2004 AIME II Exam Problems
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1.
A chord of a circle is perpendicular to a radius at the midpoint of the radius. The ratio of the area of the larger of the two regions into which the chord divides the circle to the smaller can be expressed in the form where and are positive integers, and are relatively prime, and neither nor is divisible by the square of any prime. Find the remainder when the product is divided by
Answer: 592
Difficulty rating: 2050
Solution:
Scale so the radius is The chord lies at distance from the center, so each radius to an endpoint of the chord makes a angle with the bisected radius, and the two endpoint radii form a central angle of The isosceles triangle they cut off has area and the whole disk has area
The smaller region is the sector minus the triangle, and the larger region is the rest, The ratio is which has the required form with
The product is whose remainder upon division by is
2.
A jar has red candies and blue candies. Terry picks two candies at random, then Mary picks two of the remaining candies at random. Given that the probability that they get the same color combination, irrespective of order, is where and are relatively prime positive integers, find
Answer: 441
Difficulty rating: 2180
Solution:
The combinations match exactly when both draw two reds, both draw two blues, or both draw one candy of each color. The probability that Terry draws two reds is after which reds and blues remain, so Mary draws two reds with probability That case has probability and by symmetry two blues each is also
For mixed draws, Terry succeeds with probability leaving of each color, and Mary with probability for
The total is Since and the fraction is in lowest terms, and
3.
A solid rectangular block is formed by gluing together congruent -cm cubes face to face. When the block is viewed so that three of its faces are visible, exactly of the -cm cubes cannot be seen. Find the smallest possible value of
Answer: 384
Difficulty rating: 2110
Solution:
Let the block measure A cube is hidden exactly when it touches none of the three visible faces, so the hidden cubes form a block, giving
The ways to write as a product of three positive integers are and giving blocks and with volumes and
The smallest is
4.
How many positive integers less than have at most two different digits?
Answer: 927
Difficulty rating: 2300
Solution:
All positive integers below qualify. A qualifying -digit number is either a repdigit ( of them) or uses a leading digit together with a second value in some of the last two positions: patterns, each realized in ways ( choices for then for ), for numbers.
Similarly a qualifying -digit number is a repdigit () or has appearing in a nonempty subset of the last three positions: patterns, each in ways, for numbers.
The total is
5.
In order to complete a large job, workers were hired, just enough to complete the job on schedule. All the workers stayed on the job while the first quarter of the work was done, so the first quarter of the work was completed on schedule. Then workers were laid off, so the second quarter of the work was completed behind schedule. Then an additional workers were laid off, so the third quarter of the work was completed still further behind schedule. Given that all workers work at the same rate, what is the minimum number of additional workers, beyond the workers still on the job at the end of the third quarter, that must be hired after three-quarters of the work has been completed so that the entire project can be completed on schedule or before?
Answer: 766
Difficulty rating: 2390
Solution:
Measure time so that workers complete a quarter of the job in unit; the schedule allows units in all. With workers the second quarter takes units, and with workers the third quarter takes units. The time used so far is leaving of a unit for the last quarter.
The last quarter requires worker-units of labor, so the workforce must satisfy that is so at least workers are needed.
Since remain on the job, at least additional workers must be hired.
6.
Three clever monkeys divide a pile of bananas. The first monkey takes some bananas from the pile, keeps three-fourths of them, and divides the rest equally between the other two. The second monkey takes some bananas from the pile, keeps one-fourth of them, and divides the rest equally between the other two. The third monkey takes the remaining bananas from the pile, keeps one-twelfth of them, and divides the rest equally between the other two. Given that each monkey receives a whole number of bananas whenever the bananas are divided, and the numbers of bananas the first, second, and third monkeys have at the end of the process are in the ratio what is the least possible total for the number of bananas?
Answer: 408
Difficulty rating: 2400
Solution:
Say the first monkey takes bananas, keeping and giving to each of the others; the second takes keeping and giving to each; the third takes keeping and giving to each. All divisions are whole numbers exactly when are positive integers. The final amounts are and
The ratio says the first amount is triple the third and the second is double the third: Substituting into the first equation gives so Thus and for a positive integer and then
The total is least when the answer is
7.
is a rectangular sheet of paper that has been folded so that corner is matched with point on edge The crease is where is on and is on The dimensions and are given. The perimeter of rectangle is where and are relatively prime positive integers. Find
Answer: 293
Difficulty rating: 2650
Solution:
Folding reflects to across the crease, so In right triangle and Place
Points on the crease are equidistant from and so is perpendicular to Since has slope the crease through has slope and it meets the line (at height ) at The condition gives
The perimeter is so
8.
