Twenty-fifth Annual UNC Math Contest First Round Fall, 2016 Rules: 90 minutes; no electronic devices. The positive integers are 1, 2, 3, 4,...

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1 Twenty-fifth Annual UNC Math Contest First Round Fall, 01 Rules: 90 minutes; no electronic devices. The positive integers are 1,,,, A snail crawls on a vertical rock face that is 5 feet high. The snail climbs up feet in a day and then rests through the night. Each night it slides down feet while it rests. If it starts at the bottom on the morning of September 1, on what day of the month does it first reach the top of the rock face?!! 0! 1/! 1/! /! 1/! 1/! /! 1!!!. (a) A tape measure s short, medium, and long marks indicate quarter-inches, half-inches, and whole inches, respectively. How many short marks fall between the 1-inch mark and the 7-inch mark? (b) How many short marks fall between the 11-inch mark and the 1-inch mark? 0º 0º What is the perimeter (that is, the sum of the lengths of the sides) of the figure? The lengths of the lower sides are 17, 1,, 5, and 11. The two acute angles at the top each measure 0 degrees. There are four right angled corners in the figure, indicated by the small squares A hog trading team sells two hogs for $10 each. They sell one of the hogs for 15% of the price they paid for the hog. They sell the other hog for 80% of the price they paid for it. Do they make a profit or a loss overall, and how much, in dollars, is that profit or loss? 5. A bo of 500 balls contains balls numbered 1,,, in each of five different colors. Without ever looking at any of the balls, you are to choose balls at random from the bo and put them into a bag. If you must be sure that when you finish, the bag contains at least one set of five balls with identical numbers, then what is the smallest number of balls that you can put in the bag? TURN PAGE OVER

2 . Tutu, Jada, and Faith eat lunch together. Tutu contributes 9 sausages, Jada contributes 8 sausages, and the three girls divide the sausages equally. Faith has brought no food, but gives the other two girls 17 wupiupi coins in echange for her share of the sausages. How many of the coins should Tutu get? 7. What is the first time after midnight at which the hour hand and minute hand on an ordinary clock face are perpendicular to one another? Epress the time in the format Hour, Minute, Second, with your answer rounded to the nearest second. Assume the clock is a 1 hour clock with hands that move at uniform speeds. 8. How many integers greater than 0 and less than 100, 000 are palindromes? An integer is a palindrome if its digits are the same when read left to right and right to left. For instance, 11 and 5 are palindromes; so are 1001, 99, 5, and 1. Reminder: do not count the number A polynomial P() satisfies the equation P(P() 1) = What is P()? (The epression P(P() 1) on the left side of the equation means "plug P() 1 into P." The parentheses in this case do not indicate multiplication.) 10. How many different case-sensitive passwords can be created with at most 8 keystrokes, if each keystroke may touch either the "caps lock" key or any of the 10 alphabetic keys on the top row of the keyboard: Q W E R T Y U I O P? Assume that password entry always begins with "caps lock" in lowercase mode, and assume that a password must contain at least one letter. Tapping the "caps lock" key toggles the mode of the keyboard between lowercase and uppercase. Assume that holding a key down does not produce multiple copies of a letter; that is, in the password field, holding a key down has no effect. END OF CONTEST

3 University of Northern Colorado Mathematics Contest Problems and Solutions of First Round 1. A snail crawls on a vertical rock face that is 5 feet high. The snail climbs up feet in a day and then rests through the night. Each night it slides down feet while it rests. If it starts at the bottom on the morning of September 1, on what day of the month does it first reach the top of the rock face? Answer: September At the end of the day of September 1, it will be at feet away from the bottom. In the morning of September, it will be at 1 foot away from the bottom. At the end of the day of September, it will be at feet away from the bottom. In the morning of September, it will be at feet away from the bottom. At the end of the day of September, it will be at 5 feet away from the bottom. The snail is at the top. The answer is September.. (a) A tape measure s short, medium, and long marks indicate quarter-inches, half-inches, and whole inches, respectively. How many short marks fall between the 1-inch mark and the 7-inch mark? (b) How many short marks fall between the 11-inch mark and the 1-inch mark? Answer: (a) 1; (b) 80 In one whole inch there are two short marks. (a) From the 1-in mark to the 7-inch mark there are 7 1 whole inches. The answer is 1. (b) From the 11-in mark to the 1-inch mark there are whole inches. The answer is Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

