$5n$ ends with digit $0$ if and only if $n$ is even.
So the statement is not true for every natural number. It is true exactly for even $n$ (and also for $n=0$ if your convention includes 0 in naturals).
Why
A number ends in $0$ exactly when it is divisible by $10$.
So we need: $$ 10 \mid 5n. $$ Since $10 = 2 \cdot 5$, this means $5n$ must have a factor $2$. The factor $5$ is already present, so this happens exactly when $n$ is even.
That is the whole test. Multiplying by $5$ guarantees a factor of $5$, but a number only ends in $0$ when it has both a factor of $5$ and a factor of $2$. If $n$ is odd, then $5n$ has the factor $5$ but still lacks the factor $2$, so the last digit is $5$ instead of $0$.
What "ends with 0" really means
In base 10, the last digit tells us the remainder after division by $10$. A number ends with $0$ exactly when its remainder is $0$: $$ m \text{ ends with } 0 \Longleftrightarrow m \equiv 0 \pmod{10}. $$
For this question, set $m = 5n$. We want: $$ 5n \equiv 0 \pmod{10}. $$
There are only two possible last digits for $5n$:
| Type of $n$ | Example | $5n$ | Last digit |
|---|---|---|---|
| Odd | 3 | 15 | 5 |
| Even | 4 | 20 | 0 |
So the last digit alternates between $5$ and $0$ as $n$ moves from odd to even.
Quick check
| $n$ | $5n$ | Last digit |
|---|---|---|
| 1 | 5 | 5 |
| 2 | 10 | 0 |
| 7 | 35 | 5 |
| 12 | 60 | 0 |
The examples are not random. They show the two cases:
- $n = 1$ and $n = 7$ are odd, so $5n$ ends in $5$.
- $n = 2$ and $n = 12$ are even, so $5n$ ends in $0$.
The common mistake
The common mistake is to think that because $5n$ has a factor of $5$, it must end in $0$. That is not enough. Numbers like $5$, $15$, $25$, and $35$ all have a factor of $5$, but they do not end in $0$.
To end in $0$, the number must be a multiple of $10$. A multiple of $10$ needs:
- one factor of $5$,
- one factor of $2$.
The expression $5n$ supplies the factor of $5$. The value of $n$ must supply the factor of $2$. That happens exactly when $n$ is even.
Modular arithmetic version
If $n$ is even, then $n = 2k$ for some integer $k$. Then: $$ 5n = 5(2k) = 10k. $$ Since $10k$ is divisible by $10$, it ends in $0$.
If $n$ is odd, then $n = 2k + 1$. Then: $$ 5n = 5(2k + 1) = 10k + 5. $$ That has remainder $5$ when divided by $10$, so it ends in $5$.
Visual pattern
Therefore, $5n$ can end with $0$, but only when $n$ is even.
More examples
| $n$ | Even or odd? | $5n$ | Does it end in $0$? |
|---|---|---|---|
| 5 | Odd | 25 | No |
| 6 | Even | 30 | Yes |
| 19 | Odd | 95 | No |
| 20 | Even | 100 | Yes |
| 101 | Odd | 505 | No |
| 102 | Even | 510 | Yes |
The size of $n$ does not matter. Only its parity matters. Once you know whether $n$ is even or odd, you know the last digit of $5n$.
General rule
For any integer $n$:
- if $n$ is even, then $5n$ ends in $0$;
- if $n$ is odd, then $5n$ ends in $5$.
This also explains why multiples of $5$ alternate between endings of $5$ and $0$: $$ 5, 10, 15, 20, 25, 30, \ldots $$
Every second multiple of $5$ is also a multiple of $10$.
Practice checks
Try deciding before multiplying:
| Question | Answer |
|---|---|
| Does $5 \cdot 44$ end in $0$? | Yes, because $44$ is even. |
| Does $5 \cdot 73$ end in $0$? | No, because $73$ is odd. |
| Does $5 \cdot 1001$ end in $0$? | No, because $1001$ is odd. |
| Does $5 \cdot 2048$ end in $0$? | Yes, because $2048$ is even. |
The fastest method is not to multiply. Just check whether $n$ is even.
FAQ
Can $5n$ ever end in $0$?
Yes. It ends in $0$ whenever $n$ is even, such as $n = 2, 4, 6, 20,$ or $102$.
Does $5n$ always end in $0$?
No. If $n$ is odd, $5n$ ends in $5$. For example, $5 \cdot 7 = 35$.
What if $n = 0$?
If your definition of natural numbers includes $0$, then $5 \cdot 0 = 0$, so it ends in $0$. If your definition starts natural numbers at $1$, then the rule for positive natural numbers is still: even $n$ gives last digit $0$, odd $n$ gives last digit $5$.
Related Reading
If you want a practical engineering walkthrough, read: From Zero to Production AI Agent with Claude.