23.3. We have 10 natural numbers whose sum is 1001. What is the greatest value that the greatest common divisor of these numbers can take?

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23.3. It is clear that the greatest common divisor of these ten numbers will also be a divisor of the number 1001. Let's consider the divisors of this number. We have: $1001=7 \cdot 11 \cdot 13$. The sought greatest common divisor cannot be equal to $11 \cdot 13=143$, since $143 \cdot 10>1001$. However, it can be equal to $7 \cdot 13=91$, if nine of the ten given numbers are taken to be 91, and the tenth is 182.