# Happy days alphametic

Every letter a decimal digit, different letters for different digits:

    HAPPY
HAPPY
HAPPY
+ DAYS
----------
AHEAD


Which digit does each letter represent?
(Please present the full analysis how these digits can be determined.)

## 2 Answers

Denote $$C_i$$ the carry over from the sum of the digits in column $$i$$ and $$S_i$$ be the sum of the digits in column $$i$$ plus $$C_{i-1}$$. $$C_i$$ is simply $$S_i \div 10$$ rounded down. Lastly, denot $$D_i$$ to be the ones digit of $$S_i$$. In other words, $$D_i = S_i \mod 10$$.

Since the sum of all the numbers is still a 5 digit number, we know that $$H \in \{1,2,3\}$$. If $$H=3$$, then $$A=9$$. But then in the 4th column you have 3 $$A$$s, which would mean $$C_4 \ge 2$$, making a six digit number. Therefore, $$H \in \{1,2\}$$.

This means, $$A \ge 3$$. Since $$D \ne 0$$, $$A=3 \implies C_4=1$$, which results in $$A=4$$. Thus, $$A \ne 3$$.

If $$A=4$$, then $$C_4=1$$ and $$H=1$$. Then $$S_4=12+D+C_3$$ and $$D_4=H=1$$. This can only happen if $$D+C_3=9 \implies C_4=2$$. Thus, $$A \ne 4$$.

If $$A=6$$, then $$C_4=0$$ and $$H=2$$, or $$C_4=3$$ and $$H=1$$. $$S_4=18+D+C_3=H$$. If $$H=1$$, then $$D+C_3=13$$, which only works if $$C_3$$ could be 4, but this is only possible when you are summing five or more numbers. Thus, $$H=2$$ and $$C_4=0$$. But $$S_4 \gt 3 \times A = 18 \implies C_4 \gt 0$$. Thus $$A \ne 6$$.

If $$A=7$$, then $$C_4=1$$ and $$H=2$$, or $$C_4=4$$ and $$H=1$$. Since $$S_4=3\times 7+D+C_3=21+D+C_3$$, we know that $$C_4$$ can only be 4 when $$D+C_3 \ge 19$$ (which can't happen), and is at least $$2$$. Thus, $$C_4=1$$ and $$H=2$$. But $$S_4\ge3\times7=21 \implies C_4\ge2$$. Thus $$A \ne 7$$.

If $$A=8$$, then $$C_4=2$$ and $$H=2$$. $$S_4=3 \times 8 + D + C_3=24 + D +C_3$$. We know that $$D_4=H=2$$ only works when $$D+C_3=8 \implies C_4=3$$. Thus, $$A \ne 8$$.

After all this, we know that $$A \in \{5, 9\}$$ and $$H \in \{1, 2\}$$.

### Assume $$A=5$$, so $$H=1$$

If $$A=5$$, $$S_4=15+D+C_3$$ and $$D_4=H=1$$. Thus, $$D+C_3=6$$ and $$C_4=2$$.

If $$C_3=0$$, then $$P=1$$, which is already taken by $$H$$. If $$C_3 = 1$$, then $$S_4=3 \times 5 + D + C_3 = 15 + D + 1 = 16 + D$$. We know that $$D_4=H=1$$ only works if $$D=5$$, which is already taken by $$A$$. We know that $$C_3 \ne 4$$ because the highest $$S_3$$ can be is when $$P=9$$ and $$C_1=3$$, which makes $$S_3=35$$.

Thus $$C_3 \in \{2,3\}$$ and $$D \in \{4,3\}$$.

Lets look at $$S_3=3 \times P + 5 + C_2$$ and $$S_2=3 \times P + Y + C_1$$.

If $$P \in \{2,3,4\}$$, then $$C_3=1$$. Thus, $$P \ge 6$$.

If $$P=6$$, then $$S_2=18+Y+C_1$$ and $$D_2=5$$. So $$Y+C_1=7 \implies C_2=2$$. Thus, $$S_3=12+5+2=19 \implies E=9$$ and $$C_3=1$$.

If $$P=7$$, then $$S_2=21+Y+C_1$$ and $$D_2=5$$. So $$Y+C_1=4$$. Thus, $$Y \in \{2,3\}$$. If $$Y=2$$, then $$S_1=6+S$$ and $$D_1=D$$. If $$D=4$$ then $$S=7$$ which is a contradiction. Thus, $$S=6$$ and $$D=3$$ and $$C_3=3$$. $$S_3=21+5+C_2=26+C_2 \implies C_2=4$$. This is impossible when $$P=7$$, so $$Y=3$$. Then $$C_1=1$$ and $$S_1=9+S$$ and $$D_1=D \in \{3,4\}$$. If $$D=4$$ then $$S=3$$.

