Pista forgot her friend's phone number. She remembers that the first digit is 7, the fifth is 2. She knows that the number is six digits long, odd, and when divided by 3, 4, 7, 9, 11, and 13, it gives the same remainder. What is the phone number?
Let $T$ denote the sought telephone number. $T$ when divided by 3 can leave a remainder of 0, 1, or 2. We also know that $T$ is odd, so when divided by 4, it leaves an odd remainder. Therefore, the common remainder in question is 1.

From the condition, it follows that $T-1$ is divisible by 3, 4, 7, 9, 11, and 13. For now, we only use the fact that $T-1$ is divisible by $7 \cdot 11 \cdot 13 = 1001$, since the multiples of 1001 are of the form $A B C A B C$. Thus, $A=7, B=2$, so

$$
T-1=72 C 72 C
$$

Due to divisibility by 9, the sum of the digits of $T-1$ is also divisible by 9. Since $7+2+C+7+2+C=18+2C$, $2C$ must also be a multiple of 9. This is only possible if $C=0$ or $C=9$, but $T-1$ is also divisible by 14, so $C$ must be even. If $C=0$, then all conditions are satisfied, i.e., $T=720721$.

Remark. The solution can also be completed in another way. From the fact that $T-1$ is divisible by the listed numbers, it follows that it is divisible by their least common multiple, which is 36036. On the other hand, $700020 \leq T \leq 799929$,

so $700019 \leq T-1 \leq 799928$, hence

$$
19<\frac{700019}{36036} \leq \frac{T-1}{36036} \leq \frac{799928}{36036}<23
$$

Thus, the value of $\frac{T-1}{36036}$ is 20, 21, or 22:

$$
\begin{aligned}
& 20 \cdot 36036+1=720721 \\
& 21 \cdot 36036+1=756757 \\
& 22 \cdot 36036+1=792793
\end{aligned}
$$

The fifth digit is only 2 in the first case, so the sought telephone number is 720 721. (Sz. Cs.)