4 . What is the perimeter (that is, the sum of the lengths of the sides) of the figure? The lengths of the lower sides are 17, 1,, 5, and 11. The two acute angles at the top each measure 0 degrees. There are four right angled corners in the figure, indicated by the small squares. Answer: 8 Connect the two top points. Since 17 11, the top dotted line is parallel to the bottom line. The length of the top dotted segment is The triangle is equilateral with the given two angles of 0. Therefore, the perimeter of the original polygon is A hog trading team sells two hogs for $10 each. They sell one of the hogs for 15% of the price they paid for the hog. They sell the other hog for 80% of the price they paid for it. Do they make a profit or a loss overall, and how much, in dollars, is that profit or loss? Answer: loss of $. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

5 10 10 The team paid dollars for the two hogs They made 10 0 dollars by selling the two hogs. They lost $. 5. A bo of 500 balls contains balls numbered 1,,,, 100 in each of five different colors. Without ever looking at any of the balls, you are to choose balls at random from the bo and put them into a bag. If you must be sure that when you finish, the bag contains at least one set of five balls with identical numbers, then what is the smallest number of balls that you can put in the bag? Answer: 01 In the worst case we don t have 5 balls with identical numbers if we already have balls with balls of each number. However, we will have 5 balls of some number if we take one more ball. The answer is Practice Problem: A bo of 500 balls contains balls numbered 1,,,, 100 in each of five different colors. Without ever looking at any of the balls, you are to choose balls at random from the bo and put them into a bag. If you must be sure that when you finish, the bag contains at least one set of five balls of the same color, then what is the smallest number of balls that you can put in the bag? Answer: 1 In the worst case we don t have 5 balls of the same color if we already have 5 0 balls with balls of each color. However, we will have 5 balls of some color if we take one more ball. The answer is Tutu, Jada, and Faith eat lunch together. Tutu contributes 9 sausages, Jada contributes 8 sausages, and the three girls divide the sausages equally. Faith has brought no food, but gives the other two girls 17 wupiupi coins in echange for her share of the sausages. How many of the coins should Tutu get? Answer: 10 wupiupi coins The total number of sausages is Each gets sausages. So Tutu gives 9 to 17 7 Faith, and Jada gives 8 to Faith. They should distribute Faith s 17 wupiupi coins in this Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 5

6 10 7 ratio: : 10 : 7. Therefore, Tutu gets 10 wupiupi coins. 7. What is the first time after midnight at which the hour hand and minute hand on an ordinary clock face are perpendicular to one another? Epress the time in the format Hour, Minute, Second, with your answer rounded to the nearest second. Assume the clock is a 1 hour clock with hands that move at uniform speeds. Answer: 1:1: Let O be the center of the clock. OA is the ray pointing to 1 o clock. OB is the ray on which the hour hand lies, and OC is the ray on which the minute hand lies. Let the desired time be minutes past 1 o clock. Note that minutes = 0 hours. From 1 o clock the minute hand moves for minutes. The minute hand moves per minute. So AOC in degrees. From 1 o clock the hour hand moves for hours. The hour hand moves 0 per hour. 0 So AOB 0 in degrees. 0 We have the following equation 180 Solving for we obtain Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

7 180 minutes 1 minutes and seconds rounding to the nearest second. 11 The desired time is 1:1:. Practice Problem 1: Between 1 o clock and o clock the hour hand and minute hand coincide as shown. What is the eact time? X XI XII I II IX III VIII VII VI V IV Answer: 5 5 minutes past one 11 Let the two hands coincide at minutes past one. Note that minutes = 0 hours. X XI A XII I II B IX III VIII VII VI V IV From 1 o clock the minute hand moves for minutes. The minute hand moves per minute. So AOB in degrees. From 1 o clock the hour hand moves for 1 hours. The hour hand moves 0 per hour. 0 So AOB in degrees. 0 We have the equation Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 7

