2024年1月10日发(作者：高中数学试卷选修是什么)

2019 Math League Summer Tournament

MATH LEAGUE PRESS

Math League Press, PO Box 17, Tenafly, New Jersey 07670-0017 USA

P.O. Box 17, Tenafly, New Jersey 07670-0017

July, 2019 – Grades

4 &

5

Individual Questions – Part 2 of 2

Each question is worth 10 points. Calculators are PROHIBITED.

#2-1. (Time Limit: 7 minutes) Different letters in the word “MathLeague” represent different digits.

What is the greatest possible value of the sum of the value of all 10 letters?

题目翻译: MathLeague这个单词中的每一个不同的字母对应一个不同的数字(0...9这十个数字中的一个数字)，请问这十个字母所对应的数字的和的最大值是多少?

Solution:

Since the letters “a” and “e” appear twice, let one of them have a value of 9 and the other one

a value of 8. Each of the six other letters appears only once, so their values should be 7, 6, 5,

4, 3, and 2. Assigning these values to the letters, the greatest possible sum is (18 + 16 + 7 + 6 +

5 + 4 + 3 + 2) =

61.

#2-2. (Time Limit: 7 minutes) The number N = 2

666

666 . . . 666

666 consists of a “2” followed by 2019

“6”s. What is the remainder when N is divided by 11?

题目翻译: N = 266666…666666666，数字2后面有2019个6，N除以11的余数是多少？

Solution 1:

66 is a multiple of 11, so is any even number of 6 in a row. Let N1 = N – 66…66 (2018 “6”s) =

2600…00 (2018 “0”s). The remainder when N is divided by 11 is equal to the remainder

when N1 is divided by 11.

Let N2 = N1 – 00 (2018 “0”s) = 400…00 (2018 “0”s). The remainder when N1 is

divided by 11 is equal to the remainder when N2 is divided by 11.

Let N3 = N2 – 4 * 99…99 (2018 “9”s) = 4. Since 99…99 (2018 “9”s) is a multiple of 11, the

remainder when N2 is divided by 11 is equal to the remainder when N3 is divided by 11. So

the answer is

4.

Solution 2:

There are 2020 digits in N. As we divide, we see that the quotient is 242424. . ., with the partial

remainders after each step of the division alternating between 4 and 2. Since there are 2019

6s, the last remainder will be a

4.

#2-3. (Time Limit: 7 minutes) If 2 watermelons can be exchanged for 7 apples and 3 apples can be

exchanged for 4 bananas, for how many bananas can 12 watermelons be exchanged?

题目翻译: 如果2个西瓜可以换7个苹果，3个苹果可以换4个香蕉，那么12个西瓜可以换多少个香蕉？

Solution:

If 2 watermelons can be exchanged for 7 apples, 12 watermelons can be exchanged for 42

apples. Similarly, if 3 apples can be exchanged for 4 bananas, then 42 apples can be

exchanged for

56

bananas.

#2-4. (Time Limit: 7 minutes) Six years ago, Steve was six times as old as Rui. Six years from now,

Steve will be 10 years older than Rui. How old is Rui?

题目翻译: 6年前，Steve的年纪是Rui的年纪的6倍。6年后，Steve的年纪比Rui的年纪大10岁。Rui今年多大?

Solution:

Let r = Rui’s current age. Then six years ago, Steve’s age was 6(r – 6) = 6r – 36. Therefore, his

current age is 6r – 30. Since Steve is 10 years older than Rui, 6r – 30 = r + 10 and r =

8.

#2-5. (Time Limit: 7 minutes) The lengths of three sides of a rectangle are x – 5, 2x + 7, and 3x – 17.

What is the greatest possible numerical value of the perimeter of this rectangle?

题目翻译: 一个长方形的三条边的边长分别是 x - 5, 2x + 7, 3x - 17, 那么这个长方形的周长最大是多少？

Solution:

There are three possibilities to consider: x – 5 = 2x + 7, x – 5 = 3x – 17, and 2x + 7 = 3x – 17.

The values of x corresponding to these equations are -12, 6, and 24, respectively. The

greatest possible perimeter occurs when x = 24. The dimensions of the rectangle are 19 by

55 and its perimeter is

148.

#2-6. (Time Limit: 7 minutes) Points A, B, and C are chosen on circle O, as shown. If mÐOAC = 15°

and mÐOBC = 40°, what is mÐACB?

题目翻译: A、B、C三个点都在圆O上，如果∠OAC = 15°，∠OBC = 40°， 那么∠ACB 等于多少？

Solution:

Draw

OC, as shown. △OCB and △OAC are both isosceles triangles. Thus, mÐOCB = mÐOBC = 40°,

and mÐOCA = mÐOAC = 15°. Since mÐACB = mÐOCB – mÐOCA, mÐACB = 40° – 15° =

25°.

#2-7. (Time Limit: 7 minutes) How many positive integers less than 2019 are squares of integers

divisible by both 2 and 3?

题目翻译: 比2019小的正整数中，有多少个正整数是能够被2和3整除的完全平方数？

Solution:

Any integer divisible by both 2 and 3 is divisible by 6. All perfect squares divisible by 6

must be divisible by 36. The perfect squares less than 2019 that are divisible by 36 are 1 

36, 4  36, 9  36, 16  36, 25  36, 36  36, and 49  36. In all, there are

7

positive integers

less than 2019 that are divisible by the squares of integers divisible by both 2 and 3.


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