## Puzzle Nekst 1 2016-2017

**With the summer holidays already several weeks behind us and with the first few midterms coming up in the surprisingly near future, now seems like the perfect time for some mental fitness exercises to kick start your brain into high-performance mode. This puzzle should be plenty to get up to second gear!**

Tom and Coenraad, two young and upcoming econometrician boys, have been chatting up a beautiful girl in the TOP-week. Their shy friend Björn was obviously interested, but could not muster the courage to talk to her. Being a mysterious young woman that finds intelligence attractive above all else, she decides to take the opportunity to test the boys in an attempt to choose between them; the boys have to figure out her address. She gives the boys the following list of possibilities:

Korvelseweg 10, Korvelseweg 12, Korvelseweg 14, Akkerstraat 11, Akkerstraat 13, Tarwehof 9, Tarwehof 12, Capucijnenstraat 9, Capucijnenstraat 11, Capucijnenstraat 10.

She then tells Tom the street on which she lives, and she tells Coenraad her house number. Being as competitive as they are, the boys do not share any information. Tom says: “I do not know your address, but I am certain Coenraad does not know it either!” Coenraad thinks for a while and says: “Then I know her address,” after which Tom cries out: “Well then I know it too!” Happy with their success, the boys give each other a high five and turn around to tell the girl, only to find out that Björn had figured it out already, and has walked off with the girl.

Can you figure out on which address the girl lives? Please send your solution to nekst@Asset-Econometrics.nl before November 29. A crate of beer or a delicious pie, whichever the winner prefers, will be waiting for whoever has sent the best (partial) solution. Please note that, as before, every recipient of this magazine is eligible to send in their solution, so members of the department are invited to participate as well. Good luck!

*Dolf Oosten is the winner of the previous puzzle! He can come pick up his crate of beer or his pie at room E.10. The solution to the previous puzzle is shown below!*

## Solution puzzle 4 2015-2016

**Answer:**

APPLE = 67794, LEMON = 94832, BANANA = 162626

**Elaboration:**

We rewrite the equation to APPLE + LEMON = BANANA, and stack the words underneath each other, so that six columns are created:

APPLE

LEMON +

——

BANANA

Both APPLE and LEMON consist of five digits, while BANANA consists of six digits. Hence, this equation can only hold if B = 1. From the first and second column we can conclude that 10 + A = L + A + r must hold, where r is a carry of 0 or 1 from the equation P + E + r’ = N +10*r (r’ is again a carry of 0 or 1). The former equation can only be solved if L = 9 and r = 1. The equation for the third column now is P + E + r’ = N + 10. Because P < 9 and E < 9, it must hold that N < P and N < E.

Going all the way to the last column, we see E + N = A, or E + N = A + 10. If the latter is the case, the fifth column results in 9 + O + 1 = 10 + N, which leads to O = N, and we have a contradiction. So, E + N = A. Because A < 9, and N < E, we know that N < 4. The only remaining options are N =2 and N = 3. Let us examine this in two cases.

Case 1: N = 2. Column 5 gives 9 + O = 10 + 2, so we have O = 3. The last column gives E + 2 = A. We know that A < 9, so E < 7, and the remaining numbers for E are 4, 5 and 6. We check these numbers in three subcases. We know that the equation for the third column is P + E + r’ = N + 10 = 12, and the equation for the fourth column is P + M + 1 = A + 10*r’, with r’ either 0 or 1. A subcase is valid if we can find numbers for which these equations hold.

Subcase 1.1: E = 4. The last column gives A = 6. Column 3 now yields P + 4 + r’ = 12 and column 4 now yields P + M + 1 = 6 + 10*r’. These equations can be solved with r’ = 1, P = 7 and M = 8.We have now obtained the solution: APPLE=67794, LEMON=94832 and BANANA=162626. For completeness, we show that the other cases have no solution and this is the unique solution.

Subcase 1.2: E = 5. The last column gives A = 7. Column 3 now yields P + 4 + r’ = 12 and column 4 now yields P + M + 1 = 7 + 10*r’. If r’ = 0, there is clearly no solution, and if r’ = 1, we find P = 7 and M = 9. However, we have L = 9 so we have a contradiction.

Subcase 1.3: E = 6. The last column gives A = 8. Column 3 now yields P + 4 + r’ = 12 and column 4 now yields P + M + 1 = 7 + 10*r’. If r’ = 0, there is clearly no solution, and if r’ = 1, we find P = 7 and M = 10. However, no letter can be assigned a number larger than 9, so we have a contradiction.

Case 2: N = 3. Column 5 gives 9 + O = 10 + 3, so we have O = 4. The last column gives E + 3 = A. We know that A < 9, so E < 6. We also know that N < E, so the remaining number for E is 5. We know that the equation for the third column is P + E + r’ = N + 10, this gives P + 5 + r’ = 13. , and the equation for the fourth column is P + M + 1 = A + 10*r’, with r’ either 0 or 1. If r’ = 0 then P = 8 and it must hold that 9 + M = A, which clearly has no solution. If r’ = 1 then P = 7 and it must hold that 9 + M = A + 1, which implies A > 9 so this also has no solution.

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