8 Solving for we obtain The hour hand and minute hand coincide at Practice Problem : minutes past one. 11 Between o clock and 5 o clock there are two moments such that the hour hand and minute hand intersect at a right angle. Find the eact times for the two moments. Answer: 5 5 minutes past four and 8 minutes past four Let the time be minutes past four. After several minutes past four there is a moment when the hour hand and minute hand intersect at a right angle: X XI A XII I II C IX III VIII VII VI V IV B Look at the hour hand: AOB 0 10 in degrees. 0 Look at the minute hand: AOC in degrees. With AOB 90 AOC in degrees we have Solving for we obtain The time is 5 5 minutes past four Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 8

9 After some minutes past four and a half there is a moment when the hour hand and minute hand intersect at a right angle: A X XI XII I II IX III Look at the hour hand: AOB 0 10 in degrees. 0 Look at the minute hand: AOC in degrees. Now AOC AOB 90 in degrees we have Solving for we obtain The time is C 8 minutes past four. 11 VIII VII VI V IV B 8. How many integers greater than 0 and less than 100, 000 are palindromes? An integer is a palindrome if its digits are the same when read left to right and right to left. For instance, 11 and 5 are palindromes; so are 1001, 99, 5, and 1. Reminder: do not count the number 0. Answer: 1098 There are 9 one-digit palindromes: 1,,, 9. For two-digit palindromes, the units digit is determined by the tens digit. There are 9 choices (1,,, 9) for the tens digit. So there are 9 two-digit palindromes: 11,,, 99. For three-digit palindromes, the units digit is determined by the hundreds digit. There are 9 choices (1,,, 9) for the hundreds digit, and there are 10 choices (0, 1,, 9) for the tens digit. So there are three-digit palindromes. For four-digit palindromes, the units digit is determined by the thousands digit, and the tens digit is determined by the hundreds digit. There are 9 choices (1,,, 9) for the thousands digit, and Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 9

10 there are 10 choices (0, 1,, 9) for the hundreds digit So there are palindromes. Similarly, there are five-digit palindromes. Therefore, the number of palindromes greater than 0 and less than 100,000 is four-digit 9. A polynomial P What is Answer: 17 satisfies the equation P? (The epression PP P1 into P Rewrite the given equation: Let Q P1 P 1 P on the left side of the equation means plug. The parentheses in this case do not indicate multiplication.) P 1 P 1 1., which is also a polynomial. We have Q 1 Q Now we can easily see Q (see the appendi for the proof). Then Q1 1 Therefore, 1 17 Appendi: If P. 1 Q is a polynomial and QQ, then Proof: Let n be the degree of n n Then Q Q a a. Q. Prove it. P. n Qwhere n is a nonnegative integer. That is, a n 1 n Q. In the right side the leading term is a. We must have n 1 and a 1. Since n is a non-negative integer, n. And then a 1. Let Q b c d e We claim that b 0. Let Q b R where R c d e We have 1 Q Q b R.. Then n1 b b R R b R. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 10

11 15 0 b. So 0 b. Then Q c d e Similarly, c d e 0. Therefore, Q How many different case-sensitive passwords can be created with at most 8 keystrokes, if each keystroke may touch either the caps lock key or any of the 10 alphabetic keys on the top row of the keyboard: Q W E R T Y U I O P? Assume that password entry always begins with caps lock in lowercase mode, and assume that a password must contain at least one letter. Tapping the caps lock key toggles the mode of the keyboard between lowercase and uppercase. Assume that holding a key down does not produce multiple copies of a letter; that is, in the password field, holding a key down has no effect. Answer: 0780 My counting is based on the understanding that the series of caps lock Q, Q, Q, caps lock with 5 keystrokes produces the same password as the series of caps lock, caps lock, caps lock, Q, Q, Q with keystrokes as an eample. Case 1: passwords of length k ( k 1,,, ) Any of the k letters in the passwords can be any of the 10 letters in both cases. So there are passwords of length k ( k 1,,, ). 1 In total there are possible passwords. Case : passwords of length 5 There are 5 letters in the passwords. We may have no press of caps lock. Before any of the five letters we may insert one caps lock. We may choose two letters before each of which we insert one caps lock. We may choose three letters before each of which we insert one caps lock The number of the possible passwords in this case is Case : passwords of length Similarly, the number of the possible passwords in this case is Case : passwords of length 7 7 The number of the possible passwords in this case is Case 5: passwords of length 8 The number of the possible passwords in this case is The total number of passwords which may be created is k 0 Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 11

