AIME Problems for college kids Steven Davis CSULA

AIME Problems for college kids Steven Davis, CSULA

2010 AMC 10 A #8 Which of the following numbers is a perfect square?

AIME 2001 Problem 3 Find the sum of all the roots, real and non-real, of the equation x 2001 + ( ½ - x)2001 = 0, given that there are no multiple roots.

Solution How many roots does this polynomial have? [Fundamental Theorem of Algebra] A polynomial of degree n has exactly n roots, including multiplicities, real, and complex roots. What is the degree of this polynomial?

Solution How many roots does this polynomial have? [Fundamental Theorem of Algebra] A polynomial of degree n has exactly n roots, including multiplicities, real, and complex roots. What is the degree of this polynomial? Answer: 2000. That’s a lot of roots!

Solution How many roots does this polynomial have? [Fundamental Theorem of Algebra] A polynomial of degree n has exactly n roots, including multiplicities, real, and complex roots. What is the degree of this polynomial? Answer: 2000. That’s a lot of roots! Key Idea: If r is a root, then so is ½ - r. Group the roots in pairs that sum to ½. There are 1000 pairs, so the sum is 500.

2018 AIME 1 Problem 2 The number n can be written in base 14 as a b c, can be written in base 15 as a c b, and can be written in base 6 as a c, where a > 0. Find the base 10 representation of n.

Solution This problem can be solved in one of three ways. Each way involves a Diophantine Equation that may be easier or harder to solve than another equation. Which of these three ways is best? • abc 14 = acb 15 • abc 14 = acac 6 • acb 15 = acac 6

acb 15 = acac 6 225 a + 15 c +b = 216 a + 36 c +6 a +c from which it follows 3 a + b = 22 c. Immediately c has to be 1. If a = 3 and b = 13 then 225 a + 15 c +b =703, 216 a + 36 c +6 a +c = 703, but 196 a + 14 b + c =771. If a = 4 and b = 10 then 225 a + 15 c +b =925, 216 a + 36 c +6 a +c = 925, and 196 a +14 b + c =925. So the answer is 925.

AIME 2004 Problem 8 Define a regular n-pointed star to be the union of n line segments P 1 P 2, P 2 P 3, . . . , Pn. P 1 such that • the points P 1, P 2, . . . , Pn are coplanar and no three of them are collinear, • each of the n line segments intersects at least one of the other line segments at a point other than an endpoint, • all of the angles at P 1, P 2, . . . , Pn are congruent, • all of the n line segments P 1 P 2, P 2 P 3, . . . , Pn. P 1 are congruent, and • the path P 1 P 2. . . Pn. P 1 turns counterclockwise at an angle of less than 180◦ at each vertex.

Polya’s 4 Step Approach to Problem Solving Understand the Problem • Devise a plan • Carry out the plan • Check and Review •

How many 1, 000 -pointed stars are there?

How many 1, 000 -pointed stars are there?

There are no regular 3 -pointed, 4 -pointed, or 6 -pointed stars. All regular 5 - pointed stars are similar, but there are two non-similar regular 7 -pointed stars. How many non-similar regular 1000 pointed stars are there?

How many 1, 000 -pointed stars are there?

How many 1, 000 -pointed stars are there? j=2 or j=3

How many 1, 000 -pointed stars are there? Activity Sheet #1

How many 1, 000 -pointed stars are there? n 2 3 4 5 6 7 8 9 10 11 12 Pairs of j values producing stars Number of n-pointed stars

How many 1, 000 -pointed stars are there? 2 Pairs of j values producing stars N/A Number of n-pointed stars 0 3 N/A 0 4 N/A 0 5 (2, 3) 1 6 N/A 0 7 (2, 5) and (3, 4) 2 8 (3, 5) 1 9 (2, 7) and (4, 5) 2 10 (3, 7) 1 (2, 9) and (3, 8) and (4, 7) and (5, 6) (5, 7) 4 n 11 12 1

How many 1, 000 -pointed stars are there? Activity Sheet #3 #1) Circle the numbers below that are relatively prime to n = 20 and determine the number of stars with 20 points. #2) Circle the numbers below that are relatively prime to n = 45 and determine the number of stars with 45 points. #3) How many numbers from 1 to 100 contain a factor of 2 ? _______ How many numbers from 1 to 100 contain a factor of 5 ? _______ How many numbers from 1 to 100 contain a factor of both 2 and 5 ? _______ How many numbers from 1 to 100 are relatively prime to 100 ? How many 100 -pointed stars are there? __________ #4) How many numbers from 1 to 1000 contain a factor of 2 ? ______ How many numbers from 1 to 1000 contain a factor of 5 ? ______ How many numbers from 1 to 1000 contain a factor of both 2 and 5 ? ______ How many numbers from 1 to 1000 are relatively prime to 1000 ? How many 1000 -pointed stars are there? __________

