Assignment 1, due Monday September 19, 2005

Transcription

1 Assignment 1, due Monday September 19, 2005 Problem 1. Four people are being pursued by a menacing beast. It is nighttime, and they need to cross a bridge to reach safety. It is pitch black, and only two can cross at once. They need to carry a lamp to light their way. The first person can cross the bridge in no less than 10 minutes, the second in 5 minutes, the third in 2 minutes, and the fourth in 1 minute. If two cross together, the couple is only as fast as the slowest person. (That is, a fast person can t carry a slower person to save time, for example. If the 10-minute person and the 1-minute person cross the bridge together, it will take them 10 minutes.) The person or couple crossing the bridge needs the lamp for the entire crossing, and the lamp must be carried back and forth across the bridge (no throwing, etc.) If they don t all get completely across in under minutes, who ever is on the bridge or left behind will be eaten by the beast. Is it possible for all of them to get across? What is the best possible solution? Solution: One solution is as follows: Label the people 1, 2, 5, and 10, in accordance with the time it takes each to cross the bridger, then 1 and 2 cross (2 minutes) 2 returns with the lantern (2 minutes) 5 and 10 cross (10 minutes) 1 returns with the lantern (1 minute) 1 and 2 cross (2 minutes) This takes a total of 17 minutes, and it cannot be done in shorter time. crossings is as follows: It is clear that the pattern of Two people cross to the safe side, one returns, then two more cross to the safe side and one returns, and then the final two cross to the safe side. Each person has to cross the bridge an odd number of times, so either once or three times. Since 10 has to cross at least once, this uses up 10 minutes, and crossing 3 times would exceed 17 minutes. If 5 crosses three times, then one of the times would be with 10, and so this would total , or 20 minutes, again exceeding 17 minutes. Thus, 10 and 5 each cross once, and this means that 2 and 1 must cross to the safe side together at least once. If 10 and 5 do not cross together, then this accounts for just for the crossing to the safe side, and the return trips then bring the total to more than 17. Thus, 10 and 5 must cross together. They cannot be the first pair to cross, else one of them would have to return with the lantern. They cannot be the last pair to cross, since one person in the last pair has to have returned from the safe side. So 10 and 5 must be the second pair to cross to the safe side. Therefore, 1 and 2 cross first, and one of them returns. Then 10 and 5 cross to the safe side, and the other member of the 1 and 2 pair returns. This uses up 15 minutes total. Then 1 and 2 cross to the safe side.

2 Problem 2. The OPEC inspector found that a wholesaler of crude oil was swindling his retailers by including one barrel of oil with a high sulphur content to every nine barrels of top grade oil he sold them. Each barrel contained 55 gallons of oil, and the external appearance of both the low grade and top grade oil was identical. However, the substandard barrels were each 1 kilogram too heavy. The retailers were informed of this discrepancy. The barrels all arrived in pallets of ten barrels, each with one substandard barrel but which one? Halle Burton, the CEO of one of the largest oil distributors in the mideast had just taken delivery of a large order and needed to identify the substandard barrels quickly. She soon found a way to do this using a pair of scales (not pan balances) which required only one weighing on each scale for each batch of ten barrels. She had at her disposal two empty tin cans, one of which held exactly 1 gallon, and the other of which held exactly 1 cup. How did she do it? Note that she did not need to know what a barrel of oil should weigh. Solution: Halle Burton puts the following on scale #1: one cup of oil from barrel 1, two cups of oil from barrel 2, three cups of oil from barrel 3, four cups of oil from barrel 4, five cups of oil from barrel 5. After noting the weight on scale #1, she puts the following on scale #2: one cup of oil from barrel 6, two cups of oil from barrel 7, three cups of oil from barrel 8, four cups of oil from barrel 9, five cups of oil from barrel 10. Since a barrel of oil contains 55 gallons, there are 16 cups to a gallon, and a barrel of the low-grade oil weighs 1 kilogram more than a gallon of the high-grade oil, then 1 cup of the low-grade oil weighs W = 1 kilogram/barrel 1/55 barrels/gallon 1/16 gallons/cup = kilograms/cup more than a cup of the high-grade oil. She notes the weights on both scales, and this tells her which barrel has the substandard oil. For example if the difference in weight on scale #2 and scale #1 is 3 W, then barrel #8 contains the substandard oil, while if the difference in weight on scale #1 and scale #2 is 3 W, then barrel #3 contains the substandard oil.