12 Practice Problem: How many different case-sensitive passwords can be created with eactly 8 keystrokes, if each keystroke may touch either the caps lock key or any of the 10 alphabetic keys on the top row of the keyboard: Q W E R T Y U I O P? Assume that password entry always begins with "caps lock" in lowercase mode, and assume that a password must contain at least one letter and a password must begin with a capital letter. Tapping the caps lock key toggles the mode of the keyboard between lowercase and uppercase. Assume that holding a key down does not produce multiple copies of a letter; that is, in the password field, holding a key down has no effect. The order of caps lock matters. For eample, the series of 8 presses: caps lock, caps lock, caps lock, Q, Q, Q, caps lock, caps lock produces a different password as a series of caps lock, Q, Q, Q, caps lock, caps lock, caps lock, caps lock. Answer: 1780 The first press (keystroke) must be the caps lock. The second press has 11 choices: any of 10 letters and the caps lock. For any second press with a letter, the following keystrokes can be any of 11 possible presses. So there are 1011 passwords. If the second press is the caps lock, it comes to the lowercase mode. To produce a password, the third press must be the caps lock. Then the fourth press has 11 choices: any of 10 letters and the caps lock. For any fourth press with a letter, the following keystrokes can be any of 11 possible presses. So there are 1011 passwords. Continue this pattern. The total number of passwords is Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 1

13 Twenty-fifth Annual UNC Math Contest Final Round January 1, 017 Rules: Three hours; no electronic devices. The positive integers are 1,,,,... A prime is an integer strictly greater than one that is evenly divisible by no integers other than itself and 1. The primes are,, 5, 7, 11, 1, 17, A circle has radius, a second circle has radius 15, and the centers of the two circles are 5 units apart. A line tangent to both circles crosses the line connecting the two centers at a point P between the two centers. How much farther is P from the center of the bigger circle than it is from the center of the smaller circle?. Find the ratio of the area of a regular heagon circumscribed around a circle to the area of a regular heagon inscribed inside the same circle. (A polygon is called regular if all its sides are the same length and all its corner angles have the same measure. A heagon is a polygon with si sides.). Prime mates Find the largest 9 digit integer in which no two digits are the same and the sum of each pair of adjacent digits is prime. That is, the sum of the first two digits is prime, the sum of the second and third digits is prime, the sum of the third and fourth digits is prime, and so on.. Monkey business Harold writes an integer; its right-most digit is. When Curious George moves that digit to the far left, the new number is four times the integer that Harold wrote. What is the smallest possible positive integer that Harold could have written? 5. Double encryption (a) Find a substitution code on the seven letters A, B, C, D, E, F, and G that has the property that if you apply it twice in a row (that is, encrypt the encryption), the message ABCDEFG becomes ECBFAGD. Describe your answer by giving the message that results when encryption is applied once to the message ABCDEFG. (b) Find another such code, if there is one.. The spider 0 s divider On a regular pentagon, a spider forms segments that connect one endpoint of each side to n different non-verte points on the side adjacent to the other endpoint of that side, going around clockwise, as shown. Into how many non-overlapping regions do the segments divide the pentagon? Your answer should be a formula involving n. (In the diagram, n = and the pentagon is divided into 1 regions.) TURN PAGE OVER

