# ZENO'S COFFEEHOUSE Results #9 ## The Ninth Coffeehouse Challenge

### Here is the problem, as presented. along with the responses. Congratulations to the winners! I have listed an example of a clearly-described solution to the B&B riddle. Enjoy!

#### Here's Sue McCalden's reasoning---good work!```The address of the B&B is 31486. The following will show how I arrived at this number. Since each letter represents a different number (0-9), then each number may only be used once. Following the representation of the words: FORTY we can start! +TEN +TEN ------ SIXTY (1) We have to determine what N and E represent. With Y+2N=Y, N could be 5 or 0. If N=5, there will be a carryover to the next line. However, this does not work since T+2E=T would then include a carryover. Therefore, N must equal 0. (We can leave this line for now, for whatever number is left, it will equal Y) We know that N=0, therefore, E=5. (2) There is a problem with O=I, and F=S since this obviously cannot be the case. Therefore, for both O and F, there must be a carryover in order for O+carryover to equal I (the same must apply for F). Therefore, R+2E must be > than 20. O=9 otherwise there could be no carryover to F. (3) Why is this the case? F+carryover has to be < than 4 in order for F+ carryover to equal S. Since R+2T>20, than R and T will have to be > than 5. (4) So lets now plug in some numbers. If T=6, and R=7, than R+2T=20. Well this obviously does not work since we know that N=0. If we make T=8 and R=6, than R+2E=23. This also doesnot work for if I=3, than the carryover plus whatever F represents will not give us S. Why is this the case? We know that F < than 4. If F=4 +1(carryover) than S=5. We know this cannot work since E=5. Therefore, T=8, and R=7. (5) We know that N=0. We know that T=8, and E=5. Therefore, 8+(2)5=18. We now carryover the 1 and T=8. We know that R=7, and T=8, therefore, 7+(2)8=23. We now carryover the 1, and X=4. We know that O=9 + the 2 carryover equals 11. Therefore, I=1. Now, F has to be 2. Therefore, 2+1carryover=3. Therefore S=3. (6) Therefore: F=2 T=8 O=9 E=5 R=7 N=0 T=8 Y=6 (remember, whatever # was left over, it would be Y) Therefore: S=3 I=1 X=4 T=8 Y=6 So the address of the B&B is 31486!!!! ```

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From Linda@doughty.source.co.uk Wed Oct 23 06:14 EDT 1996
Dear Ron
The address is 31486.  (TEN = 850 and FORTY = 29786)

Best wishes,

Linda Sayle

From deacon@merlin.net.au Sun Oct 27 01:36 EST 1996
Hope this is correct
SIXTY = 31486

From walkerj@cadvision.com Mon Oct 28 13:11 EST 1996
Dan

From suresh@cns.nyu.edu Tue Oct 29 20:04 EST 1996
R
0  1  2  3  4  5  6  7  8  9
N  I  F  S  X  E  Y  R  T  O

Time taken: 2 and a half minutes for complete solution.

From suresh@cns.nyu.edu Tue Oct 29 20:41 EST 1996
> BTW, want to explain your strategy for solving the problem in this short
> time? I'd like to post it!
> Thanks,
> Ron Barnette
>

Sure. This is a rather simple problem in this genre, more difficult ones
usually have the base number system ( decimal in this case ) as a further
variable to be determined. That is, one doesnt know in advance which base
is being used (eg: binary or decimal or hexadecimal etc).

Anyways, for this problem, it is clear that since T + 2 E gives T in the
unit's place, and Y + 2 N gives Y in the unit's place too, both 2*E and 2*N
have 0 in the unit's place, that is , E and N are one of 0 and 5. Now,
since T + 2*E gives T in the unit's place, there is nothing being carried
over from the previous addition, so N = 0, and E = 5. Now some
consideration of the problem will give :

1 + F = S    2 + O = I, since 2 and 1 are the only numbers that can be
carried over by summing 3 digits ( 9 * 3 = 27 being the maximum sum).
It now follows that since N is already zero, I = 1 and O = 9.

So 1 + R + 2 T gives 2 in the ten's place, so T has to be one of 7 or 8,
assuming maximum R. If T = 7, R = 8 ( O is already 9), but that gives X =
3, and 1 + F = S cannot be satisfied anymore. So T = 8, and R = 7, R = 6
being disqualified because that again leads to X  = 3. We now also have X
= 4.

This gives F = 2 , S =3, and Y = 6, the only number remaining.

I know this sounds complicated, but really, if someone has practice in
doing these kind of problems , it should take a minute or so. I made an
error in filling up my number to letter matrix, and it took me some time
to realize the error......

Cool concept for a page though. My compliments. The level of difficulty
could be far higher, however.

