amc10b2013problemsandsolutions

1 AMC 10 The American Mathematics Competitions Instructions 1. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. 2. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. 3. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). 4. Figures are not necessarily drawn to scale. 5. You will have 75 minutes working time to complete the test. 2013 AMC 10B Problems Problem 1 What is ? Problem 2 Mr. Green measures his rectangular garden by walking two of the sides and finding that it is steps by steps. Each of Mr. Green's steps is feet long. Mr. Green expects a half a pound of potatoes per square foot from his garden. How many pounds of potatoes does Mr. Green expect from his garden? 2 Problem 3 On a particular January day, the high temperature in Lincoln, Nebraska, was degrees higher than the low temperature, and the average of the high and the low temperatures was . In degrees, what was the low temperature in Lincoln that day? Problem 4 When counting from to , is the number counted. When counting backwards from to , is the number counted. What is ? Problem 5 Positive integers and are each less than . What is the smallest possible value for ? Problem 6 The average age of 33 fifth-graders is 11. The average age of 55 of their parents is 33. What is the average age of all of these parents and fifth-graders? Problem 7 Six points are equally spaced around a circle of radius 1. Three of these points are the vertices of a triangle that is neither equilateral nor isosceles. What is the area of this triangle? 3 Problem 8 Ray's car averages 40 miles per gallon of gasoline, and Tom's car averages 10 miles per gallon of gasoline. Ray and Tom each drive the same number of miles. What is the cars' combined rate of miles per gallon of gasoline? Problem 9 Three positive integers are each greater than , have a product of , and are pairwise relatively prime. What is their sum? Problem 10 A basketball team's players were successful on 50% of their two-point shots and 40% of their three-point shots, which resulted in 54 points. They attempted 50% more two-point shots than three-point shots. How many three-point shots did they attempt? Problem 11 Real numbers and satisfy the equation . What is ? Problem 12 Let be the set of sides and diagonals of a regular pentagon. A pair of elements of are selected at random without replacement. What is the probability that the two chosen segments have the same length? 4 Problem 13 Jo and Blair take turns counting from to one more than the last number said by the other person. Jo starts by saying " ", so Blair follows by saying " " . Jo then says " " , and so on. What is the 53rd number said? Problem 14 Define . Which of the following describes the set of points for which ? Problem 15 A wire is cut into two pieces, one of length and the other of length . The piece of length is bent to form an equilateral triangle, and the piece of length is bent to form a regular hexagon. The triangle and the hexagon have equal area. What is ? Problem 16 In triangle , medians and intersect at , , , and . What is the area of ? Problem 17 Alex has red tokens and blue tokens. There is a booth where Alex can give two red tokens and recieve in return a silver token and a blue token, and another booth where Alex can give three blue tokens and recieve in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end? 5 Problem 18 The number has the property that its units digit is the sum of its other digits, that is . How many integers less than but greater than share this property? Problem 19 The real numbers form an arithmetic sequence with . The quadratic has exactly one root. What is this root? Problem 20 The number is expressed in the form where and are positive integers and is as small as possible. What is ? Problem 21 Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallest possible value of N? 6 Problem 22 The regular octagon has its center at . Each of the vertices and the center are to be associated with one of the digits through , with each digit used once, in such a way that the sums of the numbers on the lines , , , and are all equal. In how many ways can this be done? Problem 23 In triangle , , , and . Distinct points , , and lie on segments , , and , respectively, such that , , and . The length of segment can be written as , where and are relatively prime positive integers. What is ? Problem 24 A positive integer is nice if there is a positive integer with exactly four positive divisors (including and ) such that the sum of the four divisors is equal to . How many numbers in the set are nice? 7 Problem 25 Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer . For example, if , Bernardo writes the numbers and , and LeRoy obtains the sum . For how many choices of are the two rightmost digits of , in order, the same as those of ? AMC 12 The American Mathematics Competitions Instructions 6. