Feb 2019 (Mathcounts)

I will post seven new problems every weekend, one for each day of the week. The idea is that you work on your own during the week and I will post answers the following Saturday. For those of you who practiced with me last year, the process is similar.

For instructions on submitting weekly answers, click on the “Logistics” tab.
For previously posted problems, click on the “Archive” tab.
For previous prizes, click on the “News” tab.

 


Week of Feb 24 – March 2.

Click GoogleForm to submit answers by Sat, March 2.

Feb 24: A basket contains 5 green marbles and 7 red marbles. A marble is taken from the basket at random; its color is recorded, then the marble is returned to the basket. A second marble is then taken from the basket at random, and its color is recorded. What is the probability that the same color is recorded both times?

Answer: The probability of getting green and green is 5/12 * 5/12; the probability of getting red and red is 7/12*7/12. So the total probability is (25+49)/144 = 37/72.

Feb 25: A basket contains 5 green marbles and 7 red marbles. Two marbles are taken from the basket at random without replacement (i.e., the first marbles is not put back before the second marble is drawn). What is the probability that both marbles are the same color?

Answer: The probability of getting green and green is 5/12 * 4/11; the probability of getting red and red is 7/12 * 6/11. So the total probability is (20+42)/132 = 31/66.

Feb 26: There are 2 senators from each of the states. We wish to make a 3-senator committee in which no two of the members are from the same state. In how many ways
can we do it?

Answer: {50 \choose 3} \cdot 2\cdot 2\cdot 2

Feb 27: 8 chairs are arranged evenly around a round table. In how many ways can 8 people be seated in the chairs, if Al and Ben insist on being seated diametrically opposite each other?

Answer: Al has 8 seats in which he can sit. Once Al is seated, Ben’s seat is fixed. The remaining 6 people have 6! ways to be seated. So the answer is 8*6!.

Feb 28: Now suppose that not only must Al and Ben be diametrically opposite, but Chuck and Dan also demand to be diametrically opposite. How many seatings of the 8 people are possible?

Answer: Al has 8 seats in which he can sit. Once Al is seated, Ben’s seat is fixed. Chuck has 6 seats in which he can sit. Once Chuck is seated, Dan’s seat is fixed. The remaining 4 people have 4! ways to be seated. So the answer is 8*6*4!.

March 1: How many ways are there to put 4 balls into 3 boxes, given that the balls can all be distinguished and so can the boxes? (For instance, perhaps each ball is a different color, and each box is a different color as well.)

Answer: Each ball is 3 boxes to choose from. So the answer is 3*3*3*3. 

March 2: How many ways are there to put 4 balls into 3 boxes, given that the balls are not distinguished and neither are the boxes?

Answer: Case 1: no box is empty. There is only one way to do it: 1 box has 2 balls, and the other two boxes each has 1 ball. Case 2: one box is empty. There are 2 ways to do it: 3 balls in 1 box and 1 ball in the other; or 2 balls in both boxes. Case 3: two boxes are empty. There is one way to do it. So in total 4 ways. 4-0-0, 3-1-0, 2-2-0, 2-1-1.


Week of Feb 17 – Feb 23.

Click GoogleForm to submit answers by Sat, Feb 23.

Feb 10: Rectangle ABCD has AB=6 and BC=3. Point M is chosen on side AB so that ∠AMD =∠CMD. What is the degree measure of ∠AMD?

Answer: Let H be a point on MC such that DH is perpendicular to CM. Triangle CDH and triangle MCB are congruent. Therefore CM=CD=6. ∠BMC=30 and ∠AMD=75.

Feb 11: In trapezoid ABCD, AB and CD are perpendicular to AD,  with AB+CD=BC,  AB<CD, and AD=7. What is AB⋅CD?

Answer: (CD-AB)^2 + 7^2=BC^2 = (AB+CD)^2, which implies AB \cdot CD = 49/4.

Feb 12: In trapezoid ABCD we have AB parallel to DC, E as the midpoint of BC, and F as the midpoint of DA. The area of ABEF is twice the area of FECD. What is the length of AB divided by the length of CD?

Answer: AB+EF = 2(CD+EF) and EF = (AB+CD)/2. Therefore, AB=5CD and AB/CD = 5. 

Feb 13: A regular octagon ABCDEFGH has an area of one square unit. What is the area of the rectangle ABEF?

Answer: 1/2

Feb 14: What is the number of degrees in the smaller angle formed by the hands of a clock at 5:44?

Answer: {16\over 60}\cdot{1\over 12}\cdot 360 + {14\over 60}\cdot 360 = 92.

Feb 15: When the measures of the angles of a triangle are placed in order, the difference between the middle angle and smallest angle is equal to the difference between the middle angle and largest angle. If one of the angles of the triangle has measure 23∘, then what is the measure in degrees of the largest angle of the triangle?

Answer: The middle angle is 60∘, the smallest is 23∘, and the largest is 97∘

Feb 16: Find the area of triangle ABC if AB=BC=12 and ∠ABC = 120∘?

Answer: 36\sqrt{3}

Feb 17: Find the length of the shortest altitude of a triangle with sides of lengths 10, 24, and 26.

