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Find three consecutive numbers, greater than 10 and none with a digit 0 in it, each of which is such that the product of its digits is a multiple of the sum of its digits.

What about four or more such numbers?

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1 Answer 1

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Three numbers

Observe that:

The product $P$ can only have prime factors less than $10$, meaning $2$, $3$, $5$ and $7$. Therefore, the same has to be true for the sum $S$.

Moreover:

The sums $S_1, S_2, S_3$ have to be consecutive, or else one of the numbers would end in a $0$. In fact, if $n_1 < n_2 < n_3$ are the integers, this argument also shows that $S_1 < S_2 < S_3$ are in the same increasing order.

So we can start by finding...

... three consecutive integers $S_1, S_2, S_3$ all of whose prime divisors are among $2$, $3$, $5$ and $7$.
It is also clear that the $S_i$ cannot be themselves one of the numbers $2, 3, 5, 7$, because the condition that $S_i$ divides $P_i$ would force one of the digits to equal $S_i$, and then the number is less than $10$, which is not allowed.

At this point it is worth noting that:

By Størmer's Theorem, even for two such consecutive numbers $S_1, S_2$, there are only finitely many possibilities, which can in principle be found algorithmically.

So we search and find that

The first such triplets of consecutive integers $S_1, S_2, S_3$ is $14,15,16$.

Now observe that:

The smaller number $n_1$, whose digit sum is $S_1 = 14$, has to have a digit equal to $7$. The number $n_2$ has to have a digit equal to $5$ and a digit divisible by $3$. If the last digit of $n_1$ were a $7$, then we would have $S_1 \geq 3+5+7$ so this is not possible. If another digit were a $7$, then we get $S_2 \geq 7 + 3 + 5$ which is fine, and shows that $n_2$ must be consist of the digits $3, 5, 7$. But then $P_3$ is not divisible by $16$.

So:

We keep looking and find that the next triplet is $48, 49, 50$.

With this much flexibility for choosing digits, some trial and error gives

$$\begin{align*}n_1 &= 5578896\\ n_2 &= 5578897\\ n_3 &= 5578898\end{align*}$$

Four or more numbers

Using the remark above:

We can use an algorithm to compute all the pairs $S_1, S_2$ of consecutive integers that are only divisible by primes $2, 3, 5, 7$. I used the code here: https://11011110.github.io/blog/2007/03/19/strmers-algorithm.html

We obtain:

1, 2
2, 3
3, 4
4, 5
5, 6
6, 7
7, 8
8, 9
9, 10
14, 15
15, 16
20, 21
24, 25
27, 28
35, 36
48, 49
49, 50
63, 64
80, 81
125, 126
224, 225
2400, 2401
4374, 4375
In particular, the only such triples of consecutive numbers greater than $10$ are (14, 15, 16) and (48, 49, 50), and there are no consecutive quadruples!

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