Two-Digit Number Whose Digit Product Equals Half Itself
Find a two-digit number with the following property: when you multiply its two digits together, the result equals half the original number.
For example, for the number $36$, the digits are $3$ and $6$, and $3 \times 6 = 18 = 36/2$. Does $36$ satisfy the condition? Verify, and determine if there are other solutions.
Hints
- Write the two-digit number as $10x + y$ where $x$ is the tens digit and $y$ is the ones digit, then translate the condition into an equation.
- After clearing denominators you get $2xy = 10x + y$. Solve for $y$ in terms of $x$: $y = 10x/(2x-1)$.
- For $y$ to be an integer, $(2x-1)$ must divide $10x$. Use the identity $10x = 5(2x-1) + 5$ to show $(2x-1)$ must divide $5$, which has only two positive odd divisors.
Worked Solution
How to Think About It: Write the two-digit number algebraically: if the tens digit is $x$ and the ones digit is $y$, the number equals $10x + y$. The condition becomes $xy = (10x + y)/2$. From here it is a small Diophantine search -- clear denominators, rearrange, and use divisibility constraints to narrow down candidates.
Quick Estimate: The product $xy$ grows quadratically, while the number $10x + y$ grows linearly in the digits. For large digits like $x = y = 9$, the product ($81$) far exceeds half the number ($49.5$) -- so solutions, if any, will involve moderate digit sizes. For $x = 1$, $y = 8$: product $= 8$, half of $18 = 9$ -- close but no. For $x = 3$, $y = 6$: product $= 18$, half of $36 = 18$ -- yes. So $36$ works. Expect at most a couple solutions.
Approach: Set up the Diophantine equation and use divisibility arguments to enumerate all solutions.
Formal Solution:
Let the tens digit be $x$ ($1 \leq x \leq 9$) and the ones digit be $y$ ($0 \leq y \leq 9$), so the number is $10x + y$. The condition is: $$xy = \frac{10x + y}{2} \implies 2xy = 10x + y \implies y(2x - 1) = 10x \implies y = \frac{10x}{2x-1}$$
For $y$ to be a positive integer, $(2x-1)$ must divide $10x$. Write: $$10x = 5(2x-1) + 5 \implies y = 5 + \frac{5}{2x-1}$$
So $(2x-1)$ must divide $5$. The positive odd divisors of $5$ are $1$ and $5$:
- $2x - 1 = 1 \implies x = 1$: $y = 5 + 5 = 10$. Not a valid digit.
- $2x - 1 = 5 \implies x = 3$: $y = 5 + 1 = 6$. Valid. Number: $36$.
Verification: $3 \times 6 = 18 = 36/2$. Confirmed.
Answer: The unique two-digit solution is $\boxed{36}$.
Intuition
The key technique here is representing a multi-digit number algebraically as a sum of its digits times powers of $10$. Once you have $10x + y$, number theory takes over: you need an integer solution to a linear Diophantine-style equation, which you attack with divisibility. Writing $10x = 5(2x-1) + 5$ is an instance of polynomial division -- you find the remainder to isolate the divisibility constraint. This trick generalizes: whenever you need an integer solution to $y = f(x)/g(x)$, write $f(x) = q \cdot g(x) + r$ and check when $g(x)$ divides the constant remainder $r$.
In interviews, these problems often have a unique or very small number of solutions, and the divisibility argument is the clean path. Brute-forcing all $90$ two-digit numbers would also work here, but spotting the algebraic structure and bounding the search to two cases is what distinguishes a strong answer.