advanced-solutions - SMT 2013 Advanced Topics Test...

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SMT 2013 Advanced Topics Test Solutions February 2, 2013 1. How many positive three-digit integers a b c can represent a valid date in 2013, where either a corresponds to a month and b c corresponds to the day in that month, or a b corresponds to a month and c corresponds to the day? For example, 202 is a valid representation for February 2nd, and 121 could represent either January 21st or December 1st. Answer: 273 Solution: The integers which are valid have a 1-1 correspondence to days in the first 9 months – this is straightforward to see for all positive integers that do not have a 1 in the hundreds place and just requires careful inspection of the case where 1 is in the hundreds place. There are 365 - 31 - 30 - 31 = 273 such days. 2. Consider the numbers { 24 , 27 , 55 , 64 , x } . Given that the mean of these five numbers is prime and the median is a multiple of 3, compute the sum of all possible positive integral values of x . Answer: 60 Solution: The restriction on the median means either x 27 or 3 | x and x < 55. Hence, the sum of all five numbers is 24 + 27 + 55 + 64 + x = 170 + x < 225, so the average is > 170 5 = 34 and < 225 5 = 45. The only prime numbers in this range are 37, 41, and 43, which yield x = 15, x = 35, or x = 45. 35 is greater than 27 but not a multiple of 3, so it doesn’t work. Hence, the answer is 15 + 45 = 60 . 3. Nick has a terrible sleep schedule. He randomly picks a time between 4 AM and 6 AM to fall asleep, and wakes up at a random time between 11 AM and 1 PM of the same day. What is the probability that Nick gets between 6 and 7 hours of sleep? Answer: 3 8 Solution: Consider the rectangle with lower-left corner at (4 , 11) and upper-right corner at (6 , 13). We want to compute the probability that a randomly generated point inside the rectangle falls above the line y = x + 6 and below the line y = x + 7. These lines cut out a trapezoid of area 3 2 , so therefore the probability that a randomly generated point in the rectangle falls in this trapezoid is 3 2 4 = 3 8 . 4. Given the digits 1 through 7, one can form 7! = 5040 numbers by forming different permutations of the 7 digits (for example, 1234567 and 6321475 are two such permutations). If the 5040 numbers obtained are then placed in ascending order, what is the 2013 th number? Answer: 3657214 Solution: When the numbers are ordered, the first 6! = 720 numbers all have 1 in the millions place value. The next 720 numbers all have 2 in the millions place value. The 2013 th number must then lie in the next batch, with 3 as the millions place value digit. Within the third batch of 720 numbers, the first 5! = 120 have a 1 in the hundred thousands place value, the next 120 have a 2 in the hundred thousands place, the next 120 have a 4, and so on. Continuing in the same manner, we can deduce that the 2013 th number is 3657214 . 5. An unfair coin lands heads with probability 1 17 and tails with probability 16 17 . Matt flips the coin repeatedly until he flips at least one head and at least one tail. What is the expected number of times that Matt flips the coin?
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  • Spring '16
  • WU
  • Calculus, Integers, Problem, Prime number, Nick, small candles, Topics Test Solutions

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