Cryptic Algebra V.2

We can rewrite $\overline{STOP}-\overline{TOPS} = \overline{POTS}$ as:

$\begin{aligned} 1000S+100T+10O+P-1000T-100O-10P-S & = 1000P+100O+10T+S \\ \implies 998 S & = 1009P + 910T + 190O \end{aligned}$

From the above equation, we note that:

1. $S>P$
2. Since the LHS is even, the RHS must also be even and $P$ is even. And since $P \ne 0$ , $P = 2$ , $4$ , $6$ , or $8$ .
3. Taking modulo $10$ on both sides, we have: $8S \equiv 9P \text{ (mod 10)}$ . Then if $\\ \begin{cases} P = 2 & \implies S = 1 \red {< P} & \small \red{\text{Unacceptable}} \\ P = 4 & \implies S = 2 \red {< P} & \small \red{\text{Unacceptable}} \\ P = 6 & \implies S = 8 \blue{> P} & \small \red{\text{Acceptable}} \\ P = 8 & \implies S = 9 \blue{> P} & \small \red{\text{Acceptable}} \end{cases}$
4. If $P=6$ and $S=8$ , then $7984 = 6054 + 910T + 190O \implies 91T + 19O = 193$ . There is no solution.
5. If $P=8$ and $S=9$ , then $8982 = 8072 + 910T + 190O \implies 910T + 190O = 910$ , $\implies T=1$ and $O=0$ .

Therefore, $O=0$ , $P=8$ , $S=9$ , and $T=1$ and $\dfrac {\overline{POT}\times \overline{TOO}}{\overline{SO}} = \dfrac {801 \times 100}{90} = \boxed{890}$ .

---

Reference: Modular arithmetic

We can easily see that

(i) $P$ must be even.

(ii) $S$ must be as large as possible. In particular, $S>5$ .

(iii) $S+S=10+P$ , $P+T+1=10+O$ , $O+O+1=T$ or $10+T$ , $P+T=S$ or $P+T+1=S$ .

Considering all these we get $T=1, O=0, P=8, S=9$ . So the required value of the given fraction is $\dfrac{801\times 100}{90}=\boxed {890}$ .