How many 1, 000 -pointed stars are there? Activity Sheet #3 #1) Circle the numbers below that are relatively prime to n = 20 and determine the number of stars with 20 points. ANSWER: 1, 3, 7, 9, 11, 13, 17, 19 3 stars with 20 points #2) Circle the numbers below that are relatively prime to n = 45 and determine the number of stars with 45 points. ANSWER: 1, 2, 4, 7, 8, 11, 13, 14, 16, 17, 19, 22, 23, 26, 28, 29, 31, 32, 34, 37, 38, 41, 43, 44 11 stars with 45 points #3) How many numbers from 1 to 100 contain a factor of 2 ? _____50____ How many numbers from 1 to 100 contain a factor of 5 ? ______20_______ How many numbers from 1 to 100 contain a factor of both 2 and 5 ? _____10____ How many numbers from 1 to 100 are relatively prime to 100 ? How many 100 -pointed stars are there? ____19______40_______ #4) How many numbers from 1 to 1000 contain a factor of 2 ? ______500______ How many numbers from 1 to 1000 contain a factor of 5 ? ______200_____ How many numbers from 1 to 1000 contain a factor of both 2 and 5 ? ____100____ How many numbers from 1 to 1000 are relatively prime to 1000 ? How many 1000 -pointed stars are there? ____199______400_______

AIME 2 2008 Problem 1 Let N = 1002 + 992 – 982 – 972 + 962 +…+ 42 + 32 – 22 – 12, where the additions and subtractions alternate in pairs. Find the remainder when N is divided by 1000. How to best attack this problem?

This is the easiest way; we rewrite it as such: N = 1002 + (992 – 982) – (972 – 962) + (952– 942) –…+ (32 – 22) – 12 = 1002 + (99 – 98)(99+98) – (97 – 96)(97 + 96) + (95 – 94)(95 + 94) – … + (3 – 2)(3 + 2) – 12 = 1002 + (99 + 98 – 97 – 96) + (95 + 94 – 93 – 92) + … + (3 + 2 – 1 – 0) = 1002 + 4*25 = 1002 + 100 = 10100 ANSWER: 100

AIME 2 2006 Problem 4 Let (a 1, a 2, a 3, . . . , a 12) be a permutation of (1, 2, 3, . . . , 12) for which a 1 > a 2 > a 3 > a 4 > a 5 > a 6 and a 6 < a 7 < a 8 < a 9 < a 10 < a 11 < a 12. An example of such a permutation is (6, 5, 4, 3, 2, 1, 7, 8, 9, 10, 11, 12). Find the number of such permutations.

Solution From a 1 > a 2 > a 3 > a 4 > a 5 > a 6 < a 7 < a 8 < a 9 < a 10 < a 11 < a 12 , Do we know specifically any of the values?

Solution From a 1 > a 2 > a 3 > a 4 > a 5 > a 6 < a 7 < a 8 < a 9 < a 10 < a 11 < a 12 , Do we know specifically any of the values? YES: a 6 =1.

Solution From a 1 > a 2 > a 3 > a 4 > a 5 > a 6 < a 7 < a 8 < a 9 < a 10 < a 11 < a 12 , Do we know specifically any of the values? YES: a 6 =1. From the remaining 11 numbers, choose 5 of them to be assigned to a 1, a 2, a 3, a 4, a 5. The remaining numbers are then forced to be assigned a 7, a 8, a 9, a 10, a 11 , a 12.

Solution From a 1 > a 2 > a 3 > a 4 > a 5 > a 6 < a 7 < a 8 < a 9 < a 10 < a 11 < a 12 , Do we know specifically any of the values? YES: a 6 =1. From the remaining 11 numbers, choose 5 of them to be assigned to a 1, a 2, a 3, a 4, a 5. The remaining numbers are then forced to be assigned a 7, a 8, a 9, a 10, a 11 , a 12. ANSWER: C(11, 5) = 462.

2017 AMC 12 B Problem 11 Call a positive number monotonous if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. For example, 3, 23578, 987620 are monotonous, but 88, 7434, and 23557 are not. How many monotonous positive integers are there? A) 1024 B) 1524 C) 1533 D) 1536 E) 2048

Question 1. What is the maximum number of digits a monotonous number can have? Question 2. Can we apply the same technique as the previous problem or is there a better way?

AIME 2 2011 Problem 3 The degree measures of the angels in a convex 18 -sided polygon form an increasing arithmetic sequence with integer values. Find the degree measure of the smallest angle. This problem can be solved using formulas for an arithmetic sequence and noting the series sums to 16*180=2880, but that leaves us with one equation and two unknowns. What’s the fastest way to solve it?

2014 AMC 10 A Angle of Reflection �

A) 1: 2 B) 3: 5 C) 2: 3 D) 3: 4 E) 4: 5

Solution