If $$P=8$$, then $$S_3=24+5+C_2=29+C_2$$ and $$C_3=3$$ and $$D=3$$. From $$S_2=24+Y+C_1$$ and $$D_2=5$$, we know that $$Y+C_1 \in \{1,11\}$$. If $$Y=0$$, then $$C_1=0$$, and $$Y \ne 1$$ since that is already taken by $$H$$. Thus, $$Y+C_1=11 \implies Y=9$$ and $$C_1=2$$. But $$S_1=27+S$$ and $$D_1=D=3 \implies S=6$$ and $$C_1=3$$

If $$P=9$$, then $$S_3=27+5+C_2=32+C_2$$ and $$C_3=3$$ and $$D=3$$. From $$S_2=27+Y+C_1$$ and $$D_2=5$$, we know that $$Y+C_1=8$$. If $$C_1=1$$, then $$Y=7$$ and $$C_1=2$$. Thus, $$C_1=2$$ and $$Y=6$$. This makes $$S=5$$ which is already taken by $$A$$.

## Thus $$A=9$$ and $$H=2$$

$$S_4=27+D+C_3 \implies D+C_3=5$$ and $$C_4=3$$. Since $$D \ne 2$$, $$C_3 \ne 3$$. Any value of $$P$$ will result in $$C_3 \ge 1$$, so $$C_3 \in \{1,2\}$$ and $$D \in \{4,3\}$$.

If $$P=8$$, then $$S_3=32+C_2 \implies C_3=3$$.

If $$P=7$$, then $$C_2 \ge 2 \implies S_3=29+C_2\ge31 \implies C_3=3$$

If $$P=1$$, then $$S_3=12+C_2 \implies C_3=1 \implies D=4$$. $$S_2=3+Y+C_1$$ and $$D_2=A=9 \implies Y+C_1=6$$ and $$C_2=0$$. Thus $$S_3=12$$ so $$D_3=E=2$$ which is already taken.

If $$P=3$$, then $$S_3=18+C_2$$. If $$C_2\ge2$$, then $$C_3=2$$ and $$D=3$$, which is already taken by $$P$$. If $$C_2=1$$ then $$D_3=E=9$$ which is already taken. If $$C_2=0$$, then $$S_2=9+Y+C1 \implies Y+C_1=0$$. Thus, $$Y=0$$ and $$S=D$$, which is not allowed.

If $$P=4$$, then $$S_3=21+C_2 \implies C_3=2 \implies D=3$$. $$S_2=12+Y+C_1 \implies Y+C_1=7$$ so that $$D_2=A=9$$. But then $$C_2=1$$ and $$D_3=2$$, which is already taken by $$H$$.

If $$P=5$$, then $$S_3=24+C_2 \implies C_3=2 \implies D=3$$. $$S_2=15+Y+C_1 \implies Y+C_1=4$$ so that $$D_2=A=9$$. But then $$C_2=1$$ and $$D_3=5$$ which is already taken by $$P$$.

## Therefore $$P=6$$

From $$S_2=18+Y+C_1$$ and $$D_2=A=9$$, we know that $$Y+C_1 \in \{1,11\}$$. If $$Y+C_1=11$$, then $$Y=8$$ and $$C_1=3$$. But then $$S_1=24+S$$. No value of $$S$$ can make $$C_1=3$$ and $$D\in \{4,3\}$$. Therefore, $$Y+C_1=1$$, and $$S_2=19$$, so $$C_2=1$$. If $$Y=0$$, then $$C_1=0$$ as well, which is a contradiction, so $$Y=1$$.

This means that $$S_3=18+9+C_2=28$$, so $$D_3=E=8$$ and $$C_3=2$$. Thus, $$D=3$$.

Lastly, $$S_1=3+S=D=3 \implies S=0$$.

Full solution:

$$H=2$$ $$A=9$$ $$P=6$$ $$Y=1$$ $$E=8$$ $$D=3$$ $$S=0$$

The corresponding numbers are:

A = 9
D = 3
E = 8
H = 2
P = 6
S = 0
Y = 1

The summation then becomes:

29661
29661
29661
+ 3910
------
92893