




免費(fèi)預(yù)覽已結(jié)束,剩余12頁可下載查看
下載本文檔
版權(quán)說明:本文檔由用戶提供并上傳,收益歸屬內(nèi)容提供方,若內(nèi)容存在侵權(quán),請進(jìn)行舉報(bào)或認(rèn)領(lǐng)
文檔簡介
2002 AMC 12B Problems and SolutionProblem 1 The arithmetic mean of the nine numbers in the set is a -digit number , all of whose digits are distinct. The number does not contain the digit Solution We wish to find , or . This does not have the digit 0, so the answer is Problem 2 What is the value of when ? Solution By the distributive property, Problem 3 For how many positive integers is a prime number? Solution Factoring, we get . Exactly of and must be and the other a prime number. If , then , and , which is not prime. On the other hand, if , then , and , which is a prime number. The answer is . Problem 4 Let be a positive integer such that is an integer. Which of the following statements is not true: Solution Since , From which it follows that and . Thus the answer is . Problem 5 Let and be the degree measures of the five angles of a pentagon. Suppose that and and form an arithmetic sequence. Find the value of . Solution The sum of the degree measures of the angles of a pentagon (as a pentagon can be split into triangles) is . If we let , it follows that Note that since is the middle term of an arithmetic sequence with an odd number of terms, it is simply the average of the sequence. Problem 6 Suppose that and are nonzero real numbers, and that the equation has solutions and . Then the pair is Solution Solution 1 Since , it follows by comparing coefficients that and that . Since is nonzero, , and . Thus . Solution 2Another method is to use Vietas formulas. The sum of the solutions to this polynomial is equal to the opposite of the coefficient, since the leading coefficient is 1; in other words, and the product of the solutions is equal to the constant term (i.e, ). Since is nonzero, it follows that and therefore (from the first equation), . Hence, Problem 7 The product of three consecutive positive integers is times their sum. What is the sum of their squares? Solution Let the three consecutive positive integers be , , and . So, . Rearranging and factoring, , so . Hence, the sum of the squares is . Problem 8 Suppose July of year has five Mondays. Which of the following must occur five times in August of year ? (Note: Both months have 31 days.) SolutionIf there are five Mondays, there are only three possibilities for their dates: , , and . In the first case August starts on a Thursday, and there are five Thursdays, Fridays, and Saturdays in August. In the second case August starts on a Wednesday, and there are five Wednesdays, Thursdays, and Fridays in August. In the third case August starts on a Tuesday, and there are five Tuesdays, Wednesdays, and Thursdays in August. The only day of the week that is guaranteed to appear five times is therefore . Problem 9 If are positive real numbers such that form an increasing arithmetic sequence and form a geometric sequence, then is SolutionSolution 1 We can let a=1, b=2, c=3, and d=4. Solution 2 As is a geometric sequence, let and for some . Now, is an arithmetic sequence. Its difference is . Thus . Comparing the two expressions for we get . The positive solution is , and . Solution 3 Letting be the common difference of the arithmetic progression, we have , , . We are given that = , or Cross-multiplying, we get So . Problem 10 How many different integers can be expressed as the sum of three distinct members of the set ? SolutionEach number in the set is congruent to 1 modulo 3. Therefore, the sum of any three numbers is a multiple of 3. We can make all multiples of three between 1+4+7=12 (the minimum sum) and 13+16+19=48 (the maximum sum), inclusive. There are integers we can form. Problem 11 The positive integers and are all prime numbers. The sum of these four primes is Solution Since and must have the same parity, and since there is only one even prime number, it follows that and are both odd. Thus one of is odd and the other even. Since , it follows that (as a prime greater than ) is odd. Thus , and are consecutive odd primes. At least one of is divisible by , from which it follows that and . The sum of these numbers is thus , a prime, so the answer is . Problem 12 For how many integers is the square of an integer? Solution Solution 1 Let , with (note that the solutions do not give any additional solutions for ). Then rewriting, . Since , it follows that divides . Listing the factors of , we find that are the only solutions (respectively yielding ). Solution 2 For and the fraction is negative, for it is not defined, and for it is between 0 and 1. Thus we only need to examine and . For and we obviously get the squares and respectively. For prime the fraction will not be an integer, as the denominator will not contain the prime in the numerator. This leaves , and a quick substitution shows that out of these only and yield a square. Problem 13 The sum of consecutive positive integers is a perfect square. The smallest possible value of