How many positive integer divisors of are divisible by exactly positive integers?
Answer: 54
Difficulty rating: 2450
Solution:
Since we have so its divisors are with and Such an has divisors, so we need
Every ordered triple of positive integers with product yields admissible exponents, since each factor is at most Counting prime by prime: the exponent of the prime is split among the three factors in ways by stars and bars, and each of the primes and goes to one of the factors.
The count is
9.
A sequence of positive integers with and is formed so that the first three terms are in geometric progression, the second, third, and fourth terms are in arithmetic progression, and, in general, for all the terms and are in geometric progression, and the terms and are in arithmetic progression. Let be the greatest term in this sequence that is less than Find
Answer: 973
Difficulty rating: 2840
Solution:
Let The geometric condition gives the arithmetic condition gives then and so on: inductively In particular and so Expanding gives which factors as so
With we get so and the sequence is increasing. Since while the greatest term below is
Therefore
10.
Let be the set of integers between and whose binary expansions have exactly two 's. If a number is chosen at random from the probability that it is divisible by is where and are relatively prime positive integers. Find
Answer: 913
Difficulty rating: 2920
Solution:
The set consists of the numbers with Since is coprime to we have exactly when The powers of modulo cycle through with period so exactly when
For each difference there are pairs, so the number of multiples of in is
The probability is and since while it is in lowest terms. Thus
11.
A right circular cone has a base with radius and height A fly starts at a point on the surface of the cone whose distance from the vertex of the cone is and crawls along the surface of the cone to a point on the exact opposite side of the cone whose distance from the vertex is Find the least distance that the fly could have crawled.
Answer: 625
Difficulty rating: 2990
Solution:
The slant height is Cutting the cone along the ruling through the starting point and unrolling gives a sector of radius whose arc has the base circumference since a full circle of radius has circumference the central angle is A point on the exact opposite side of the cone is halfway around, which in the unrolled sector is away.
The shortest crawl is the straight segment between the two points, at radii and with a angle between them. By the law of cosines,
Thus
12.
Let be an isosceles trapezoid, whose dimensions are and Draw circles of radius centered at and and circles of radius centered at and A circle contained within the trapezoid is tangent to all four of these circles. Its radius is where and are positive integers, is not divisible by the square of any prime, and and are relatively prime. Find
Answer: 134
Difficulty rating: 3060
Solution:
Dropping perpendiculars from and shows each leg of length spans a horizontal offset of so the height of the trapezoid is By symmetry the inner circle's center lies on the vertical axis through the midpoints of and of If its radius is external tangency gives and so with and
Since moving one radical across and squaring gives and squaring again yields that is
The positive root is so
13.
Let be a convex pentagon with and Given that the ratio between the area of triangle and the area of triangle is where and are relatively prime positive integers, find
Answer: 484
Difficulty rating: 3160
Solution:
By the law of cosines, so Let be the intersection of and Since and quadrilateral is a parallelogram, so lies at the same distance from line as on the opposite side, where
Since triangles and are similar with ratio so the distance from to line is with on the far side of from The distance from to line is therefore giving
Thus which is in lowest terms since The answer is
14.
Consider a string of 's, into which signs are inserted to produce an arithmetic expression. For example, could be obtained from eight 's in this way. For how many values of is it possible to insert signs so that the resulting expression has value
Answer: 108
Difficulty rating: 3270
Solution:
Dividing by turns the summands into or (no longer summand fits, since ) and the target into If count the summands of each size, then and Subtracting gives so the possible correspond exactly to the attainable values of
The constraints are and (then follows). For the value ranges over the intervals and (when only fits). Their union is every integer from to together with only is unattainable.
So takes values, and takes values.
15.
A long thin strip of paper is units in length, unit in width, and is divided into unit squares. The paper is folded in half repeatedly. For the first fold, the right end of the paper is folded over to coincide with and lie on top of the left end. The result is a by strip of double thickness. Next, the right end of this strip is folded over to coincide with and lie on top of the left end, resulting in a by strip of quadruple thickness. This process is repeated more times. After the last fold, the strip has become a stack of unit squares. How many of these squares lie below the square that was originally the nd square counting from the left?
Answer: 593
Difficulty rating: 3500
Solution:
After folds the strip is squares long and layers thick, so the positions from the left and from the right satisfy and the positions from the bottom and from the top satisfy When the right half is folded over onto the left, a square in the left half keeps its and while a square in the right half is flipped: its new is its old and its new is its old
The nd square starts at Applying the rule through the ten folds gives For example, at the fourth fold the strip has length and so the new is and the new is the old making
In the final stack of squares, this square sits at height from the bottom, so squares lie below it.