Find two-digit numbers such that their ratio equals $0.254902$.

Question

If I have the quotient $0.254902$ and I know that it was obtained by dividing a two-digit dividend by a two-digit divisor, how do I find the values of each?

Answer 1

The rational number you give can be expressed as a fraction of two integers as $$0.254902=\frac{254902}{1000000}=\frac{127451}{500000},$$ where the latter cannot be simplified any further.

To find integers less than $100$ whose ratio is close to $0.254902$, note that $$0.254902>\frac{1}{4}\qquad\text{ and }\qquad0.004902<\frac{1}{200},$$ where $0.004902=.254902-\tfrac{1}{4}$. It follows that $$\frac{50}{200}<0.254902<\frac{51}{200}.$$ So for a fraction to approximate the given value, it must be of the form $\tfrac{x}{4x-1}$ for some $x\leq25$. This is a strictly decreasing function in $x$, allowing us to 'close in' on the desired value of $x$ by computing a few values. We find that for $x=13$ we obtain the repeating decimal expansion $$\frac{13}{51}=0.\overline{2549019607843137},$$ which rounds to the given value.

Feeding $0.254902$ to Wolfram Alpha also returns $\frac{13}{51}$ as a possible closed form.

Answer 2

For two digits, you can just try them. I find $\frac {13}{51}=0.\overline{2549019607843137}$. For larger numbers of digits you can use continued fractions to find the best rational approximation

Answer 3

Your number is not the quotient of two 2-digit numbers. It is very similar to $13/51=.254901961$, so maybe whoever asked the question made a mistake by rounding.

If you multiply $0.254902$ by a number of the form $10a+b$ (with $a,b$ digits) and you want the result to be an integer (which would be the case if it were a quotient by a 2-digit integer), then $b=5$ (because otherwise the last decimal digit of the result will not be zero).

Now $5\times0.254902=1.27451$. This, plus $10a\times0.254902=a\times 2.54902$ will have to be an integer. The last digit of $(10a+5)\times0.254902$ will then be $1+2a$, which can never be zero. In other words, no choice of $a,b$ will make the product an integer.