AMC 12 Solutions 2001

Solutions 2001 52nd AMC 12 2 (E) Suppose the two numbers are a and b. Then the desired sum is 2(a+3) + 2(b+3) = 2(a+b) + 12 = 2S + 12. (E) Suppose N = 10a + b. Then 10b + a = ab + (a + b). It follows that 9a = ab, which implies that b = 9, since a ≠ 0. (B) If Kristin's annual income is x ≥ 28,000 dollars, then (p/100) * 28000 + (p + 2)/100 * (x - 28000) = (p + 0.25)/100 * x. Multiplying by 100 and expanding yields 28000p + px + 2x - 28000p - 56000 = px + 0.25x. So, 1.75x = (7/4)x = 56000 and x = 32000. (D) Since the median is 5, we can write the three numbers as x, 5, and y, where (1/3)(x + 5 + y) = x + 10 and (1/3)(x + 5 + y) + 15 = y. If we add these equations, we get (2/3)(x + 5 + y) + 15 = x + y + 10 and solving for x + y gives x + y = 25. Hence the sum of the numbers x + y + 5 = 30.   OR Let m be the mean of the three numbers. Then the least of the numbers is m - 10 and the greatest is m + 15. The middle of the three numbers is the median, 5. So (1/3)((m - 10) + 5 + (m + 15)) = m and m = 10. Hence, the sum of the three numbers is 3(10) = 30.   (D) Note that 1 * 3 * ... * 9999 = (1 * 2 * 3 * ... * 9999 * 10000) / (2 * 4 * ... * 10000) = 10000! / (2^5000 * 1 * 2 * ... * 5000) = 10000! / (2^5000 * 5000!) (E) The last four digits (GHLJ) are either 9753 or 7531, and the remaining odd digit (either 1 or 9) is A, B, or C. Since A + B + C = 9, the odd digit among A, B, and C must be 1. Thus the sum of the two even digits in ABC is 8. The three digits in DEF are 864, 642, or 420, leaving the pairs 2 and 0, 8 and 0, or 8 and 6, respectively, as the two even digits in ABC. Of those, only the pair 8 and 0 has sum 8, so ABC is 810, and the required first digit is 8. The only such telephone number is 810-642-9753.   Solutions 2001 52nd AMC 12 3 (A) Let n be the number of full-price tickets and p be the price of each in dollars. Then (np + (140-n) \frac{p}{2} = 2001) so (p(n+140) = 4002). Thus n + 140 must be a factor of 4002 = 2 * 3 * 23 * 29. Since 0 ≤ n ≤ 140, we have 140 ≤ n + 140 ≤ 280, and the only factor of 4002 that is in the required range for n + 140 is 174 = 2 * 3 * 29. Therefore, n + 140 = 174, so n = 34 and p = 23. The money raised by the full-price tickets is 34 * 23 = 782 dollars.   (C) The slant height of the cone is 10, the radius of the sector. The circumference of the base of the cone is the same as the length of the sector's arc. This is (252/360) = 7/10 of the circumference, 20π, of the circle from which the sector is cut. The base circumference of the cone is 14π, so its radius is 7.   (C) Note that (f(600) = f(500 * \frac{6}{5}) = \frac{5}{6} f(500) = \frac{5}{6} * 3 = \frac{5}{2}) OR For all positive x, (f(x) = f(1 * x) = \frac{f(1)}{x}) so x * f(x) is the constant f(1). Therefore, 600 * f(600) = 500 * f(500) = 500 * 3 = 1500, so (f(600) = \frac{1500}{600} = \frac{5}{2}). Note: f(x) = 1500/x is the unique function satisfying the given conditions.   (D) The pattern shown at left is repeated in the plane. In fact, nine repetitions of it are shown in the statement of the problem. Note that four of the nine squares in the three-by-three square are not in the four pentagons that make up the three-by-three square. Therefore, the percentage of the plane that is enclosed by pentagons is (1 - \frac{4}{9} = \frac{5}{9} = 55\frac{5}{9}%)   Solutions 2001 52nd AMC 12 4 (D) Think of continuing the drawing until all five chips are removed from the box. There are ten possible orderings of the colors: RRRWW, RRWRW, RWRRW, WRRRW, RRWWR, RWRWR, WRRWR, RWWRR, WRWRR, and WWRRR. The six orderings that end in R represent drawings that would have ended when the second white chip was drawn.   OR Imagine drawing until only one chip remains. If the remaining chip is red, then that draw would have ended one chip before the second white chip was removed. The last chip will be red with probability 3/5.   (B) For integers not exceeding 2001, there are 2001/3 = 667 multiples of 3 and 2001/4 = 500 multiples of 4. The total, 1167, counts the 2001/12 = 166 multiples of 12 twice, so there are 1167 - 166 = 1001 multiples of 3 or 4. From these we exclude the 2001/5 = 400 multiples of 5. However, this excludes the 2001/60 = 33 multiples of 60 twice, so we must re-include these. The number of integers satisfying the conditions is 1001 - 400 + 33 = 634.   (E) The equation of the first parabola can be written in the form y = a(x - h)^2 + k = ax^2 - 2ahx + ah^2 + k, and the equation for the second (having the same shape and vertex, but opening in the opposite direction) can be written in the form y = a(x - h)^2 + k = ax^2 + 2ahx - ah^2 + k.   Hence, a + b + c + d + e + f = a + (-2ah) + (ah^2 + k) + (-a) + (2ah) + (-ah^2 + k) = 2k.