# Alphametic as protest #2

Riddles for all.
Since my latest alphametic as protest got a good feedback but my riddle as protest did not, I decided to stick to what works and added a new alphametic as a protest for the riddle sandbox:

 RIDDLE +
RIDDLE +
RIDDLE =
--------
FORALL


Same rules apply.

• each letter is a different digit.
• the leading digits cannot be 0.
• the numbers are in base 10.
• an answer without a reasoning is worth nothing.
• first to answer with a good reasoning gets the checkmark.

L is 4 or 9, because (3L+[0,1,2])%10 = L and L+[0,10,20]/3=E, L!=E eliminating 0. E+E+E=20+L
L = 4, E = 8
D+D+D+1=A
D+D+D+(0 or 2) = R
D!=0
R+R+R=F
R<=3

Assume R=0
D=3
A=0
R and A can't both be 1, contradiction Assume R=1
D=0
A=0
A and D can't both be 0, contradiction Assume R=2
D=7
A=2
R=3
R can't be both 2 and 3, contradiction R=3
D=7
A=2
F=9
leaving I=2 and
0=5

For a total of

317748*3 = 953244

• You can skip the assumption of R=0 because the leading digit cannot be 0. – Marius Aug 29 '16 at 14:25
• if you explain the first line L is 4 or 9, the checkmark is yours. – Marius Aug 29 '16 at 14:40

Immediately, we know that

$R \in \{1,2,3\}$ since the result is also a 6 digit number.

Also, we know from the ones and tens column that there is a close relationship with $E$ and $L$. A quick run through of all the possibilities shows that

the only choice for $E$ that works is $E=8$. Thus, $L=4$.

Since

$3 \times L + 2 = 14$, we have a carry over of 1 into the hundreds column. So $3 \times D + 1 \implies A$. If $D$ is large, then the carry over into the next column could be 2. If $D$ is small, then the carry over can be 0. But if $D$ is middling, then the carry over will be 1 again, and we cannot have $A=R$. Thus, $D$ must be large or small to get a different carry over. As a consequence, we know that $A$ and $R$ differ by 1. Since we already know $R\in \{1,2,3\}$, we know that $A \in \{0,1,2,3,4\}$.

Running through possibilities for $D$, the only one that doesn't cause a contradiction with valid values is

$D=7$. Thus, $A=2$ and $R=3$. Therefore, $F=9$.

Also, we know that $I$ must be

small enough not to cause a carry over. Thus, $I \in \{0,1\}$. The carry over into the column of $I$s is 2, so if $I=0$, then $O=2$ which is already taken. Thus, $I=1$ and $O=5$.

Final solution is:

317748
317748
317748
----------
953244

• This is much better explained than mine, good work. – Sconibulus Aug 29 '16 at 15:05
• @Sconibulus Thanks! But I wouldn't say much better. In fact, if I had seen your answer before I added mine, I probably wouldn't have gone through the effort. – Trenin Aug 29 '16 at 15:07