The product of digits equal to the sum of digits How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?
 A: You can narrow your search rapidly: 


*

*no digits $0$;

*at least one digit $1$ (otherwise the product exceeds the sum easily);

*at least two digits greater than $1$ (otherwise the sum now exceeds the product);

*exactly two digits greater than $1$ (the product of three such digits would exceed their  sum by at least $2$).


So we're looking for pairs of digits in $\{2,3,\ldots,9\}$ whose product exceeds their sum by exactly $2$ (the number of digits $1$ we need to throw in). If one of them is $2$, the other must be $4$. If the smallest of the pair is at least $3$, then their product exceeds their sum by at least $3$, so this cannot happen.
So all in all there is essentially one solution, but since you asked for numbers , the $12$ permutations of the digits of $1124$ give you all solutions.
A: First of all, let's observe that all of the digits of such a number cannot be the same. You can just manually check that numbers $1111$, $2222$ and so on don't suit us. It is also clear that all of the digits should be non-zero.
Now suppose that we have such a number. Let $a,\,b,\,c,\,d$ be its digits written in non-ascending order: $a \geqslant b \geqslant c \geqslant d$. Then we have
$$
abcd = a + b + c + d.
$$
From this we have an inequality:
$$
a\cdot bcd < 4a.
$$
This inequality is strict, because at least one of $b, c, d$ is strictly smaller than a. So we have:
$$
bcd < 4,
$$
which is the same as saying
$$
bcd \leqslant 3.
$$
This only leaves us with 3 possible combinations for $(b, c, d)$: $(1, 1, 1)$, $(2, 1, 1)$ and $(3, 1, 1)$.
If $b=c=d=1$, then $a\cdot 1 \cdot 1 \cdot 1 = a + 1 + 1 + 1$, which can't be true.
If $b=2$ and $c=d=1$, then $a \cdot 2 \cdot 1 \cdot 1 = a + 2 + 1 + 1$, which means that $a=4$. This gives us one possible solution: $a=4, b=2, c=d=1$.
If $b=3$ and $c=d=1$, then $a \cdot 3 \cdot 1 \cdot 1 = a + 3 + 1 + 1$, which is impossible.
So, the only solution is $a=4$, $b=2$, $c=d=1$. There are $12$ numbers with such digits.
A: i observed a pattern in these numbers.
22
123
1124
11125
111126
1111127
and derived a formula for this.
if the last two digits are assumed to be a and b and for an n digit number
there will be n-2 1's and a and b are to be found out using the below formula.
a=(b+n-2)/(b-1)
where a and b are from 2 to 9 which need to be evaluated manually for b=2 to 9
A: Must be various combinations of $1,1,2,4$.
I don't think there are any other combinations, I looked at all small numbers...
