A two-digit number can be divided by 2, while divided by 3 will remain 1, divided by 4 will remain 2, divided by 5 will remain 3. What is the number?

I've tried to think like this:

all numbers that can be divided by 5 have a last digit of either 0 or 5. So, the second digit of n must be 3 or 8. Since n is completely divisible by 2, it is an even number, therefore the second digit of n must be 8.

n divided by 3 will remain 1, since the second digit is 8, the number that can still be divided by 3 must have a second digit of 7. The possible answers are 27, 57 and 87. So n can be 28, 58 or 88.

n divided by 4 will remain 2, the number that can still be divided by 4 must have a second digit of 6. The possible answers are 16, 36, 56, 76 and 96. So n can be 18, 38, 58, 78 or 98.

Since 58 is the only number appears twice, the answer is 58.

This is kinda messy and requires long calculation. So I was wondering if anyone can show me some more elegant way of solving it?