this sum is Solution Let be the consecutive positive integers. Their sum, , is a perfect square. Since is a perfect square, it follows that is a perfect square. The smallest possible such perfect square is when , and the sum is . Problem 14 Four distinct circles are drawn in a plane. What is the maximum number of points where at least two of the circles intersect? Solution For any given pair of circles, they can intersect at most times. Since there are pairs of circles, the maximum number of possible intersections is . We can construct such a situation as below, so the answer is . Problem 15 How many four-digit numbers have the property that the three-digit number obtained by removing the leftmost digit is one ninth of ? Solution Let , such that . Then . Since , from we have three-digit solutions, and the answer is . Problem 16 Juan rolls a fair regular octahedral die marked with the numbers through . Then Amal rolls a fair six-sided die. What is the probability that the product of the two rolls is a multiple of 3? Solution Solution 1 On both dice, only the faces with the numbers are divisible by . Let be the probability that Juan rolls a or a , and that Amal does. By the Principle of Inclusion-Exclusion, Alternatively, the probability that Juan rolls a multiple of is , and the probability that Juan does not roll a multiple of but Amal does is . Thus the total probability is . Solution 2 The probability that neither Juan nor Amal rolls a multiple of is ; using complementary counting, the probability that at least one does is . Problem 17 Andys lawn has twice as much area as Beths lawn and three times as much area as Carlos lawn. Carlos lawn mower cuts half as fast as Beths mower and one third as fast as Andys mower. If they all start to mow their lawns at the same time, who will finish first? Solution We say Andys lawn has an area of . Beths lawn thus has an area of , and Carloss lawn has an area of . We say Andys lawn mower cuts at a speed of . Carloss cuts at a speed of , and Beths cuts at a speed . Each persons lawn is cut at a speed of , so Andys is cut in time, as is Carloss. Beths is cut in , so the first one to finish is . Problem 18 A point is randomly selected from the rectangular region with vertices . What is the probability that is closer to the origin than it is to the point ? Solution The region containing the points closer to than to is bounded by the perpendicular bisector of the segment with endpoints . The perpendicular bisector passes through midpoint of , which is , the center of the unit square with coordinates . Thus, it cuts the unit square into two equal halves of area . The total area of the rectangle is , so the area closer to the origin than to and in the rectangle is . The probability is . Problem 19 If and are positive real numbers such that and , then is Solution Adding up the three equations gives . Subtracting each of the above equations from this yields, respectively, . Taking their product, . Problem 20 Let be a right-angled triangle with . Let and be the midpoints of legs and , respectively. Given that and , find . Solution Let , . By the Pythagorean Theorem on respectively, Summing these gives . By the Pythagorean Theorem again, we have Problem 21 For all positive integers less than , let Calculate . Solution Since , it follows that Thus . Problem 22 For all integers greater than , define . Let and . Then equals Solution By the change of base formula, . Thus Problem 23 In , we have and . Side and the median from to have the same length. What is ? Solution Let be the foot of the median from to , and we let . Then by the Law of Cosines on , we have Since , we can add these two equati
溫馨提示
- 1. 本站所有資源如無特殊說明,都需要本地電腦安裝OFFICE2007和PDF閱讀器。圖紙軟件為CAD,CAXA,PROE,UG,SolidWorks等.壓縮文件請下載最新的WinRAR軟件解壓。
- 2. 本站的文檔不包含任何第三方提供的附件圖紙等,如果需要附件,請聯(lián)系上傳者。文件的所有權(quán)益歸上傳用戶所有。
- 3. 本站RAR壓縮包中若帶圖紙,網(wǎng)頁內(nèi)容里面會有圖紙預(yù)覽,若沒有圖紙預(yù)覽就沒有圖紙。
- 4. 未經(jīng)權(quán)益所有人同意不得將文件中的內(nèi)容挪作商業(yè)或盈利用途。
- 5. 人人文庫網(wǎng)僅提供信息存儲空間,僅對用戶上傳內(nèi)容的表現(xiàn)方式做保護(hù)處理,對用戶上傳分享的文檔內(nèi)容本身不做任何修改或編輯,并不能對任何下載內(nèi)容負(fù)責(zé)。
- 6. 下載文件中如有侵權(quán)或不適當(dāng)內(nèi)容,請與我們聯(lián)系,我們立即糾正。
- 7. 本站不保證下載資源的準(zhǔn)確性、安全性和完整性, 同時(shí)也不承擔(dān)用戶因使用這些下載資源對自己和他人造成任何形式的傷害或損失。
最新文檔
- 2025各類合同租賃合同范本(含注意事項(xiàng))
- 2025合同范本企業(yè)采購合同協(xié)議條款參考模板
- 2025年鋁制欄桿加工制作合同
- 2025屆初升高數(shù)學(xué)銜接專題講義第三講 一元二次方程的判別式與韋達(dá)定理(精練)含答案
- 郵政營業(yè)員考試試題及答案
- 2025標(biāo)準(zhǔn)合同范本參考
- 銀行競賽面試題目及答案
- 2025物流配送合同模板
- 英語期末考試試題及答案2025
- 銀行柜員面試題目及答案
- 勞動與社會保障專業(yè)大學(xué)生職業(yè)生涯發(fā)展
- 外研版(三起)小學(xué)英語三年級下冊Unit 1 Animal friends Get ready start up 課件
- 讀后續(xù)寫+原諒之花綻放在童真的田野上+講義 高一下學(xué)期7月期末英語試題
- 導(dǎo)數(shù)中的同構(gòu)問題【八大題型】解析版-2025年新高考數(shù)學(xué)一輪復(fù)習(xí)
- 2024年中國海鮮水餃?zhǔn)袌稣{(diào)查研究報(bào)告
- 腸內(nèi)外營養(yǎng)護(hù)理要點(diǎn)
- 2019版人教版新課標(biāo)高中英語選擇性必修1詞匯表帶音標(biāo)單詞表+帶音標(biāo)漢譯英默寫+無音
- 機(jī)械設(shè)備故障應(yīng)急預(yù)案與處理措施
- 一個(gè)人與公司合伙協(xié)議書范文
- 中國生殖支原體感染診療專家共識(2024年版)解讀課件
- 氣壓傳動課件 項(xiàng)目五任務(wù)三 壓印設(shè)備氣動系統(tǒng)的組裝與調(diào)試
評論
0/150
提交評論