Last Sunday’s puzzle solution:

How did you get on with last week’s puzzle? Those of you who solved it are probably pretty confident that you got it right (it was easy to check after all), but for those of you who didn’t here is how it goes:

Remember we are solving a so-called alphabetic puzzle, DATA +DATA = FACTS where the letters all stand for different digits.

Solution

Since a four-digit number added to itself has produced a five-digit answer, the leading digit of that answer must be 1, so we start with F=1.

Now look at the tens column. If $c_1$ is the carry from the units column and $c_2$ the carry into the hundreds column, then

$T+T+c_1=T+10c_2$. Subtracting $T$ from both sides gives $T+c_1=10c_2$.

Both carries can only be 0 or 1, so there are two possibilities:

or

We need to investigate both.

Could $T=0$?

Suppose $T=0,\qquad c_1=c_2=0.$ Because there is no carry from the units column, $A+A=S$ and so $2A < 10$, i.e., $A\leq 4.$

We already know $F=1$, and different letters must represent different digits, so $A\neq1$.

Now consider the hundreds column. There is no incoming carry there either, so it gives $A+A = C$, But the units column already told us $A+A=S$, so $C=S$ which is impossible because different letters must stand for different digits. So the $T=0$ possibility cannot occur.

We must therefore have $T=9,c_1=c_2=1$.

The remaining digits

The units column gives $A+A = S+10$, so S=2A-10. In particular $A\geq5$. The hundreds column gives $A+A+1=C+10c_3$where $c_3$ is the carry into the thousands column.

Since $A\geq5$ this produces a carry, so $c_3 = 1$ and $C = 2A-9$.

Now examine the thousands column: $D+D+1 = A+10$. Thus

So $A$ must be odd and the possible values of $A$ are 5, 7 or 9: it cannot be 9 because $T=9$. If $A=5$, then $2D+1=15$ so $D=7$. But the units column would give $S=0$, and the hundreds column would give $C=1$. Since $F=1$, that repeats a digit, so $A=5$ is impossible and we must have $A=7$ giving $D=8$.

The units and hundreds columns then give $S=4, C=5$ and all the digits are different as required.

Lo and behold, the only possible solution does in fact work,

  8 7 9 7
+ 8 7 9 7
---------
1 7 5 9 4

So our fraudster’s attempt at clever bookkeeping was not quite clever enough.