Find the smallest positive whole number that is equal to seven times the product of its digits

Question

I am helping a kid with the preparation for a mathematical competition. One of the training question is:

Find the smallest positive whole number that is equal to seven times the product of its digits

They do not provide the answer, but using this little python script I found out it is 735:

for i in range(1, 1000):
    digits = [int(x)for x in str(i)]
    prod = 1
    for digit in digits:
        prod = prod * digit
        
    if prod*7 == i:
        print(f"The number is: {i}")
        break

Now I want to find a way that could be resolved just with paper and pencil, as they must do in this mathematical competition.

I tried to write: $$ 100a + 10b + c = 7abc $$

and then I tried many more things, including dividing rule by 7 etc. But I couldn't find a way of solving this if not by brute-force substituting digits for a, b and c and find the values that satisfy the equation.

Thanks!

EDIT

I know it is a 3 digits number, so it must be less than 999, and since 0 is not a digit it must be greater than 111. Since the number is 7 times the product of the digits, I know that $$a*b*c ≤ 999/7$$.

I think rather than 999 we could use 994 which is the greatest nnumber divisible by 7 before 1000.

I tried to do a system of equations but I am missing one more condition to make this work. Maybe it can give you guys some idea.

Answer 1

One-digit numbers don’t work. Let us try two-digit numbers: $$10a+b=7ab,$$

where $a>0$. We get $b=a(7b-10)$. So $b\le 2$. We check $b=0,1,2$ manually.

Consider three-digit numbers now. Let us write $$10(10a+b)=(7ab-1)c,$$

where $a>0$. Consider three cases.

• Suppose $c=5$. Then $20a+2b=7ab-1.$ We see that $b$ is an odd digit such that $7b\ge 20$. So we manually check $b=3, 5, 7, 9$.

• Suppose $c=2$. Then $50a+5b=7ab-1$. Now $7b$ is greater $50$. So $b=8,9$. Check those manually.

• If $c$ is any other digit then $10\mid 7ab-1$. So $7ab=21,91,161,231,301,371,441, 511$. Or $ab=3, 13, 23, 33, 43, 53, 63, 73$. $13, 23, 33=3\cdot 11, 43, 53, 73$ are impossible. If $ab=3$ then $a=1, b=3$ or vice versa. If $ab=63$ then $a=7,b=9$ or vice versa. We check those four cases manually.

The answer is $735$.

Answer 2

I would start with a single digit, calculate seven times the sum of the digits, and continue.

Like this you start with $1$, then you get $7$, $49$, $252$, $140$ and you end up with zero.
Then you start with $2$, then you get $14$, $28$, $112$ and you end up with $14$ again.
Then you start with $3$, then you get $21$, leading again to $14$ which generates again the endless loop.

... and like this you continue until you find the solution. (I have no idea, however, if the final solution $735$ is an attractor of this dynamical system)