14 7. A bo of 8 balls contains balls numbered 1,,,..., 1 in each of four different colors. Without ever looking at any of the balls, you choose balls at random from the bo and put them in a bag. (a) If you must be sure that when you finish, the bag contains at least one set of five balls whose numbers are consecutive, then what is the smallest number of balls you can put in the bag? (For eample, a set of balls, in any combination of colors, with numbers,, 5,, and 7 is a set of five whose numbers are consecutive.) (b) If instead you must be sure that the bag contains at least one set of five balls all in the same color and with consecutive numbers, then what is the smallest number of balls you can put in the bag? Remember to justify answers for maimum credit. 8. For what integer n does + n divide into 1 1 with no remainder? That is, for what integer n is the first polynomial a factor of the second one? As always, justify your answer. 9. Suppose n points on the circumference of a circle are joined by straight line segments in all possible ways and that no point that is not one of the original n points is contained in more than two of the segments. How many triangles are formed by the segments? Count all triangles whose sides lie along the segments, including triangles that overlap with other triangles. For eample, for n = there is one triangle and for n = (shown in the diagram) there are 8 triangles. 10. Powerless progressions Find an infinite sequence of integers a 1, a, a,... that has all of these properties: (1) a n = c + dn with c and d the same for all n = 1,,,... () c and d are positive integers, and () no number in the sequence is the r th power of any integer, for any power r =,,,... Reminder: Justify answers. In particular, for maimum credit, make it clear in your presentation that your sequence possesses the third property. 11. Divide and conquer (a) How many different factorizations are there of 09 (which is the twelfth power of ) in which each factor is either a square or a cube (or both) of an integer and each factor is greater than one? Regard 8 8 and 8 8 as the same factorization: the order in which the factors are written does not matter. Regard the number itself, 09, as one of the factorizations. (b) How many different factorizations are there of,5 as a product of factors in which each factor is either a square or a cube (or both) of an integer and each factor is greater than one? As before, the order in which the factors is written does not matter, and the number itself counts as a factorization. Note that, 5 =. END OF CONTEST

15 University of Northern Colorado Mathematics Contest Solutions of Final Round 1. A circle has radius, a second circle has radius 15, and the centers of the two circles are 5 units apart. A line tangent to both circles crosses the line connecting the two centers at a point P between the two centers. How much farther is P from the center of the bigger circle than it is from the center of the smaller circle? Answer: 1 Draw the diagram. Let O1 and O be the centers of the bigger circle and the smaller circle respectively. Let A and B be the tangent points of the tangent line to the two circles respectively. Draw O 1 A and O B. Then O1 A AB and O B AB. A O 1 O P B Obviously, O 1 AP ~ O BP. We have O1P O1 A. That is, O P O B O P O P Note that O 1P OP 5. So O1P 5 and O P The answer is Find the ratio of the area of a regular heagon circumscribed around a circle to the area of a regular heagon inscribed inside the same circle. (A polygon is called regular if all its sides are the same length and all its corner angles have the same measure. A heagon is a polygon with si sides.) Answer: Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 1

16 Solution 1: Rotate the inscribed triangle clockwise by 0. Let us shade the 1 of the whole shape as shown. We see that the answer is. Solution : Let O be the center of the circle. A B C E O Let A be a verte of the larger heagon and B be a verte of the smaller heagon. We may rotate one heagon such that A, B, and O are in a line. Let AE be a side of the larger heagon. Draw OC AE with C on AE. Obviously, C is a tangent point. So ACO is a triangle. OA OA We have. Since OB OC,. OC OB Note that regular heagons are naturally similar. The area ratio of the larger heagon to the smaller is the square of the ratio of the corresponding lengths: OA OB. The answer is. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

17 . Prime mates Find the largest 9-digit integer in which no two digits are the same and the sum of each pair of adjacent digits is prime. That is, the sum of the first two digits is prime, the sum of the second and third digits is prime, the sum of the third and fourth digits is prime, and so on. Answer: We start from It is good that is a prime, but is not. We have to replace at least one of 8 and 7. Since we want the number to be greatest, we replace 7 with some digit. We can replace 7 only with an odd digit. The net greatest odd digit is 5. Let us switch 7 and 5. We have Now the number satisfies all conditions. The answer is Monkey business Harold writes an integer; its right-most digit is. When Curious George moves that digit to the far left, the new number is four times the integer that Harold wrote. What is the smallest possible positive integer that Harold could have written? Answer: 105 Solution 1: Let N be an n-digit number. Let Harold s number be N formed by placing at the right to N. The new number after George s change is N. We have We have n We must have 10 n 10 N 10N. 10 N. 9 n N N. That is, n19's to be divisible by 9. That is, 99 9 is divisible by 9. Do the division. It is easy to see that 9999 is divisible by and it is the smallest possible. With , the smallest value for N is Therefore, Harold s smallest possible number is 105. Solution : Let Harold s number be number is a n a a. 1 aa1 a n where a n,, a, a 1 are digits with a n 1. Then the new Note that an aa1 an a 1 a. So a. 1 Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

