More mean numbers

A mathematical approach .

In general, if $d_{p,i}$ is the exponent of prime number $p$ in digit $d_i$ , we must have $\sum_{i = 1}^6 d_{p,i} = 6d_{p,7}.$ This requires that $\sum_{i=1}^6 d_{p,i}$ is a multiple of six. Note also that there is only one digit that combines two different primes, namely 6.

Write $a_d$ for the number of times digit $d$ occurs among the first six digits. Naturally, $a_1 + a_2 + \cdots + a_9 = 6$ . Then $\sum_{i=1}^6 d_{2,i} = a_2 + a_6 + 2a_4 + 3a_8;\ \ \sum_{i=1}^6 d_{3,i} = a_3 + a_6 + 2a_9; \\ \sum_{i=1}^6 d_{p,i} = a_p\ \ \ \text{for}\ p = 5,7.$

We have a solution if all of these four sums are a multiple of six. Moreover, once we know values for $a_1, a_2, \dots, a_9$ , then there are $\binom{6}{a_1\ \ a_2\ \ a_3\ \ \dots\ \ a_9} = \frac{6!}{a_1! a_2!\cdots a_9!}$ numbers with the given numbers of each digit.

With this preparation, consider the possible values for the last digit, $a_7$ .

• If $d_7 = 1, 5, 7, 8, 9$ , there is only one way to fulfill the conditions: all digits should be equal. (This is easy to prove.) This gives the 5 solutions $1111111, 5555555, 7777777, 8888888, 9999999$ .

• If $d_7 = 3$ , then the number can only contain the digits 1, 3, and 9. We have $a_1 = a_9 =: a$ and $a_3 = 6 - 2a$ . Clearly, the possible values of $a$ range from 0 to 3, so we calculate $\sum_{a=0}^3 \binom{6}{a\ \ a\ \ 6-2a} = \frac{6!}{0! 0! 6!} + \frac{6!}{1! 1! 4!} + \frac{6!}{2! 2! 2!} + \frac{6!}{3! 3! 0!} = 1 + 30 + 90 + 20 = 141.$ This is the number of solutions ending in $3$ ; they are all permutations of $333333$ , $1333393$ , $1133993$ and $1119993$ .

• If $d_7 = 2$ , then the number can only contain the digits 1, 2, 4, and 8. We have $a_1 = a_4 + 2a_8$ . If $a_1 > 4$ then $a_8 > 2$ , which would make for too many digits in total; this we must consider $0 \leq a_1 \leq 4$ . Then $0 \leq 2a_8 \leq a_1$ . Thus, the count of numbers ending in 2 is $\sum_{a_1=0}^4 \sum_{a_8=0}^{\lfloor a_1/2 \rfloor} \binom{6}{a_1\ \ \ a_8\ \ \ a_1 - 2a_8\ \ \ 6 - 2 a_1 + a_8}$ $= \frac{6!}{0! 0! 0! 6!} + \frac{6!}{1! 0! 1! 4!} + \frac{6!}{2! 0! 2! 2!} + \frac{6!}{2! 1! 0! 3!} + \frac{6!}{3! 0! 3! 0!} + \frac{6!}{3! 1! 1! 1!} + \frac{6!}{4! 2! 0! 0!} = 1 + 30 + 90 + 60 + 20 + 120 + 15 = 336.$

• For $d_7 = 4$ we use a trick: "translate" all digits according to $1 \leftrightarrow 8$ and $2 \leftrightarrow 4$ , and you have the previous case. Thus, another $336$ numbers ends in 4.

• What remains is the case $d_7 = 6$ . Considering the prime factors 2 and 3 separately, we have $\begin{cases} a_1 + a_3 + a_9 = a_4 + 2a_8 =: x; \\ a_1 + a_2 + a_4 + a_8 = a_9 \end{cases} \ \ \ \therefore\ \ \ \begin{cases} a_1 + z + a_3 = a_8 =: x; \\ z + a_4 = a_9 - x := y \\ a_1 + a_2 := z \end{cases}$ Thus, the count of numbers ending in 6 is equal to $\sum_{x=0}^2 \sum_{y = 0}^{3 - x} \sum_{z = 0}^{\text{min}(x,y)} \sum_{a_1 = 0}^{\text{min}(z, x-z)} \binom{6}{x \ \ \ x+y \ \ \ y - z \ \ \ a_1 \ \ \ z - a_1 \ \ \ x - z - a_1 \ \ \ 6 - 3x - 2y + z + a_1}.$
  We find the following terms; the right column shows examples of the numbers that are being counted. $\begin{array}{ccc|rl} x & y & z & & & \\ \hline 0 & 0 & 0 & 1 & 6666666 \\ 0 & 1 & 0 & 30 & 4666696 \\ 0 & 2 & 0 & 90 & 4466996 \\ 0 & 3 & 0 & 20 & 4449996 \\ 1 & 0 & 0 & 120 & 3666896 \\ 1 & 1 & 0 & 360 & 3468996 \\ 1 & 1 & 1 & 180 & 2668996 \\ 1 & 2 & 1 & 120 & 2489996 \\ 2 & 0 & 0 & 90 & 3388996 \\ 2 & 1 & 1 & 60 & 1889996 \\ \hline & & & 1101 & \end{array}$

Thus we find a total of $5 + 141 + 2\times 336 + 1101 = \boxed{1889}$ solutions.

Here is a solution in code .

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#include<stdio.h>

int main() {
    int i, n, nn, d[7], p1, p2, count;

    count = 0;
    for (n = 1000000; n <= 9999999; n++) {
        nn = n; p1 = 1; p2 = 1;
        for (i = 0; i < 7; i++) {
            d[i] = nn % 10; nn /= 10;
            if (i > 0) { p1 *= d[i]; p2 *= d[0]; }
        }

        if (p1 > 0 && p1 == p2) {
            printf("%d\n",n);
            count++;
        }
    }

    printf("%d\n",count); 
}

This program lists all $\boxed{1889}$ numbers that qualify.

Mathematica

Length@Select[Range[10^6,10^7],GeometricMean[(s=IntegerDigits[#])[[1;;6]]]==s[[7]]&&s~FreeQ~0&]