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct. 7. You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered if the year is before 2006, 1.5 points for each problem left unanswered if the year is after 2006, and 0 points for each incorrect answer. 8. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor and erasers (and calculators that are accepted for use on the SAT if before 2006. No problems on the test will require the use of a calculator). 9. Figures are not necessarily drawn to scale. 10. You will have 75 minutes working time to complete the test. 8 2013 AMC 10B Answer Key 9 2013 AMC 10B Problems/Problem 16 Problem In triangle , medians and intersect at , , , and . What is the area of ? Solution Solution 1 Let us use mass points: Assign mass . Thus, because is the midpoint of , also has a mass of . Similarly, has a mass of . and each have a mass of because they are between and and and respectively. Note that the mass of is twice the mass of , so AP must be twice as long as . PD has length , so has length and has length . Similarly, is twice and , so and . Now note that triangle is a right triangle with the right angle . This means that the quadrilateral is a kite. The area of a kite is half the product of the diagonals, and . Recall that they are and respectively, so the area of is Solution 2 Note that triangle is a right triangle, and that the four angles that have point are all right angles. Using the fact that the centroid ( ) divides each median in a ratio, and . Quadrilateral is now just four right triangles. The area is 2013 AMC 12B Problems/Problem 10 (Redirected from 2013 AMC 10B Problems/Problem 17) The following problem is from both the 2013 AMC 12B #10 and 2013 AMC 10B #17, so both problems redirect to this page. Problem Alex has red tokens and blue tokens. There is a booth where Alex can give two red tokens and receive in return a silver token and a blue token, and another booth where Alex can give three blue tokens and receive in return a silver token and a red token. Alex continues to exchange tokens until no more exchanges are possible. How many silver tokens will Alex have at the end? 10 Solution We can approach this problem by assuming he goes to the red booth first. You start with and and at the end of the first booth, you will have and and . We now move to the blue booth, and working through each booth until we have none left, we will end up with: , and . So, the answer is 2013 AMC 10B Problems/Problem 18 Problem The number has the property that its units digit is the sum of its other digits, that is . How many integers less than but greater than share this property? Solution First, note that the only integer is . Now let's look at all numbers where Let the hundreds digit be . Then, the tens and units digit can be , which is possibilities. We notice as the hundreds digit goes up by one, the number of possibilities goes down by one. Thus, the number of integers is 2013 AMC 10B Problems/Problem 19 Problem The real numbers form an arithmetic sequence with The quadratic has exactly one root. What is this root? Solution Solution 1 It is given that has 1 real root, so the discriminant is zero, or . Because a, b, c are in arithmetic progression, , or . We need to find the unique root, or (discriminant is 0). From , we have . Ignoring the negatives, we have . Fortunately, finding is not very hard. Plug in to , we have , or , and dividing by 11 gives , so . But , violating the assumption that . Therefore, . Plugging this in, we have . But we need the negative of this, so the answer is . Solution 2 Note that we can divide the polynomial by to make the leading coefficient 1 since dividing does not change the roots or the fact that the coefficients are in an arithmetic sequence. Also, we know that there is exactly one root so this equation is must be of the form where . We now use the fact that the coefficients are in an arithmetic sequence. Note that in any arithmetic sequence, the average is equal to the median. Thus, and . Since , we easily see that has to be between 1 and 0. Thus, we can eliminate and are left with as the answer. Solution 3 Given that has only 1 real root, we know that the discriminant must equal 0, or that . Because the discriminant equals 0, we have that the root of the quadratic is . We are also given that the coefficients of the quadratic are in arithmetic progression, where . Letting the arbitrary difference equal variable , we have that and that . Plugging those two equations into , we have which yields . Isolating , we have . Substituting that in for in , we get . Once again, substituting that in for in , we have . The answer is . 