11.1. Since the remainder is seven times greater than the quotient, it is divisible by seven. At the same time, when divided by 23, the remainder does not exceed 22. These two conditions for the remainder are satisfied by three numbers: $7,14,21$. Then for the quotient, we get the respective numbers: $1,2,3$. Thus, the problem has three solutions:

1) $23 \cdot 1+7=30$
2) $23 \cdot 2+14=60$
3) $23 \cdot 3+21=90$
Answer. $30,60,90$.
11.1. What number, when divided by 23, gives a remainder that is seven times greater than the quotient?

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