Poe

[cinelli]

In this alphametic all numbers are prime

.
  PIT
    N
 PEND
 ULUM
-----
***TI

Cinelli

[jfgw]

This is driving me potty

I see that there are only nine letters. Are you just using the digits 1 to 9?

I have found one solution except that the answer is not prime (the digits add up to a multiple of 3).

If I get time, I will go through my workings tomorrow night.

Julian F. G. W.

[UncleEbenezer]

Did someone say Poe? His corvidian subject spoke of the possibility of odd numbers featuring in:

.
QUOTH
  THE
-----
RAVEN

[UncleEbenezer]

This is driving me potty

I see that there are only nine letters. Are you just using the digits 1 to 9?

I have found one solution except that the answer is not prime (the digits add up to a multiple of 3).

I feel no motivation to try this one. Is the tenth letter (0 or otherwise) not hiding in the asterisks?

[cinelli]

I see that there are only nine letters. Are you just using the digits 1 to 9?

I don't think this is too much of a clue to say that the tenth digit is one of the asterisks.

Cinelli

[GoSeigen]

In this alphametic all numbers are prime

.
  PIT
    N
 PEND
 ULUM
-----
***TI

Cinelli

Nice alphametic, thanks!

Spoiler:

.
  613
    2
 6427
 8089
-----
15131

Rough working:

.
  PIT
    N
 PEND
 ULUM
-----
***TI
  kc

D,I,M,T=[1379] (odd)
N=2 because I is odd

c,k are carries
c=[12]
A. if c=1, u is odd
1. if T,D,M=[139] then I=5, U=7
thus T=5 which is same as I, so c=2 by contradiction
2. other T,D,M don't give a valid I.

B. if c=2, u is even, U=[468]
1. if T,D,M=[379] then I=1, thus U=8, T=3, D,M=[79]
[123789] have been used, thus P=[456], now:
Top line: 413 is divisible by 7; 513 by 3; so P=6, 613 is prime
[1236789] have been used, thus E,L = [045]
a. Now If D=7, M=9
Third line: 6027 is divisible by 3, 6427 is prime, 6527=61*107 is not, and L=[05] then:
Fourth line: 8089 is prime, and so is 613+2+6427+8089=15131 QED


GS

[Gengulphus]

Did someone say Poe? His corvidian subject spoke of the possibility of odd numbers featuring in:

.
QUOTH
  THE
-----
RAVEN

Possibly-partial spoiler...

If there were no carry out of the middle column into the two columns to its left, QU would be identical to RA, so there is such a carry. Similarly, if there were no carry out of the second leftmost column into the leftmost, Q would be identical to R, so there is also such a carry. Therefore U+1 = A+10, which can only happen for digits U and A if U=9 and A=0. Also, R = Q+1, so (Q,R) is one of the pairs (1,2), (2,3), (3,4), (4,5), (5,6), (6,7) and (7,8).

The Edgar Allen Poe quote is "Quoth the Raven, Nevermore". I'm rather inclined to take that together with the wording of the puzzle as implying that none of QUOTH, THE and RAVEN are odd - not certain of it, but I may return to the possibility of one or more of them being odd later. For the moment, I'll assume that they're all even. That implies that H, E and N are all even digits, and since we already have A being the even digit 0 and one of Q and R being even, we must have accounted for all five even digits. Therefore all of O, T, and V must be odd.

The second rightmost column digit sum is then T [odd] + H [even] + (carry in from rightmost column) [0 or 1] = 10*(carry out to middle column) + E [even]. On evenness/oddness grounds, that can only work if the carry in from the rightmost column is a 1, i.e. if H+E >= 10. The maximum possible value of two even digits is 6+8 = 14, so H+E is 10, 12 or 14, and H+E = 10 would imply N=0. So H+E is 12 or 14, which implies that the rightmost column digits (H,E,N) are (4,8,2), (8,4,2), (6,8,4) or (8,6,4).

The middle column digit sum is O [odd] + T [odd] + (carry in from second rightmost column) [0 or 1] = 10 + V [odd], since we know the middle column produces a carry out of 1. On evenness/oddness grounds, that implies that the carry from the second rightmost column to the middle column is 1, and therefore that the second rightmost digit column sum is T+H+1 = 10+E, from which it follows that T = 9+E-H. If E > H, this makes T >=10 and so not a digit, which allows us to eliminate two of the possibilities, and T = 9+E-H also allows us to extend the remaining two to (H,E,N,T) = (8,4,2,5) and (8,6,4,7).

