B1. Determine the smallest possible value of the expression $\left|25^{m}-36^{n}\right|$, where $m$ and $n$ are natural numbers.
B1. By considering $25=5^{2}$ and $36=6^{2}$, the expression can be factored as

$$
\left|25^{m}-36^{n}\right|=\left|5^{2 m}-6^{2 n}\right|=\left|\left(5^{m}\right)^{2}-\left(6^{n}\right)^{2}\right|=\left|\left(5^{m}+6^{n}\right)\left(5^{m}-6^{n}\right)\right|=\left(5^{m}+6^{n}\right)\left|\left(5^{m}-6^{n}\right)\right|
$$

Since $m$ and $n$ are natural numbers, $5^{m}+6^{n} \geq 5+6=11$. The number $5^{m}-6^{n}$ is an integer and clearly cannot be equal to 0, so $\left|5^{m}-6^{n}\right| \geq 1$. It follows that $\left|25^{m}-36^{n}\right| \geq 11 \cdot 1=11$. The value 11 is achieved when $m=n=1$.

2. Another way. The last digits of the expressions $25^{m}$ and $36^{n}$ are 5 and 6, respectively. If $36^{n}>25^{m}$, the last digit of the expression $\left|25^{m}-36^{n}\right|=36^{n}-25^{m}$ is 1. If $36^{n}-25^{m}=1$, we would have $36^{n}-1=25^{m}$ and $\left(6^{n}-1\right)\left(6^{n}+1\right)=25^{m}$. However, the number $6^{n}+1$ is not a multiple of 5, as its last digit is 7. Therefore, $36^{n}-25^{m} \geq 11$ and the value 11 is achieved when $m=n=1$. If $25^{m}>36^{n}$, the last digit of the expression $\left|25^{m}-36^{n}\right|=25^{m}-36^{n}$ is 9. If $25^{m}-36^{n}=9$, we would have $25^{m}=36^{n}+9$, which is impossible since the right side is divisible by 3, but the left side is not. Therefore, in this case, $25^{m}-6^{n} \geq 19$. The smallest value of the expression $\left|25^{m}-36^{n}\right|$ is thus 11.

## 1. Method

Clear justification that the answer is $\leq 11$ or the claim that the minimum of the expression is achieved at $m=n=1$ 2 points

Writing the formula $\left|25^{m}-36^{n}\right|=\left(5^{m}+6^{n}\right) \cdot\left|5^{m}-6^{n}\right|$ 2 points

Justification that $5^{m}+6^{n} \geq 11$ 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-06.jpg?height=51&width=1725&top_left_y=2499&top_left_x=177)
Conclusion that $\left(5^{m}+6^{n}\right) \cdot\left|5^{m}-6^{n}\right| \geq 11$, since both factors are appropriately bounded 1 point

2. Method

Clear justification that the answer is $\leq 11$ or the claim that the minimum of the expression is achieved at $m=n=1$ 2 points

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-07.jpg?height=60&width=1742&top_left_y=278&top_left_x=171)
Factorization $\left(6^{n}-1\right)\left(6^{n}+1\right)=25^{m} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ Justification that the last digit of $6^{n}+1$ is 7 and no $m, n$ satisfy the previous factorization 1 point

![](https://cdn.mathpix.com/cropped/2024_06_07_95af3fbbcea2fb458cf3g-07.jpg?height=60&width=1739&top_left_y=478&top_left_x=176)
Justification that $25^{m}=36^{n}+9$ is impossible 1 point