18 We see an a an a. We see a 5. We have an a5 an 5 a. Then a an a5 an 5 a a 0 an a505 an 505 a a5 1 an a105 an a105. Now we must have a. Since a matches the leading digit of the new number, we stop here to obtain the smallest possible number. The multiplication is Therefore, Harold s smallest possible number is Double encryption (a) Find a substitution code on the seven letters A, B, C, D, E, F, and G that has the property that if you apply it twice in a row (that is, encrypt the encryption), the message ABCDEFG becomes ECBFAGD. Describe your answer by giving the message that results when encryption is applied once to the message ABCDEFG. (b) Find another such code, if there is one. Answer: BEAGCDF and CAEGBDF We use 1,, to replace A, B,. After two operations we see The rule for the combined two operations is: Look at the group with ,, 7. Since there are only three numbers, it is easy to see For the group with 1-5 and the group with - we have two choices: after 1 st operation after nd operation after operations after 1 st operation after nd operation or after 1 st operation after nd operation Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

19 We have two solutions: after one operation and after one operation Make them in the right order: after one operation and after one operation Translate 1,, back to A, B,. The two solutions are A B C D E F G after one operation B E A G C D F and A B C D E F G after one operation C A E G B D F We have an elegant way to decode if there are more numbers. For this eample, look at the group with after 1 st operation??? after nd operation 7 The key observation is that the sequence goes back to --7 after operations. 7 after 1 st operation??? after nd operation 7 after rd operation 7 We find the one-operation rule from the result of the nd operation to the result of the rd operation 7. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 5

20 . The spider s divider On a regular pentagon, a spider forms segments that connect one endpoint of each side to n different non-verte points on the side adjacent to the other endpoint of that side, going around clockwise, as shown. Into how many non-overlapping regions do the segments divide the pentagon? Your answer should be a formula involving n. (In the diagram, n and the pentagon is divided into 1 regions.) Answer: 5n 5n 1 At the beginning we have one region the pentagon. From every verte draw the 1 st line. It will increase the number of regions by 10. By drawing the second line from every verte, the number of regions will be increased by 0. By drawing the third line from every verte, the number of regions will be increased by 0. In general, by drawing the n th line from every verte, the number of regions will be increased by 10n. So the number of regions after n lines are drawn from every verte is n 1 5nn 1 5n 5n 1. How do we see that the number of regions is increased by 10n when we draw the n th line from every verte? segments between 5 points segments between 5 points We generate n n n ( + = 8 in the diagram) segments in each newly drawn line when we draw the n th line ( th line in the diagram) from every verte. Each segment divides a region into two regions making one more region. So the total number of more regions produced by drawing the n th line is n 5 10n. 7. A bo of 8 balls contains balls numbered 1,,,..., 1 in each of four different colors. Without ever looking at any of the balls, you choose balls at random from the bo and put them in a bag. (a) If you must be sure that when you finish, the bag contains at least one set of five balls whose numbers are consecutive, then what is the smallest number of balls you can put in the bag? (For eample, a set of balls, in any combination of colors, with numbers,, 5,, and 7 is a set of five whose numbers are consecutive.) Problems are duplicated and solved by Ming Song (msongmath@yahoo.com)

21 (b) If instead you must be sure that the bag contains at least one set of five balls all in the same color and with consecutive numbers, then what is the smallest number of balls you can put in the bag? Remember to justify answers for maimum credit. Answer: (a) 1; (b) 1 The solution is for both (a) and (b). You can put all the balls ecept the balls with numbers 5 and 10 in your bag and you will have 0 balls with no runs of five consecutive numbers. Therefore, the answer is larger than 0. Now we show that 1 is enough. Suppose that you have 1 balls. At least in one color you have 11 or more balls by the pigeonhole principle. In that color, you will have 5 balls with five consecutive numbers. The answer is 1 for both For what integer n does n divide into 1 with no remainder? That is, for what integer n is the first polynomial a factor of the second one? As always, justify your answer. Answer: 1 Solution 1: Note that 1 and are the Fibonacci numbers, and note that 1 has the golden ratio 1 5 as a root. We know how the Fibonacci sequence is related to the golden ratio. We guess n 1. By knowing the answer, we can easily achieve the factorization: The answer is 1. Solution : 1 Let 0. We have n 1. Let 1. We have n 7. The greatest common factor of 1 and 7 is 8. Then n 8. So n can be 1,,, 8. Let 1. We have n 88. It leaves, 1, for n. Let. We have n 758. Now we have 1, for n. Let. We have n Only 1 The answer is 1. is possible for n Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 7

