I'm writing solutions for students who are taking a competition exam and I took problems from old purple comet competition problems. This problem is the last one from the 2004 middle school contest and says:
In the addition problem WHITE+WATER=PICNIC each distinct letter represents a different digit. Find the number represented by the answer PICNIC.
I have looked for solutions online and their answers have been insufficient; usually stating the only possibilities for WHITE and WATER (of which there are only two) and then giving what PICNIC must be.
I have tried my hand at this problem using strategies for solving cryptoarithmetic puzzles. I started with recognizing that P is 1 and then working with the W being at least 5, but this has always devolved into a myriad of cases. This is a problem intended for middle school students so I'm a doubtful that the intended solution is to check so many cases.
Is there some insight for this puzzle to better organize the cases?
I tried and failed. Below is my strategy:
We are now solving the following digit puzzle:
$$ \begin{array} \\ & & W & H & I & T & E \\ +) & & W & A & T & E & R \\ \hline & P & I & C & N & I & C \end{array} $$
As usual, we assume that $P \neq 0$. Since $W \le 9$ and $9 + 9 = 18$, we immediately see that $P = 1$.
$I \ge 0$, so $2 W \ge 10$; hence, we derive that $W \ge 5$.
If $W = 9$, then $2 W = 18$, so $I = 8$ since $I \neq W$;
if $W = 8$, then $2 W = 16$, so $I \in \{ 6, 7 \}$;
if $W = 7$, then $2 W = 14$, so $I \in \{ 4, 5 \}$;
if $W = 6$, then $2 W = 12$, so $I \in \{ 2, 3 \}$;
if $W = 5$, then $2 W = 10$, so $I = 0$. $I \neq 1$ since $I \neq P$.
Now, we have got many possibilities.
We want to narrow down our search for the solution. To this end, we look at some different digits. If $T + E = I$ and $I + T = N$, then $2 T + E = N$. Since $T \neq 0$ (otherwise, $E = N$; this is not allowed) and $E \neq 0$ (otherwise, $T = I$), we know that $T \ge 1$ and that $E \ge 1$. But remember that $P = 1$.
So $T \ge 2$ and $E \ge 2$. Since $T \neq E$, $N \ge 7$.
If $N = 7$, then $T = 2$ (if $T = 3$, then $E = 1$). So $E = 3$; accordingly, $I = 5$. But then $W = 7$; this is not allowed.
If $N = 8$, then $T \in \{ 2, 3 \}$ (if $T = 4$, then $E = 0$). If $T = 2$, then $E = 4$, so $I = 6$, but then $W = 8$. If $T = 3$, then $E = 2$, so $I = 5$ and $W = 7$.
If $N = 9$, then $T \in \{ 2, 3 \}$ (if $T = 4$, then $E = 1$). If $T = 2$, then $E = 5$, so $I = 7$ and $W = 8$. If $T = 3$, then $E = 3$; this is obviously not allowed.
So, in summary, if $T + E = I$ and $I + T = N$, we have the following cases:
If $N = 8$, then $T = 3$, $E = 2$, $I = 5$, and $W = 7$. Remember that $P = 1$. So we are left with $\{ 0, 4, 6, 9 \}$, which is our search space. $2 + R = C$. Since $R \neq 0$ (otherwise, $C = E$; this is not allowed), $R \neq 6$ (otherwise, $C = N$), and $R \neq 9$ (since $C \le 9$), we are left with only one possibility: $R = 4$. If so, then $C = 6$; this seems good.
If $N = 9$, then $T = 2$, $E = 5$, $I = 7$, and $W = 8$. Again, $P = 1$. So our search space is $\{ 0, 3, 4, 6 \}$. $5 + R = C$, so $R \neq 0$ (otherwise, $C = E$), $R \neq 3$ (otherwise, $C = W$), $R \neq 4$ (otherwise, $C = N$), and $R \neq 6$ (since $C \le 9$). But then, we have effectively rejected all possible digits. So $N \neq 9$.
In summary, $N = 8$. Our remaining search space is $\{ 0, 9 \}$.
If $T + E = I$ and $I + T = N$, then, since $\left( 10 + I \right) - 2 W = 1$, we need $H + A \ge 10$. But $9 + 0 < 10$. Oh, no! So
$T + E \neq I$ or
$I + T \neq N$.
I am stopping here as I believe there is a very smart way, much, much more efficient. I am posting this just for others to know that my strategy probabily wouldn't work.