THE PROBLEM

Let $f(n)$ be a function that reverses the digits of $n$. For example, $f(112)=211$ and $f(100)=001=1$. Find the number of positive integers $m\le100$ such that $f(m)+m$ is divisible by $10$.

Difficulty: Olympiad Prequalifier · ★★☆☆☆
Source: Olympiad Prequalifier Practice Set
Topics: Number Theory · Digit Manipulation · Modular Arithmetic
Answer: $10$

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THE KEY IDEA

You never need to actually reverse the number. The last digit of $f(m)$ is simply the first digit of $m$. Since divisibility by $10$ depends only on the units digit, only the first and last digits of $m$ matter — everything in between is irrelevant.

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THE SOLUTION

Let the first digit of $m$ be $a$ and its last digit be $b$.

When we reverse the digits, the digit $a$ moves to the units place. So the units digit of $f(m)$ is $a$, while the units digit of $m$ is $b$. Therefore,

$$m+f(m)\equiv a+b\pmod{10}.$$

For $m+f(m)$ to be divisible by $10$, we need$$a+b\equiv0\pmod{10}.$$

Since $a\ge1$, $b\ge0$, and both are digits,$$1\le a+b\le18.$$

The only multiple of $10$ in this range is $10$, so$$a+b=10.$$

Now count by number of digits.

Case 1: One-digit numbers

For $m=1,2,\dots,9$, we have $a=b=m$. Thus,$$a+b=2m.$$

This equals $10$ only when $m=5$.

So there is 1 such number.

Case 2: Two-digit numbers

For $m=10,11,\dots,99$, we need$$a+b=10,$$

where$$a\in\{1,\dots,9\},\qquad b\in\{0,\dots,9\}.$$

For every choice of $a$, there is exactly one valid choice$$b=10-a.$$

Thus there are 9 such numbers.

Case 3: $m=100$

Here $a=1$ and $b=0$, so$$a+b=1\ne10.$$

Thus there are 0 such numbers.

Adding the cases,$$1+9+0=10.$$

Therefore,$$\boxed{10}$$

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WHY THE OBVIOUS APPROACH IS SLOW

The natural instinct is to compute $f(m)$ for every $m=1,2,\dots,100$, add it to $m$, and check divisibility by $10$. That is 100 reversals and 100 additions — slow, error-prone, and completely unnecessary.

The problem only asks about divisibility by $10$, so only the units digit of the sum matters. And the units digit of $f(m)$ is controlled entirely by the first digit of $m$. The middle digits never enter the calculation.

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THE INSIGHT

When a problem involves divisibility by $10$, immediately isolate the units digit. If a complicated operation such as digit reversal appears, ask:

Which digits of the original number actually affect that units digit?

Usually, most of the number is irrelevant. This same instinct applies to any “last digit” or “mod $10$” problem — reduce the problem to the digits that matter before computing anything.

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GENERALIZATION

What if we replaced $10$ with $9$?

Then we would need$$a+b\equiv0\pmod9,$$

and the count would change because $a+b$ can now equal $9$ or $18$. The structure of the argument is identical — only the arithmetic of the digit sum changes.

This works because$$9\mid10-1,$$

so digit sums control divisibility by $9$. For a general modulus $k$, the units digit alone is usually not enough; the useful digits depend on the modulus.

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CHALLENGE

Find the number of positive integers $m < 10^4$ such that $m + f(m)$ is divisible by 9.