3 Problem 3. Let x, y, and z be positive integers such that z 2 = x 2 + y 2, where x, y, and z have no common divisors. Show that exactly one of x, y, or z is divisible by 5. Hint: First show that if an integer x is not divisible by 5, then x 2 must be congruent to 1 or 4 modulo 5. Solution: If the positive integers x, y, and z have no common divisors, then at most one of them can be divisible by 5. First note that if x is not divisible by 5, then it must be congruent to 1, 2, 3, or 4 modulo 5, and if x = 5k + 1, then x 2 = 25k k + 1, and x 2 is congruent to 1 modulo 5 if x = 5k + 2, then x 2 = 25k k + 4, and x 2 is congruent to 4 modulo 5 if x = 5k + 3, then x 2 = 25k k , and x 2 is congruent to 4 modulo 5 if x = 5k + 4, then x 2 = 25k k and x 2 is congruent to 1 modulo 5 Therefore, if neither x nor y is divisible by 5, then x 2 and y 2 must be congruent to 1 or 4 modulo 5. This means that x 2 + y 2 must be congruent to one of (mod 5) (mod 5) (mod 5) (mod 5) However, as we saw above, a perfect square cannot be congruent to 2 or 3 modulo 5, and since x 2 + y 2 = z 2 is a perfect square, we cannot have z 2 2 (mod 5) or z 2 3 (mod 5). Therefore z 2 0 (mod 5). Also, from above, if z 2 is divisible by 5, this can only happen if z is divisible by 5, that is, z 0 (mod 5), and therefore, if neither x nor y is divisible by 5, then z must be divisible by 5. Problem 4. Find two counterfeits among nine coins using the smallest number of weighings. The seven genuine coins all weigh the same, and the counterfeits are slightly heavier but both are the same weight. You have a three-pan balance at your disposal, but no other weights except for the coins themselves. An adaptive solution is requested. In a three-pan balance, the heaviest pan is lowest, the lightest pan is highest, and pans of the same weight are at the same level. Solution: Put 2 coins in each of the three pans and leave three aside. case 1: all pans balance: then the counterfeits are among the three left aside and one more weighing suffices. Case 2: one pan is light, the other two are both heavy. Then there are 4 coins containing the two fakes, and one more weighing suffices. (put one of the 4 in each pan, leaving one aside). case 3: one pan is heavy. Then either there are two fakes in the heavy pan, or one in the heavy pan, and one among the three left aside. For the second weighing, put each of the coins from the heavy pan into separate pans, and one of the other coins (from the three left aside) into the remaining pan. This identifies at least one of the fakes, and if necessary, a third weighing will find the second fake.

4 Note (as one of the students pointed out) if you know it takes at least 3 weighings, then a nonadaptive solution is trivial: divide the nine coins into three groups of three, and compare each group of three coins. Here is a proof that three weighings are needed: With one weighing, you can find 2 fakes among 4 coins at most. Therefore, if after one weighing, all you can guarantee is that the two fakes are among 5 or more coins, then you will require at least 2 more weighings. With nine coins and a 3 pan balance, you can partition the set of coins into 1,1,1,6, or 2,2,2,3 or 3,3,3. In all cases there is a worst-case scenario that leaves you knowing that the fakes are among 5 or 6 coins. Problem 5. Show that among any n + 1 positive integers a 1, a 2,..., a n+1, not exceeding 2n, there must be an integer that divides one of the other integers. Hint: Any positive integer m can be written uniquely as a power of two times an odd integer, that is, m = 2 k q, where k is a nonnegative integer and q is an odd integer. Now write a j = 2 kj q j, for i = 1, 2,..., n, n + 1, then each q j is an odd integer less than 2n. Think pigeons!. Solution: For j = 1, 2,..., n + 1, we have a j = 2 kj q j, where each q j is an odd integer less than 2n; but there are only n distinct odd integers between 1 and 2n, so that two of the q j s must be the same, say q i = q j. But then either a i divides a j, or a j divides a i. Problem 6. Show that none of the integers: is the square of another integer. 11, 111, 1111, 11111,... Hint: First show that all such integers can be represented as N +11 where N = 0, 100, 1100, 11100, ,... and each N is divisible by 4, so that each such integer is congruent to 3 modulo 4. Solution: From the hint. we see that each of the integers is congruent to 3 modulo 4. However, if x is any integer, then from the division algorithm, x must be congruent to 0, 1, 2, or 3 modulo 4, and then if x = 4k, then x 2 = 16k 2 and x 2 is congruent to 0 modulo 4 if x = 4k + 1, then x 2 = 16k 2 + 8k + 1 and x 2 is congruent to 1 modulo 4 if x = 4k + 2, then x 2 = 16k k + 4 and x 2 is congruent to 0 modulo 4 if x = 4k + 3, then x 2 = 16k k and x 2 is congruent to 1 modulo 4 so that the square of an integer can only be congruent to 0 or 1 modulo 4, not 3. Problem 7. Five darts are thrown at a square target measuring 14 inches on a side. Prove that two of them must be at a distance no more than 10 inches apart. Solution: Divide the target into 4 squares, each measuring 7 inches on a side, and throw 5 darts at the target. By the pigeonhole principle, one of the smaller squares must contain at least 2 of the darts, and the distance between these two darts must be less than or equal to = = 98 < 10. Thus, there are at least two darts which are at a distance no more than 10 inches apart.

5 Problem 8. On a two-pan balance, there are 9 small marbles on one pan and 2 large marbles on the other pan. You have an unlimited number of marbles of both sizes available. Each of the large marbles weighs 4 grams and each of the small marbles weighs 3 grams. What is the minimum number of marbles, and their weights, that must be added or removed to make the pans balance. Solution: Add 4 large marbles to the pan with the large marbles, and remove one small marble from the pan with the small marbles. Problem 9. Write the digits from 1 to 9 (in that order). Place 3 of the basic arithmetic operation symbols +,,, between the digits so that the result is 100. The order of the digits cannot be changed. The operations may be repeated but the total number of symbols must be exactly 3. For example, = Solution: One answer is = 100. Problem 10. There is only one way to open the safe below. 4D 4D 1L 3L Open 2R 1D 1U 2L 4L 4R 1L 2D 1U 2L 4R 2R 2L 1D 2U 4R 1U 1U 4U 4U You must press each button exactly once in the correct order in order to reach Open. Each button is marked with a direction: U is up, L is left, R is right, and D is down. The number of spaces to move is also marked on each button. Which button is the first one you must press? Solution: The first button you must push is 1 U in row 3, column 4.