12 2013 AMC 12B Problems/Problem 15 (Redirected from 2013 AMC 10B Problems/Problem 20) The following problem is from both the 2013 AMC 12B #15 and 2013 AMC 10B #20, so both problems redirect to this page. Problem The number is expressed in the form , where and are positive integers and is as small as possible. What is ? Solution The prime factorization of is . To have a factor of in the numerator, must equal . Now we notice that there can be no prime which is not a factor of 2013 such that because this prime will not be represented in the denominator, but will be represented in the numerator. The highest less than is , so there must be a factor of in the denominator. It follows that , so the answer is , which is . One possible way to express is Problem 14 (Redirected from 2013 AMC 10B Problems/Problem 21) The following problem is from both the 2013 AMC 12B #14 and 2013 AMC 10B #21, so both problems redirect to this page. Problem Two non-decreasing sequences of nonnegative integers have different first terms. Each sequence has the property that each term beginning with the third is the sum of the previous two terms, and the seventh term of each sequence is . What is the smallest possible value of ? Solution Let the first two terms of the first sequence be and and the first two of the second sequence be and . Computing the seventh term, we see that . Note that this means that and must have the same value modulo 8. To minimize, let one of them be 0; WLOG 13 assume that . Thus, the smallest possible value of is ; since the sequences are non- decreasing . To minimize, let . Thus, . Problem 22 The regular octagon has its center at . Each of the vertices and the center are to be associated with one of the digits through , with each digit used once, in such a way that the sums of the numbers on the lines , , , and are all equal. In how many ways can this be done? Solution First of all, note that must be , , or to preserve symmetry. We also notice that . WLOG assume that . Thus the pairs of vertices must be and , and , and , and and . There are ways to assign these to the vertices. Furthermore, there are ways to switch them (i.e. do instead of ). Thus, there are ways for each possible J value. There are possible J values that still preserve symmetry: 14 Problem 19 (Redirected from 2013 AMC 10B Problems/Problem 23) Problem In triangle , , , and . Distinct points , , and lie on segments , , and , respectively, such that , , and . The length of segment can be written as , where and are relatively prime positive integers. What is ? Solution Since , quadrilateral is cyclic. It follows that . In addition, since , triangles and are similar. It follows that . By Ptolemy, we have . Cancelling , the rest is easy. We obtain Problem 24 A positive integer is nice if there is a positive integer with exactly four positive divisors (including and ) such that the sum of the four divisors is equal to . How many numbers in the set are nice? Solution A positive integer with only four positive divisors has its prime factorization in the form of , where and are both prime positive integers or where is a prime. One can easily deduce that none of the numbers are even near a cube so that case is finished. We now look at the case of . The four factors of this number would be , , , and . The sum of these would be , which can be factored into the form . Easily we can see that Now we can take cases again. Case 1: Either or is 2. If this is true then we have to have that one of or is odd and that one is 3. The other is still even. So we have that in this case the only numbers that work are even multiples of 3 which are 2010 and 2016. So we just have to check if either or is a prime. We see that is the only one that works in this case. Case 2: Both and are odd primes. 15 This implies that both and are even which implies that in this case the number must be divisible by . This leaves only and . We know works so it suffices to check whether works. so we have that a factor of must go to both and . So we have that and equal the numbers , but this contradicts our assumption for the case. Thus the answer is as is the only solution.   Note: = - = = * , so why did we get the right answer?**   Note: In Case 1, isn't one of or $(b+1) even and one is 3, as one of a and b is 2 and one is an odd prime?** Problem 25 Problem Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer . For example, if , Bernardo writes the numbers and , and LeRoy obtains the sum . For how many choices of are the two rightmost digits of , in order, the same as those of ? Solution First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities. Say that also that After some inspection, it can be seen that , and , so , , , Therefore, can be written as and can be written as Keep in mind that can be , five choices; Also, we have already found which digits of will add up into the units digits of . Now, examine the tens digit, by using and to find the tens digit (units digits can be disregarded because will always work) Then we see that and take it and to find the last two digits in the base and representation. Both of those must add up to ( ) 16 Now, since will always work if works, then we can treat as a units digit instead of a tens digit in the respective bases and decrease the mods so that is now the units digit. Say that (m is between 0-6, n is 0-4 because of constraints on x) Then and this simplifies to From inspection, when This gives you choices for , and choices for , so the answer is