101. Append three digits to 523 ... so that the resulting six-digit number is divisible by 7, 8, and 9.
101. The desired six-digit number, starting with the digits 523 and divisible without remainder by $7 \cdot 8 \cdot 9=504$, can be represented as $523000+X$, where $X$ is a three-digit number. Direct division gives $523000=504 \cdot 1037+352$, i.e., 523000 gives a remainder of 352 when divided by 504. Since the sum of the number 523000 and the three-digit number $X$ must be divisible by 504, it follows that $X$ can be either

$$
504-352=152
$$

or

$$
2 \cdot 504-352=656
$$

(since the number $3 \cdot 504$ - 352 is already a four-digit number). Thus, the two numbers that satisfy the condition of the problem are 523152 and 523656.