Thanks, Suresh

-------------------------------------------------------------------------
B. Suresh Krishna,
Graduate Fellow,		      247 Montgomery Street, # 4,
Center for Neural Science,            Jersey City, NJ 07302
Room 809, 4 Washington Place,	      Ph: 201-333-7041.
New York, NY 10003-6621.
Ph:    212-998-3901
Fax:   212-995-4011
Email: suresh@cns.nyu.edu  URL: http://www.cns.nyu.edu/home/suresh/suresh.html
-------------------------------------------------------------------------

From webspin@clever.net Thu Oct 31 16:12 EST 1996

I'm doing a project on Zeno and his paradoxes at the moment, and after
typing 'Zeno' into Alta Vista, I was given Zeno's Coffeehouse as one of
the choices.  Being a Cyber-Cafe owner myself, i had to duck in and
check it out.  Very cool.  And your slogan is awesome.  (Incidently, how
many people get the joke?)

Thanks!
-Kevin Beimers
-webspin@clever.net
-http://clever.net/webspin/spinoff/seattle.html

PS.  By the way: The order of letters representing 0 through 9 is
N,I,F,S,X,E,Y,R,T,O and the address of the B&B is 31486.  Later.

From qulog@mail.idt.net Fri Nov  1 15:50 EST 1996

29786
850
850

31486

From mccalden@unixg.ubc.ca Sun Nov  3 16:17 EST 1996

Hi Ron,
Well, it is interesting that I was able to come up with a solution for
the new challenge having the words set up differently.  Actually, on my
computer, the set up was shown as FORTY
TEN
TEN
--------
SIXTY
Anyways, by doing it the proper way, the B&B address is 31486.  I
wanted
to make sure with you first to see if in fact I am correct this time
before I send you the information on how I arrived at this number since
it takes awhile to type it out.
So, it would be greatly appreciated if you could e-mail me back yet again
to let me know if my answer is correct.
Thanks again for your time and patience,
Best regards,
Sue McCalden

From kend@ecst.csuchico.edu Fri Nov  8 12:52 EST 1996

SIXTY=31486

This problem was used at "Ken's Puzzle of the Week" for August 13, 1996:
http://www.ecst.csuchico.edu/~kend/potw/index.html

Here is the solution provided there:
29786
850
+ 850
-----
31486

Column 1: N is either 0 or 5. If N=5, then the carry to Col2 is 1 and there is
no possible value for E,
so N=0.
Column 2: E is ether 0 or 5, but N=0, so E=5, and the carry to Col3 is 1.
Column 4: I is either 0 or 1, but N=0, so I=1, O=9, carry from Col3 is 2, and
the carry to Col5 is 1.
Column 3: For 1+R+T+T >= 20, R and T must be in {6,7,8}.
Column 5: F+1=S, so {F,S} is either {2,3} or {3,4}, so X is not 3.
Column 3: 1+R+T+T = 22 or 24, so R is odd and must be 7, 2T=14 or 16, T is 7 or
8, but R=7, so
T=8, X=4.
Column 5: F=2, S=3.
Column 1: Y=6.

From mccalden@unixg.ubc.ca Tue Nov 12 03:01 EST 1996

The address of the B&B is 31486.  The following will show how I arrived
at this number.  Since each letter represents a different number (0-9),
then each number may only be used once.  Following the representation of
the words: FORTY we can start!
+TEN
+TEN
------
SIXTY

(1) We have to determine what N and E represent.  With Y+2N=Y, N could be
5 or 0.  If N=5, there will be a carryover to the next line.  However,
this does not work since T+2E=T would then include a carryover.
Therefore, N must equal 0. (We can leave this line for now, for whatever
number is left, it will equal Y)
We know that N=0, therefore, E=5.

(2) There is a problem with O=I, and F=S since this obviously cannot be
the case.  Therefore, for both O and F, there must be a carryover in
order for O+carryover to equal I (the same must apply for F).  Therefore,
R+2E must be > than 20. O=9 otherwise there could be no carryover to F.

(3) Why is this the case? F+carryover has to be < than 4
in order for F+ carryover to equal S.  Since R+2T>20, than R
and T will have to be > than 5.

(4) So lets now plug in some numbers. If T=6, and R=7, than R+2T=20. Well
this obviously does not work since we know that N=0.  If we make T=8 and
R=6, than R+2E=23.  This also doesnot work for if I=3, than the carryover
plus whatever F represents will not give us S.  Why is this the case?
We know that F < than 4. If F=4 +1(carryover) than S=5.  We
know this cannot work since E=5.  Therefore, T=8, and R=7.

(5) We know that N=0.
We know that T=8, and E=5. Therefore, 8+(2)5=18.
We now carryover the 1 and T=8.
We know that R=7, and T=8, therefore, 7+(2)8=23.
We now carryover the 1, and X=4.
We know that O=9 + the 2 carryover equals 11.
Therefore, I=1.
Now, F has to be 2.  Therefore, 2+1carryover=3.
Therefore S=3.

