Solving for the digits in WHITE+WATER=PICNIC

Question

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?

Answer 1

I tried and failed. Below is my strategy:

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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.

Answer 2

$$ \begin{array} \\ & & W & H & I & T & E \\ +) & & W & A & T & E & R \\ \hline & P & I & C & N & I & C \end{array} $$

All $10$ digits are represented, so let's figure out where $0$ is first. Obviously, $W$ and $P$ are out and none of $T,E,R,I,H,A$ can be $0.$ So there are two cases.

First, let $C = 0.$ Then $E+R=10$ and $T+E+1 = I$ or $10+I.$ If the former, then $I\geq 1+2+3=6$ and thus $W\geq 8.$ If $I=6,$ then it forces $W=8$ but that contradicts the carryover from $C = 0.$ If $I=7,$ then $W=8.$ Now we have $T,E = \{2,4\}$ and $E,R = \{4,6\}$ due to $1,8,7$ having already been chosen. Therefore, $E=4, T=2,R=6$ and $N = I+T = 7+2=9.$ The only remaining digits are $3$ and $5$ and that won't satisfy $5+3 = 10.$

Moving on, if $I=8,$ then $W = 9$ but that again contradicts the carryover from the preceding $C = 0.$ $I=9$ is impossible as that would force $W = I= 9.$ Therefore $C\neq 0.$

The only other case is $N = 0$ and note that $W\neq 5$ for otherwise it implies $P=I=1.$ If $W=9,$ then $I=8$ and since $N=0,$ this means $T=2.$ If $T=2$, $E = I-T = 6.$ The remaining digits are $\{3,4,5,7\} = \{R,C,H,A\}.$ But $6+R = C$ implies $R=7$ which contradicts $T+E=2+6=8$ that did not have carryover.

Assume, $W=8$ and so $I\in\{6,7\}.$ If $I=6,$ then there was no carryover there and $H+A+1=C$, due to carrover from $N=0.$ Since $C\leq 1+2+3 = 6$ and $6,8$ are already taken, we have either $C = 9$ or $C = 7.$ If former, then $E+R = C=9$ with no carryover and that implies $T+E = I = 6.$ If not, then $T+E = 16$ but $C = 9$ is already taken. This implies $T=4\implies E=2\implies R=7.$ This leaves $H,A = \{5,3\}$, but that satisfies $H+A+1 = 9.$ Therefore one solution seems to be: $$ \begin{array} \\ & & 8 & 5 & 6 & 4 & 2 \\ +) & & 8 & 3 & 4 & 2 & 7 \\ \hline & 1 & 6 & 9 & 0 & 6 & 9 \end{array} $$. You can exchange $H,A$ to get another but PICNIC stays the same. Honestly, the remaining cases can be checked similarly to exhaust all cases, which are $I=6$ and $C=7$ remaining from $W=8.$ Then there are two cases $W = 6,7$, having ruled out $W = 5,8,9$ just now.