1. Task. Determine all two-digit natural numbers $\overline{a b}$ such that $a < b$ and $\overline{a b}$ is equal to the sum of all natural numbers that are not less than $a$ and not greater than $b$!
Problem 1. Determine all two-digit natural numbers $\overline{a b}$, with $a<b$, which are equal to the sum of natural numbers at least equal to $a$ and at most equal to $b$.

## Solution

According to the relationship in the statement, we have $\overline{a b}=a+(a+1)+\ldots+b$ and, since $a+(a+1)+\ldots+b \leq 1+2+\ldots+9=45$, it follows that $a \leq 4$ $3 p$

For $a=4$, we get $\overline{a b} \geq 45=1+2+\ldots+9$, so it should be that $a=1$ (and $b=9$), which does not work $\ldots \ldots$ 1p

If $a=3$ we get $3+4+\ldots+(b-1)+b=\overline{3 b}=30+b$. Subtracting $b$ and adding $1+2$ to both sides, we get $1+2+\ldots+(b-1)=32$, from which $(b-1) \cdot b=64$, an equality that does not hold for any value $b \in\{4,5, \ldots, 9\}$

If $a=2$, then $2+3+\ldots+(b-1)+b=\overline{2 b}=20+b$. Subtracting $b$ and adding 1 to both sides, we get $1+2+\ldots+(b-1)=21$, from which $(b-1) \cdot b=42$.

Since $6 \cdot 7=42$, the equality holds for $b=7$, so one number that satisfies the condition in the statement is $\overline{a b}=27$

For $a=1$ we get $1+2+\ldots+(b-1)+b=\overline{1 b}=10+b$, from which $1+2+\ldots+(b-1)=10$ or $(b-1) \cdot b=20$.

The previous equality is verified for $b=5$, so $\overline{a b}=15$ also satisfies the condition in the statement