In the first case, (Q,R) must include 6, and cannot be (5,6), so must be (6,7), while in the second case (Q,R) must include 2 and can be either (1,2) or (2,3). So we now have the three cases (H,E,N,T,Q,R) = (8,4,2,5,6,7), (8,6,4,7,1,2) and (8,6,4,7,2,3).

Returning to the middle column digit sum, it is O+T+1 = 10+V, from which it follows that O-V = 9-T, which is 4, 2 and 2 respectively in the three cases. But O and V have to be the two remaining odd digits, which are {1,3}, {3,5} and {1,5} respectively, and clearly those combine to say that only the second case works:

19578
  786
-----
20364

As I said, I may return to the alternative assumption, which would be based on the raven having commented on the possibility of the numbers being odd rather than on its impossibility: some (not me) might feel that implied that at least one of the numbers is odd. But I'm short of time to do that, so not now!

Gengulphus

[UncleEbenezer]

Possibly-partial spoiler...

Your assumption and solution is entirely correct. The unflinching finality of the raven's comment is, after all, a devastating hammer blow to the narrator.

The opposite assumption would lead to multiple solutions.

(When Cinelli used Poe as a title for an alphametic, it crossed my mind Poe's most famous quote has the right number of letters and would probably yield solutions).

[cinelli]

Sorry for the late response. GoSeigen's solution is correct and his reasoning is faultless.

Cinelli

[Gengulphus]

D,I,M,T=[1379] (odd)
N=2 because I is odd

c,k are carries
c=[12]
A. if c=1, u is odd
1. if T,D,M=[139] then I=5, U=7
thus T=5 which is same as I, so c=2 by contradiction
2. other T,D,M don't give a valid I.

B. if c=2, u is even, U=[468]
1. if T,D,M=[379] then ...

A rather quicker route through that:

So D+I+M+T = 20, and the rightmost column tells us that D+M+2+T = 10*c + I, or 10*c - 2 = D-I+M+T = (D+I+M+T) - 2*I = 20 - 2*I. Rearranging that gives 2*I = 22 - 10*c, or I = 11 - 5*c. So c=1 would imply that I=6, which isn't possible, and therefore c=2, I=1 and T,D,M=[379].

Gengulphus

[GoSeigen]

A rather quicker route through that:

So D+I+M+T = 20, and the rightmost column tells us that D+M+2+T = 10*c + I, or 10*c - 2 = D-I+M+T = (D+I+M+T) - 2*I = 20 - 2*I. Rearranging that gives 2*I = 22 - 10*c, or I = 11 - 5*c. So c=1 would imply that I=6, which isn't possible, and therefore c=2, I=1 and T,D,M=[379].

Gengulphus

Quite right. Originally I had "D,I,M,T=[13579] (odd)", i.e. a choice of four out of five odd numbers, forgetting that any number ending in five could not be prime. Later I noticed and removed the 5 from this step, but didn't rework the entire solution so failed to notice your simplification...


GS

[stewamax]

Somewhat off-thread... I (in dinner jacket) once recited The Raven dead-pan to a cabaret audience. At the point where ‘in there stepped a stately raven…’ occurs, a friend entered stage-right and crept behind me dressed as a huge pantomime chicken. He subsequently uttered ‘Nevermore’ at all the appropriate moments while I looked everywhere – including down my trousers – to see where the cry was coming from.

Poe may not have been a great classical poet (although his ‘Helen’ is an exception), but his ability – rather like Dylan Thomas - to manage unusual rhyming and alliteration schemes was extraordinary.

[UncleEbenezer]

I love The Raven. Something evocative, combined with the dazzling craftsmanship of its construction, and eclectic imagery and vocabulary. A highlight I still recollect of my student days was a brilliant and impassioned impromptu recital of it by a Canadian student.

I've played with setting it to music, but find that extraordinarily hard: the trochaic tetrameter tends to shoehorn my efforts into sounding very four-square and dull. I guess there's no shame in that when Beethoven sounds equally four-square - though of course not dull - setting poems with a similar meter (9th symphony, and even more in his choral fantasia). And for me it highlights how brilliant Coleridge-Taylor was in avoiding sounding entirely four-square with Longfellow's verse.

[stewamax]

Oddly enough, I have also recited in public (with no pantomime chickens...) a pastiche of the only other trochaic octameter* poem I have ever encountered (Tennyson's Locksley Hall): William Bromley-Davenport's Lowesby Hall. It is an extremely well-crafted pastiche and is often linked with the most famous of all foxhunting poems: Bromley-Davenport's The Dream of an Old Meltonian

It would be interesting to pursue these meters (and alliteration) further, but not perhaps on this thread.

* OK - tetrameter if you include the unwritten caesura