The sum $1+1+4$ of the digits of the number 114 divides the number itself. What is the largest number less than 900 that satisfies this property?
To find the largest number that is divisible by the sum of its digits and is also less than 900, we can start our search among numbers with the digit 8 in the hundreds place, since, at a minimum, 800 is divisible by the sum $8+0+0=8$ of its digits, and 899 does not have this property. Thus, we will examine the numbers between 800 and 899.

We want to find digits $b$ and $c$ such that $8+b+c$ divides $8bc=800+10b+c$. Remember that $8+b+c$ divides $800+10b+c$ if and only if $8+b+c$ divides $800+10b+c-(8+b+c)=792+9b$. Therefore, we need to look among the divisors of $792+9b$. For this, we assign values to $b$ in descending order, starting from 9, until we find the largest number we are looking for.

- If $b=9$, then $792+9 \times 9=873=9 \times 97$ and this number does not have a divisor $8+9+c$ between 17 ($c=0$) and 26 ($c=9$).
- If $b=8$, then $792+9 \times 8=864=2^{5} \times 3^{3}$. The largest divisor $8+b+c$ of this number between 16 and 25 is 24, that is, $c=8$.

Therefore, the number we are looking for is 888.