After a long-overdue vacation, Maggie and Charles are back, with a problem for Zeno's patrons that they haven't yet figured out. During their travels, Maggie and Charles had an occasion to stay at a bed and breakfast inn in Sacramento, California, while meeting with a newly-formed philosophy club at Sacramento State University. The inn was recently opened by a lovely couple, Sam and Susan, who happened to be logicians. Sam had recently retired from teaching, and he and Susan and their young twin children were very happy with the new venture. Especially Sam, who appeared extremely optimistic, and as Charles and Maggie were leaving he told them why. It seems that the address of the B&B (which Maggie and Charles never did know) held a special significance for Sam, who had recently become a numerologist of sorts. He explained it thusly.

"Here's why I'm excited, my good friends, even though others are less optimistic. They point out that as I am sixty years old, and my wife is only forty and our kids are only ten, we're really not the best suited for the hard task of running a B&B. But I disagree, since this was a perfect year to buy the place---let me explain.

First, consider the following representation:

FORTY

+TEN

+TEN

----------------------

SIXTY

By adding the sums of my wife's and twins' ages, you get *mine* this year. Now, I have figured out that *each different letter* in this addition problem stands for a *different digit*, 0 through 9, which is *also* a correct sum. And there is only one solution to get the corresponding numerical addition problem. And guess what? The 5-digit total of the corresponding numbers---which correctly stands for 'sixty'--- turns out to be the 5-digit *address* of our new B&B!! It's in the numbers, I tell you---at sixty, this place was meant for me this year!!"

Well, needless to say, Maggie and Charles were more than a bit skeptical, but they did begin to wonder what was the address of the B&B. Can you Zeno's patrons figure it out? What exactly are the digits that correspond to the correct addition, and what is the sum which is the address of the Sacramento B&B?

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 Linda@doughty.source.co.uk Wed Oct 23 06:14 EDT 1996 Dear Ron The address is 31486. (TEN = 850 and FORTY = 29786) But is this a paradox? 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 The correct adress is: 31486! 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 The address is 31486. 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