How many positive integers satisfy the condition that the product of their digits multiplied by the sum of their digits equals themselves? For example,
13 is not equal to (1+3)(1×3)
Here is how i proceeded I defined a number n such that $10^{k+1}$$>$$n$$>$$10^k$ Next, i will expand the number in powers of tens. Now i am trying to create a bound, but at this stage, i am unsuccessful in each of my attempts. Please help me!
To set an upper bound, note that each digit is at most $9$. If the number has $n$ digits, the computation gives at most $(9n)(9^n)=n9^{n+1}$. An $n$ digit number is at least $10^{n-1}$. Alpha tells us that for $n=85$ we have the number is greater than the sum of digits times the product of digits, so you only have to search up to $10^{84}$. Happy hunting.
Now note that most of the factors of the number need to be $2,3,5,7$ because the only way you can get a larger one is from the sum of the digits. As lulu points out you can't have both $2$ and $5$ as factors because it would end in zero and the product would be zero. The number needs to be $2^a3^b7^cd$ or $3^a5^b7^cd$ with $d \lt 9\cdot 84$. If you have a programming language that supports large ints, like Python, and a nice way to turn a number into a string you can just do the search. A little cleverness can speed it up, like noting that $e$ can't be that large unless the other numbers are large.