1
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Let me tell you a story about Alice and Bob.
When they are together it's all fun and games, we know that.
But what happens when they are not together?
I'll tell you.
Alice has an odd feeling. It might be sadness.
It is very odd for them not to have fun.

Can you find what it is?

ALICE - 
  BOB = 
----- 
NOFUN

Which digit does each letter represent? (Please present the full analysis how these digits can be determined).
There are multiple solutions to the puzzle, but I'll settle for 1 or 2.

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  • 1
    $\begingroup$ the image of "Sadness" has nothing to do with the puzzle. I just added there for visual effect. $\endgroup$ – Marius Apr 7 '16 at 14:56
  • $\begingroup$ hey...why do you think this is too broad? Because you couldn't solve it? $\endgroup$ – Marius Apr 7 '16 at 15:35
  • $\begingroup$ Is lateral thinking actually involved in answering this alphametic? $\endgroup$ – Ian MacDonald Apr 7 '16 at 21:37
  • $\begingroup$ @IanMacDonald. Some of the clues might look odd but they can help solve the problem easier. $\endgroup$ – Marius Apr 7 '16 at 21:44
  • $\begingroup$ For anyone who really enjoys this type of puzzle, you might check out Sideways Arithmetic from Wayside School. ;) $\endgroup$ – Wildcard Apr 7 '16 at 21:56
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There are 4 solutions overall.

$L = 0, O = 9, F = 8$ for all cases, the rest are:

For $E = 5, C = 6, U = 7$:
- $B = 2, N = 3, A = 4, I = 1$
- $B = 4, N = 1, A = 2, I = 3$

For $E = 7, C = 3, U = 4$:
- $B = 2, N = 5, A = 6, I = 1$
- $B = 6, N = 1, A = 2, I = 5$

Proof.

Considering only the 3rd, 4rd and 5th digits we have $ALI - B = NOF$. The only way for $A$ to change is if $L = 0$ and $B > I$. This also forces $O = 9$ and $A = N + 1$.

From $C - O = U$, we can conclude that there couldn't have been a borrowed digit in the previous subtraction or else we'd have had $C - (9 + 1) = C - 10 = C \pmod{10}$ (we got $U$ instead). Also, $U = C - 9 \pmod{10} = C + 1$. Since there was no borrowed digit for $E - B = N$, we get $E = B + N$.

Since $C - 9$ borrows a digit, $I - B$ has a carry. We also know that $I \lt B$, so $I$ must borrow $10$. Therefore, we have $F = (10 + I) - (B + 1) = 9 + I - B$.

It is implied that $ALICE$ and $NOFUN$ are odd. Therefore, $E, N = odd$. Then from $E = B + N$ we can further deduce $B = even; E \ge 3$, hence $E \in \{3, 5, 7\}$. Because of that we also get $B \in \{2, 4, 6\}; N \in \{1, 3, 5\}$. Furthermore, we require $N \ne B - 1$ or else we'd get $A = N + 1 = B$.

Brute force for $B$ and $N$, with $I \lt B$.

Let B = 2 (N can only be 3 or 5; can't be 1 because N $\ne$ B - 1)
-- Let N = 3
---- A = 4
---- E = 5
---- Left with {1, 6, 7, 8}
---- Let I = 1; F = 8, C = 6, U = 7 $\Rightarrow$ solution
-- Let N = 5
---- A = 6
---- E = 7
---- Left with {1, 3, 4, 8}
---- Let I = 1; F = 8, C = 3, U = 4 $\Rightarrow$ solution

Let B = 4 (N can only be 1; can't be 3 because N $\ne$ B - 1 and can't be 5 because B + N $\le$ 7)
-- Let N = 1
---- A = 2
---- E = 5
---- Left with {3, 6, 7, 8}
---- Let I = 3; F = 8, C = 6, U = 7 $\Rightarrow$ solution

Let B = 6 (N can only be 1; can't be 3 or 5 because B + N $\le$ 7)
-- Let N = 1
---- A = 2
---- E = 7
---- Left with {3, 4, 5, 8}
---- Let I = 3; F = 6 = B $\Rightarrow$ contradiction
---- Let I = 4; F = 7 = E $\Rightarrow$ contradiction
---- Let I = 5; F = 8, C = 3, U = 4 $\Rightarrow$ solution

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4
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I have so far found two solutions (in both cases, ALICE is odd) so I'll post them here

$(A,L,I,C,E,B,O,N,F,U) = (2,0,3,6,5,4,9,1,8,7)$ or $(2,0,5,3,7,6,9,1,8,4)$

Formatted

20365
-   494
---------
19871

and

20537
-   696
---------
19841

Reasoning

A five digit number minus a three-digit number results in a five-digit number with a different first digit

Hence, we must necessarily have
$N=A-1$,
$L=0$ and
$O=9$.

If as digits $E<B$ then the difference will carry over a $1$ and we'll necessarily get $C=U$, which is not allowed.

Hence $E>B$ and $U=C+1$ (also this means $E>N$)

The difference $C-O$ will necessarily carry a $1$ over to the hundreds place and since we need to carry over a one to the thousands place, we must have $I <B+1 \Rightarrow I < B$. Furthermore, we also must have $I<F$.

Since $A$, $E$ and $O$ are all greater than $N$, it follows that $N \le 6$.

As an example solution we can take $N=1$. Then, $A=2$, and the pairs $(B,E)$ and $(C,U)$ are consecutive digit pairs and together with $I$ and $F$ must constitute the entire set of numbers $(3,4,5,6,7,8)$. We just need to ensure $I-B-1$ gives $F (\bmod 10)$

So let's try $F=8$. Then we could have $I=3, B=4, E=5, C=6, U=7$ or perhaps $I=5, B=6, E=7, C=3, U=4$ which both seem to work. In fact, these are the only solutions in this case since $(I,B,E)$ must be consecutive as must $(C,U)$.

$F=7$ doesn't work since we need to assign two consecutive pairs to $(B,E)$ and $(C,U)$ from $(3,4,5,6)$ leaving $I=8$ which contravenes $I<B,F$

If $F=6$, then $I=3 \Rightarrow B=6=F$ (not allowed) $I=4$ means $C$ and $U$ cannot be consecutive and $I=5 \Rightarrow B=8$ and $E=9=O$ (not allowed).
It's not too hard to show that $F<5$ does not lead to any more solutions in this case ($N=1$).

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  • $\begingroup$ Both solutions are correct. (hint: there are more but it's enough to get a tick) Bu't I'm going to give it 24 hours before accepting the answer so people will not lose interest. $\endgroup$ – Marius Apr 7 '16 at 15:24
  • $\begingroup$ Yeah, that makes sense. I'll post other answers up here if I find them. $\endgroup$ – hexomino Apr 7 '16 at 15:28
0
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I think I'm correct with these hints so far:

A = N+1
L = 0
N,O,F,U are odd numbers, also B

I'm new with this, but is kinda entertaining :D (and adictive

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  • $\begingroup$ Hint: you are right only partially. $\endgroup$ – Marius Apr 7 '16 at 15:20

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