal representation, and we are left with the solution $$(a,b) = (3,6)$$

$2ab-10a-b=0$ is a simple quadratic Diophantine equation. The canonical approach in this case is the following.

You add $5$ to both sides and you obtain $$2ab-10a-b+5=5,\ \ \Rightarrow (2a-1)(b-5)=5$$ so $b-5$ or $2a-1$ must be equal to $5$ (or $-5$, but it is not possible) and the other must be equal to $1$.

But $0<b<10$ so $b-5$ can't be $5$, it must be $1$, hence $b=6$ and thus $2a-1=5$, from which $a=3$.

From what you figured out so far, $ab=5a+\frac{b}{2}$. If there is an integer solution, $b$ is even and greater than $5$, so $b$ is either $6$ or $8$. Also (see the picture), $\frac{b}{2}$ is divisible by $b-5$, which rules out $b=8$. If $b=6$, then the rightmost rectangle below is $1$ unit across and has area $\frac{6}{2}=3$, so $a=3$.