Answers and Explanations
1. Who Goes First?
Dylan should go first.
By doing this, Dylan can guarantee a win by playing to a deliberate strategy. On his first turn, he can place a quarter right on the center of the table. Because the table is symmetric, whenever Austin places a quarter on the table, Dylan simply "mirrors" his brother's placement around the center quarter when it is his turn. For example, if Austin places a quarter near a corner of the table, Dylan can place one on the opposite corner. This strategy ensures that even when Austin finds an open space, so can Dylan. As a result, Dylan gains victory, since Austin will run out of free space first!
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2. Pirate Logic 341 coins Explanation:
There are actually infinite answers to the problem, but only one number if the answer is under 1,000. This puzzle is an example of modular arithmetic and the Chinese Remainder Theorem.
The smallest solution under 1,000 for this problem is 341 coins, and the answer is found by working backwards. To find it, we first note that with 11 pirates the coins divided evenly; hence, the number of coins is in the list:
11, 22, 33, 44, 55, 66, 77, 88, 99, 110, 121, 132, 143...
What happens if we take these numbers and divide them among 12 pirates? How many coins would be left over? Well, we want 5 coins to be left over after dividing by 12. Hence, we reduce the list above to:
77, 209, 341, 473...
These numbers divide by 11 evenly and have 5 left over when divided by 12. Now we take these remaining numbers and divide them by 13 until we find the number that gives 3 extra coins left over. Hence, 341 coins.
3. Padlock Puzzle A finite number of possible solutions exist for this problem that can be listed and then crossed off based on the player's given digit. Three simple solutions exist where any player would have the combination by knowing just one single digit. These are the following: (reminder: only solutions summing to 9 and following the X<=Y<=Z property are valid)
0 0 9 (Carol and Nick would realize the answer from their given digit)0 1 8 (Nick would realize the answer if he was given 8)3 3 3 (Belle or Nick would know the combination if given this digit)
After 30 seconds have passed, each realizes none of the others knows the combination instantly. At this point Carol realizes the solution to the problem, since she has a digit where she could eliminate another player's list of possible quick solutions (i.e. everyone now knows Nick did not have a 9, 8, or 3 and Belle did not have a 3.)
If Carol's given digit was 1 she would know the only two possibilities are the following: (0, 1, 8) and (1, 1, 7). Because no player found the answer after 30 seconds, Nick did not have an 8, meaning Carol knew the combination was (1, 1, 7). A person exhausting all possible combinations and removing the obvious combinations will see this is the only set where Carol could know the answer. Therefore since Carol knew the answer, any observer could determine the combination as (1, 1, 7). As an FYI, Nick would likely know the answer a little faster than Belle (as he was able to eliminate more possibilities).
4. Weighed Down There are exactly two possible answers: Mel's weights can be 1, 5, 7, and 9 pounds, or they can be 2, 4, 6, and 10 pounds. No other combinations are possible.
Explanation Let the weights be a, b, c, and d, sorted such that a < b < c < d. We can chain inequalities to get a + b < a + c < a + d, b + c < b + d < c + d. Thus, a + b = 6, a + c = 8, b + d = 14, and c + d = 16. But we don't know if a + d = 10 and b + c = 12 or the other way around. This is how we get two solutions. If a + d = 10, we get 1, 5, 7, and 9; if b + c = 10, we get 2, 4, 6, and 10.
More on the Problem Where this problem really gets weird is that the number of solutions depends on the number of weights. For example, if Mel has three weights and knows the weight of all possible pairs, then there is only one possible solution for the individual weights. The same is true if he has five weights.
But now suppose that Mel has eight weights, and the sums of pairs are 8, 10, 12, 14, 16, 16, 18, 18, 20, 20, 22, 22, 24, 24, 24, 24, 26, 26, 28, 28, 30, 30, 32, 32, 34, 36, 38, and 40. Now what are the individual weights?
This time, there are three solutions:
5. Pie for a Professor Calc 2 at 10 in North Hall.
1) Since Julia only knows the class name, the only way she could immediately know is if it was Calc 3. Since Michael and Mary Ellen both know that Julia doesn't know, that means they know the class isn't Calc 3. Since Michael only knows the time, that means the class can't be at noon. Because Mary Ellen only knows the building, that means the building can't be West Hall. That leaves only the following possibilities:
2) Since Michael doesn't know which class it is, that means the time can't be 9 or 11. Since Mary Ellen doesn't know either, the class can't be in South Hall. That leaves only three possibilities:
3) At this point, Julia now says she knows the answer. Since there are two Calc 1 classes, it must be that Julia knows the class is Calc 2. Thus, the class is Calc 2 at 10 in North Hall.
6. Solve the Cypher The coded message reads:
"Want to know what it takes to work at NSA? Check back each Monday in May as we explore careers essential to protecting our nation."
Explanation Each "word" in the code contains 12 characters, suggesting that the spaces do not indicate word breaks. The letters in the code are simply replaced with others. The trick to determining which ones is to first note the letters in the code that are most commonly used: P, C and I. These likely correspond to the most common letters in English: E, T and A.
Leaders of the past, such as Julius Caesar, used similar codes. They and their supporters would have to mull over them for lengthy periods, trying to figure out the letters that should be substituted for those in the code. These type of substitution cyphers are rarely utilized today because they can be cracked in a flash by specially designed computer apps.
If you're up for more NSA brainteasers, try solving these crossword puzzles from the agency's magazine Cryptolog.
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