A Question about SOAP

We note that $\text{soap}(n)$ is defined as follows.

$\begin{aligned} \text{soap}(n) & = \sum_{j=1}^{n-1} j \sum_{k=j+1}^n k \\ & = \sum_{j=1}^{n-1} j \color{#3D99F6} \left(\frac {(n-j)(j+1+n)}2 \right) & \small \color{#3D99F6} \text{Sum of AP: } S = \frac {n(a+l)}2 \\ & = \frac 12 \sum_{j=1}^{n-1} j \left(n^2-j^2 + n - j\right) \\ & = \frac 12 \sum_{j=1}^{n-1} \left((n^2+n)j-j^2 - j^3\right) \\ & = \frac 12 \left(\frac {(n^2+n)n(n-1)}2-\frac {n(n-1)(2n-1)}6 - \frac {n^2(n-1)^2}4 \right) \\ & = \frac {n(n-1)}{24} \left(6n^2+6n - 4n+2 - 3n^2 + 3n \right) \\ & = \frac {n(n-1) \left(3n^2+5n +2 \right)}{24} \\ & = \frac {n(n-1)(n+1)(3n+2)}{24} \end{aligned}$

We also note that the number of product pairs of $\text{soap}(n)$ is $p(n) = {\color{#3D99F6}T_{n-1}} = \dfrac {n(n-1)}2$ , where $\color{#3D99F6} T_k$ is the triangular number. Then we have

$\begin{aligned} \frac {\text{soap}(n)}{p(n)} & = \frac {(n+1)(3n+2)}{12} & \small \color{#3D99F6} \text{For } n = 10^k \\ & = \frac {\color{#D61F06}(10^k+1)\color{#3D99F6}(3\times 10^k+2)}{12} \end{aligned}$

The factor $\color{#D61F06}10^k+1$ is always odd and has a sum of digits of 2. Therefore, it is not a multiple of 3. Therefore, for $\dfrac {\text{soap}(n)}{p(n)}$ to be an integer, $\color{#3D99F6}3\times 10^k +2$ must be divisible by 12. For $k=1$ , $3 \times 10^k+2 = 32 \bmod 12 = 8 \ne 0$ . The factor is not divisible by 12. For $k > 1$ , $3 \times 10^k + 2 \bmod 12 = 2 \ne 0$ . Therefore $10^k$ is never soapy.

Note that $\begin{aligned} \text{Soap}(n) &= \sum_{1\leq a < b \leq n} a\times b \\ &= \frac{1}{2}\sum_{1\leq a,b \leq n \atop a\neq b} a\times b \\ &= \frac{1}{2} \left(\sum_{1\leq a,b \leq n} a\times b - \sum_{1\leq a,b \leq n \atop a=b} a\times b\right) \\ &= \frac{1}{2} \left(\sum_{a=1}^n a \times \sum_{b=1}^n b - \sum_{a=1}^n a^2\right) \\ &= \frac{1}{2} \bigg(\left(1+2+3+\cdots +n\right)^2 - \left(1^2+2^2+3^2+\cdots+n^2\right)\bigg) \\ &= \frac{1}{2} \left(\frac{n^2(n+1)^2}{4} - \frac{n(n+1)(2n+1)}{6}\right) \\ &= \frac{n(n-1)(n+1)(3n+2)}{24} \end{aligned}$ and the number of pairs is $\binom{n}{2} = \frac{n(n-1)}{2}$ so the arithmetic mean of the products of pairs is $\text{Soap}(n) \times \binom{n}{2}^{-1} = \left(\frac{n(n-1)(n+1)(3n+2)}{24}\right) \times \left(\frac{2}{n(n-1)}\right) = \frac{(n+1)(3n+2)}{12}$

Now, $n$ is Soapy if and only if this is an integer, which implies $4\times \frac{(n+1)(3n+2)}{12} = \frac{(n+1)(3n+2)}{3} = (n+1)^2 - \frac{n+1}{3}$ is also an integer. That is, we have shown $n \text{ is Soapy} \implies n +1 \equiv 0 \pmod{3}$

Here, if we let $n = 10^k,$ then $10^k +1 \equiv 1^k + 1 = 2 \not\equiv 0 \pmod{3}$ so numbers of the form $10^k$ for $k\in \mathbb{N}$ are $\boxed{\text{never}}$ Soapy.

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Remark: It wasn't necessary for this problem, but if you look at the formula given for the arithmetic mean, it's not hard to show that the following is a valid definition of being Soapy: $n>1 \text{ is an integer } \implies \bigg(n \text{ is Soapy} \iff n \equiv 2 \text{ or } 11 \pmod{12}\bigg)$