1. AIME Problems for college kids
Scott Annin, CSUF
Steven Davis, CSULA

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

3. 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?

4. 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: That’s a lot of roots!

5. 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: 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.

6. 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 a c, where a > 0. Find the base 10 representation of n.

7. 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?
abc14 = acb15
abc14 = acac6
acb15 = acac6

8. acb15 = acac6
225a + 15c +b = 216a + 36c +6a +c from which it follows 3a + b = 22c.
Immediately c has to be 1.
If a = 3 and b = 13 then 225a + 15c +b =703, 216a + 36c +6a +c = 703, but 196a + 14b + c =771.
If a = 4 and b = 10 then 225a + 15c +b =925, 216a + 36c +6a +c = 925, and 196a +14b + c =925. So the answer is 925.

9. AIME 2004 Problem 8
Define a regular n-pointed star to be the union of n line segments P1P2, P2P3,..., PnP1 such that • the points P1, P2,...,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 P1, P2,...,Pn are congruent, • all of the n line segments P1P2, P2P3,..., PnP1 are congruent, and • the path P1P2 ...PnP1 turns counterclockwise at an angle of less than 180◦ at each vertex.

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

11. How many 1,000-pointed stars are there?

12. How many 1,000-pointed stars are there?

13. There are no regular 3-pointed, 4-pointed, or 6-pointed stars
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?

14. How many 1,000-pointed stars are there?

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

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

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

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

19. How many 1,000-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.   #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? ___________________

20. How many 1,000-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 ? ______40_______ How many 100-pointed stars are there? ________19___________   #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 ? ______400_______ How many 1000-pointed stars are there? ________199___________

21. AIME Problem 1
Let
N = – 982 – … – 22 –12,
where the additions and subtractions alternate in pairs. Find the remainder when N is divided by
How to best attack this problem?

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

23. AIME Problem 4
Let (a1, a2, a3, , a12) be a permutation of (1, 2, 3, , 12) for which a1 > a2 > a3 > a4 > a5 > a6 and a6 < a7 < a8 < a9 < a10 < a11 < a12. 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.

24. Solution
From a1 > a2 > a3 > a4 > a5 > a6 < a7 < a8 < a9 < a10 < a11 < a12 , Do we know specifically any of the values?

25. Solution
From a1 > a2 > a3 > a4 > a5 > a6 < a7 < a8 < a9 < a10 < a11 < a12 , Do we know specifically any of the values? YES: a6 =1.

26. Solution
From a1 > a2 > a3 > a4 > a5 > a6 < a7 < a8 < a9 < a10 < a11 < a12 , Do we know specifically any of the values? YES: a6 =1. From the remaining 11 numbers, choose 5 of them to be assigned to a1, a2, a3, a4, a5. The remaining numbers are then forced to be assigned a7, a8, a9, a10, a11 , a12.

27. ANSWER: C(11,5) = 462. Solution From
a1 > a2 > a3 > a4 > a5 > a6 < a7 < a8 < a9 < a10 < a11 < a12 ,
Do we know specifically any of the values?
YES:
a6 =1.
From the remaining 11 numbers, choose 5 of them to be assigned to a1, a2, a3, a4, a5.
The remaining numbers are then forced to be assigned a7, a8, a9, a10, a11 , a12.
ANSWER: C(11,5) = 462.

28. AIME 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?

29. AIME 2008 Problem 5
A right circular cone has base radius r and height h. The cone lies on its side on a flat table. As the cone rolls on the surface of the table without slipping, the point where the cone’s base meets the table traces a circular arc centered at the point where the vertex touches the table. The cone first returns to its original position on the table after making 17 complete rotations. The value of h/r can be written in the form m √ n, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n .

30. AIME 2009 Problem 10
The Annual Interplanetary Mathematics Examination (AIME) is written by a committee of five Earthlings, five Martians, and five Venusians. At meetings, committee members sit at a round table with chairs numbered 1 to 15 in clockwise order. Committee rules state that a Martian must occupy chair 1 and an Earthling must occupy chair Furthermore, no Earthling can sit immediately to the left of a Martian, no Martian can sit immediately to the left of a Venusian, and no Venusian can sit immediately to the left of an Earthling. The number of possible arrangements is N(5!)3. Find N.