[Combinatorics]
Problem: What is the largest two-digit number that becomes $75 \%$ greater when its digits are reversed?

Solution: Let $n$ be the original number and $N$ be the number when the digits are reversed. Since we are looking for the largest value of $n$, we assume that $n>0$.

Since we want $N$ to be $75 \%$ larger than $n$, then $N$ should be $175 \%$ of $n$, or $N=\frac{7}{4} n$.

Suppose that the tens digit of $n$ is $a$ and the units digit of $n$ is $b$. Then $n=10 a+b$.

Also, the tens digit of $N$ is $b$ and the units digit of $N$ is $a$, so $N=10 b+a$.

We want $10 b+a=\frac{7}{4}(10 a+b)$ or $4(10 b+a)=7(10 a+b)$ or $40 b+4 a=70 a+7 b$ or $33 b=66 a$, and so $b=2 a$.

This tells us that that any two-digit number $n=10 a+b$ with $b=2 a$ has the required property.

Since both $a$ and $b$ are digits then $b<10$ and so $a<5$, which means that the possible values of $n$ are 12, 24, 36, and 48 .

The largest of these numbers is 48.

Final answer: 48