## Playing with 56

Find the smallest number $n$ such that
• The digit sum of n is 56.
• The number $n$ is divisible by $56$.
• The last two digits of $n$ are $5$ and $6$

To find the smallest number $n$ we need a number with the smallest possible number of significant digits. Since the number $n$ has to end with $56$ and its digit sum has to be $56$, it will be better for us to start with $n=9999956$, however this number won't satisfy the second requirement, so we shall modify $n$ to meet all the requirements. Notice that $7\cdot 2^3=56$, so $n$ has to be divisible both by $2^3$ and $7$. Now $n$ is to be divisible by $2^3=8$ if its last digits are divisible by $8$, thus we modify as follow: $n=19999856$. Nevertheless this number won't be divisible by $7$, to fix it we need to modify $n$ in such a way that it will remain the least number possible. The price of modifying $n$ is adding $1$ to the first digit while reducing $1$ from other most significant digit, so we check $n=28999856$, but this number is not divisible by $7$. So we can reduce the instead of decreasing one from the second digit we decrease one from the third digit, we get $n=29899856$ this number is divisible by $7$ and by $8$ and therefore meets all the requirements.