The product of digits equal to the sum of digits

Question

How to find the number(or numbers ) that has $4$ digits, the product of these digits equal to the sum of these digits ?

Answer 1

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.

Answer 2

You can narrow your search rapidly:

1. no digits $0$;
2. at least one digit $1$ (otherwise the product exceeds the sum easily);
3. at least two digits greater than $1$ (otherwise the sum now exceeds the product);
4. 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.

Answer 3

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

Answer 4

Must be various combinations of $1,1,2,4$. I don't think there are any other combinations, I looked at all small numbers...