22 9. Suppose n points on the circumference of a circle are joined by straight line segments in all possible ways and that no point that is not one of the original n points is contained in more than two of the segments. How many triangles are formed by the segments? Count all triangles whose sides lie along the segments, including triangles that overlap with other triangles. For eample, for n there is one triangle and for n (shown in the diagram) there are 8 triangles. n n n n Answer: 5 5 Solution to (a): There are four kinds of triangles. Kind 1: all three vertices of a triangle are inside the circle. I shade several of this kind in the following figure. One triangle of kind 1 is determined by points on the circle, and points determine only one n triangle. So there are triangles of this kind. Kind : eactly one verte of a triangle is on the circle. I shade several of this kind as shown. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 8

23 Five points are involved in one triangle of this kind. However, when we choose 5 points, we can make 5 triangles of this kind. So there are 5 n triangles of this kind. 5 Kind : eactly two vertices of a triangle are on the circle. I shade several of this kind: Four points are involved in one triangle of this kind. However, when we choose points, we can make triangles of this kind. So there are n triangles of this kind. Kind : all three vertices of a triangle are on the circle. I shade several of this kind: Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 9

24 n Obviously there are triangles of this kind. The total number of triangles is n n n n Powerless progressions Find an infinite sequence of integers a1, a, a, that has all of these properties: (1) c dn with c and d the same for all n 1,, a n () c and d are positive integers, and () no number in the sequence is the r th power of any integer, for any power r,,. Reminder: Justify answers. In particular, for maimum credit, make it clear in your presentation that your sequence possesses the third property. Answer:,,10, (many possible different sequences) To avoid the powers of odd numbers, we will build a sequence with even number. Let k be an even number. Its r th r r r power k k ( r ) is a multiple of. So to avoid the powers of even numbers, we build a sequence without a multiple of. Then let c and d. We have a sequence,,10, satisfying all conditions. 11. Divide and conquer (a) How many different factorizations are there of 09 (which is the twelfth power of ) in which each factor is either a square or a cube (or both) of an integer and each factor is greater than one? Regard 88 and 8 8 as the same factorization: the order in which the factors are written does not matter. Regard the number itself, 09, as one of the factorizations. (b) How many different factorizations are there of,5 as a product of factors in which each factor is either a square or a cube (or both) of an integer and each factor is greater than one? As before, the order in which the factors is written does not matter, and the number itself counts as a factorization. Note that,5. Answer: (a) 1; (b) Solution to (a): We need to find the number of ways to partition 1 into a non-decreasing sequence consisting of multiples of s or s with one or more terms. Let us list the ways systematically according to the number of terms: Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 10

25 There are 1 ways. Solution to (b): There are ways to epress and there are ways to epress as a product of one or more squares and/or cubes: as a product of one or more squares and/or cubes: So there are 1 ways to put an epression from the first list and an epression from the second list together without combing any terms. Now we take a look at how many ways there are to combine the terms for each possible combination of the two epressions. (1) (a) There are three ways to combine terms: Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 11

26 (b) There are three ways to combine terms: (c) There is no way to combine terms. (d) There is one way to combine terms: In this case there are ways.. () (a) There are three ways to combine terms: (b) There are si ways to combine terms: Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 1

27 (c) (d) There is no way to combine terms. There are two ways to combine terms: In this case there are 0 11 ways... () (a) (b) (c) There is no way to combine terms. There is no way to combine terms. There are two ways to combine terms:.. (d) There is one way to combine terms: In this case there are ways.. () (a) There is one way to combine terms:. (b) There are two ways to combine terms: (c) There is one way to combine terms:. Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 1

28 (d) There is one way to combine terms:. In this case there are ways. The total number of ways is Problems are duplicated and solved by Ming Song (msongmath@yahoo.com) 1

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