Answer: Let h be the height. \sqrt{10^2-h^2} + \sqrt{24^2-h^2} =26. We get h=120/13


Week of Feb 10 – Feb 16.

Click GoogleForm to submit answers by Sat, Feb 16.

Feb 10: On a certain math exam, 10% of the students got 70 points, 25% got 80 points, 20% got 85 points, 15% got 90 points, and the rest got 95 points. What is the difference between the mean and the median score of this exam?

Answer:  The mean is 70*10% + 80*25% + 85*20% + 90*15% + 95*30% = 86. The median is 85. The differece is therefore 1. 

Feb 11: Every high school in the city of Euclid sent a team of 3 students to a math contest. Each participant in the contest received a different score. Andrea’s score was the median among all students, and hers was the highest score on her team. Andrea’s teammates Beth and Carla placed 37th and 64th, respectively. How many schools are in the city?

Answer: Andrea’s ranking is 36th or lower since she scores higher than Beth. Since the number of students is a multiple of 3 and Andrea is the unique median, the number of students is odd and at least 69. So Andrea’s ranking is at least 35. Andrea’s ranking cannot be 36 since the total number of students would be 71. So there are 69 students in total and 23 schools.

Feb 12: In 1995 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2015, with an increase of another 150 people, the population is once again a perfect square. What is the percent growth of the town’s population during this twenty-year period?

Answer: Let the population in 1995 be $\latex x^2$. The population in 2005 is x^2+150 which equals y^2+9 for some integer $\latex y$. The population in 2015 is x^2 +300 which equals z^2 for some integer z. We have )(y-x)(y+x)=141=3*37. This means either x=22, y=25 or x=70, y=71. For the latter, x^2+300 is not a square. Therefore, x=22, y=25, z=28. The percentage growth is approximately 12%.

Feb 13: Find the number of ordered pairs of integers (x, y) that satisfy 1/x+1/y=1/7.

Answer: We can rewrite the expression (x-7)(y-7) = 49. (This trick is called Simon’s Favorite Factoring Trick.) Therefore x-7 = 1, 7, 49, -1, -7, -49 and y-7 = 49, 7, 1, -49, -7, -1, for a total of 6 pairs of solutions. 

Feb 14: All even numbers from 2 to 98 inclusive, except those ending in 0, are multiplied together. What is the rightmost (units digit) of the product?

Answer: When we multiply 2, 4, 6, 8 together 10 times, the unit digit is same as the unit digit of 4^10, which is 6.

Feb 15: A charity sells 140 benefit tickets for a total of $2001. Some tickets sell for full price (a whole dollar amount), and the rest sell for half price. How much money is raised by the full-price tickets?

Answer: Let n be the number of full-priced tickets sold each at x dollars, where x and n are integers. The number of half-priced tickets is 140-n, each sold at at x/2 dollars. We have xn + (140-n)x/2 = (140+n)x/2 = 2001 = 3\cdot 23\cdot 29. This is the same as (140+n)x = 2\cdot 3\cdot 23\cdot 29. Since n<140, we must have x=23, 140+n=6\cdot 29=174. Therefore, the money raised by full-price tickets is 23\cdot 34 = 782.

Feb 16: How many positive cubes divide 3!⋅5!⋅7!?

Answer: 3! \cdot 5! \cdot 7! = 2\cdot 3 \cdot 2\cdot 3 \cdot 2^2 \cdot 5 \cdot 2\cdot 3 \cdot 2^2 \cdot 5 \cdot 2\cdot 3 \cdot 7 = 2^8 \cdot 3^4 \cdot 5^2 \cdot 7. Therefore, it contains 6 cubes:  1, 2^3, 2^6, 3^3, 2^3\cdot 3^3, 2^6\cdot 3^3.


Week of Feb 3 – Feb 9. Happy Lunar New Year!

Click GoogleForm to submit answers by Sat, Feb 9.

Feb 3: Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?

Answer: 2^6 - 2 = 62.

Feb 4: How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?

Answer: The first and last digits are both odd numbers: 5*5; the first and last numbers are both even numbers: 4*5. So the total is 45.

Feb 5: How many line segments have both their endpoints located at the vertices of a given cube?

Answer: {8 \choose 2} = 28.

Feb 6: Three tiles are marked X and two other tiles are marked O. The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX?

Answer: 3! 2! / 5! = 1/10. 

Feb 7: Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?

Answer: Four donuts have the same flavor: 3. Three donuts have the same flavor: 3*2=6. Two donuts have one flavor and the other two have the same flavor: 3. Two donuts have the same flavor and the other two have different flavors: 3. So the total is 3+6+3+3=15. 

Feb 8: How many four-digit positive integers have at least one digit that is a 2 or a 3?

Answer: The number of 4-digit integers is 9*10*10*10. Among those 7*8*8*8 have no digit that is 2 or 3. So the number that has at least one digit that is 2 or 3 is 9*10*10*10-7*8*8*8=5416.

Feb 9: A student must choose a program of four courses from a menu of courses consisting of English, Algebra, Geometry, History, Art, and Latin. This program must contain English and at least one mathematics course. In how many ways can this program be chosen?

Answer: 2*3=6.