(6) Therefore: F=2   T=8
O=9   E=5
R=7   N=0
T=8
Y=6 (remember, whatever # was left over, it would be Y)

Therefore: S=3
I=1
X=4
T=8
Y=6    So the address of the B&B is 31486!!!!

From shack@esinet.net Sun Nov 24 22:07 EST 1996

The address of the B&B is 31486.

The clue FORTY + TEN + TEN =3D SIXTY has the single solution...
29786 + 850 + 850 =3D 31486.

Proof: Since Y + 2*N =3D 10x + Y, we must have N=3D0 or N=3D5.
Since T + 2*E + (carry) =3D 10x + T, we must have E =3D 0 or E =3D 5
and no carry.   Since there is no carry, N =3D 0 and therefore E=3D5,
and 1 is carried to the 100s place.

Since S <> F, there must be a carry from the 1000s place.   Since
adding 1 + R + 2*T cannot result in a number greater than 27 (if
T=3D9, R=3D8), the carry to the 1000s place must be 1 or 2.   Since,
N =3D 0,  I cannot be 0, the carry must be 2, I =3D 1, and O =3D 9.
Also, the carry from the 1000s place is 1 and S =3D F + 1.

This leaves six letters (F, S, R, X, T, Y) and six digits (2, 3,
4, 6, 7, 8).   Since the carry to the 1000s place is 2, we must
have that 1 + R + 2*T > 19.   Indeed, since X cannot be 0 or 1,
the sum 1 + R + 2*T cannot end in 0 or 1, so we must have 1 + R +
2*T > 21.  Since R < 9, this means that 2*T > 12, or T > 6.
Thus T is either 7, or 8.

If T were 7, then 1 + R + 2*7 > 21, so R > 6.   The only digit
remaining that qualifies is 8.   If R is 8, then 1 + 8 + 2*7 =3D
23, so X could be 3.   Unfortunately this leaves only the digits
2, 4 and 6, and this cannot be, since no consecutive digits
remain and we know that S =3D F + 1 (since the carry in the 10000s
place must be 1).   Thus we have eliminated T =3D 7.

So we are left with T =3D 8.   Then 1 + R + 2*8 > 21, so R > 4.
So R must be 6 or 7.   If R =3D 6, then 1 + 6 + 2*8 =3D 23 and X =3D 3.
But this leaves only the digits 2, 4, and 7, and again since we
know S =3D F + 1 there is no solution here.

So now we know that the only possibility remaining is T =3D 8, R =3D
7.   So 1 + 7 + 2*8 =3D 24, so X =3D 4.   This leaves 2, 3, and 6, so
=46 =3D 2, and S =3D 3, the only consecutive digits.   This leaves one
letter (Y) and one digit (6) so Y =3D 6 and the result may be
written.

Shack Tom

From jbadner@helix.nih.gov Wed Nov 27 23:28 EST 1996

(I'm using Q for the letter O, since in my typeface, it's impossible
to distinguish between zero and oh.)  The problem is to find a key
for the letters E, F, I, N, Q, R, S, T, X, Y such that each is a
distinct digit, and when the equation "FQRTY + TEN + TEN = SIXTY"
is deciphered, it is a true statement.  Progressive steps toward
the solution on the left, English-language explanations of my reasoning
on the right.

FQRTY		Obviously, if Y+2N=(10m)+Y and T+2E=m+(10n)+T,
TEN		then N=0 and E=5.  Moreover, since T+2E=10+T,
TEN		(10p)+X=R+2T+1.  Since 2T<10, (10p)+X<30. And
-----		since p+Q=(10q)+I, and I=/=0, Q=9 and I=1.  Then
SIXTY		F+1=S;  but since there are no repeats of digits,
is one of <2,3>, <3,4>, <6,7>, and <7,8>.

F9RTY		Now, R+2T+1=20+X;  so either R is even and X odd,
T50		or R is odd and X even.  Now, this means that either
T50		R or X is either 3 or 7.  Suppose that R=3.  Then
-----		2T+3>21, meaning that T>9, which is not possible.
S1XTY		Suppose R=7.  Then 2T+8=20+X, or X+12=2T.  Now, X
is one of 2, 4. 6, and 8.  Plugging in, we find
F978Y		that if X=2, T=7.  This is impossible, since R=7.
850		If X=4 then T=8.  This is possible.  (I will not
850		run through the proof that the other options are not
-----		going to work, since you gave us the information that
S148Y		there is only one solution.  But I've done it.)

29786		Now, F, S, and Y are some combination of 2, 3, and 6.
850		But since F+1=S, it must be that F=2, S=3, and Y=6.
850
-----		Therefore, the address (which, before deciphering,
31486		was SIXTY) is 31486.

This one was fun!!  Thanks!

Cal Montgomery

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