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[[2023 AIME II Problems/Problem 5|Solution]] | [[2023 AIME II Problems/Problem 5|Solution]] | ||
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+ | Since 55r is greater than r, we know 55 r can be simplified. Call r <math>\frac{a}{b}</math>. This means 55r is <math>\frac{55a}{b}</math>. Since 55r can clearly be simplified, let's look at the possibilities, where b is divisible by factors of 55. | ||
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+ | Case 1: B is divisible by 55. | ||
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+ | If this is the case, the sum of the numerator and denominator of 55r is a + <math>\frac{b}{55}</math>. This clearly cannot equal a+b, so this case is invalid. | ||
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+ | Case 2: B is divisible by 11. | ||
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+ | If this is the case, the sum of the numerator and denominator of 55r is 5a + <math>\frac{b}{11}</math>. This is equal to a+b. Solving this equation gives <math>\frac{a}{b}</math> = r = <math>\frac{5}{22}</math>. | ||
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+ | Case 3: B is divisible by 5. | ||
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+ | In this case, the sum of the numerator and denominator of 55r is 11a + <math>\frac{b}{5}</math>. This is equal to a+b, and solving that equation gives <math>\frac{a}{b}</math> = <math>\frac{2}{25}</math> = r. | ||
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+ | Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give <math>\frac{169}{550}</math>, so the answer is 169 + 550 = 719. | ||
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+ | ~Anonymus | ||
==Problem 6== | ==Problem 6== |
Revision as of 16:21, 16 February 2023
2023 AIME II (Answer Key) | AoPS Contest Collections • PDF | ||
Instructions
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1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 |
Contents
[hide]Problem 1
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is Find the greatest number of apples growing on any of the six trees.
Problem 2
Recall that a palindrome is a number that reads the same forward and backward. Find the greatest integer less than that is a palindrome both when written in base ten and when written in base eight, such as
Problem 3
Let be an isosceles triangle with There exists a point inside such that and Find the area of
Problem 4
Let and be real numbers satisfying the system of equations Let be the set of possible values of Find the sum of the squares of the elements of
Problem 5
Let be the set of all positive rational numbers such that when the two numbers and are written as fractions in lowest terms, the sum of the numerator and denominator of one fraction is the same as the sum of the numerator and denominator of the other fraction. The sum of all the elements of can be expressed in the form where and are relatively prime positive integers. Find
Since 55r is greater than r, we know 55 r can be simplified. Call r . This means 55r is . Since 55r can clearly be simplified, let's look at the possibilities, where b is divisible by factors of 55.
Case 1: B is divisible by 55.
If this is the case, the sum of the numerator and denominator of 55r is a + . This clearly cannot equal a+b, so this case is invalid.
Case 2: B is divisible by 11.
If this is the case, the sum of the numerator and denominator of 55r is 5a + . This is equal to a+b. Solving this equation gives = r = .
Case 3: B is divisible by 5.
In this case, the sum of the numerator and denominator of 55r is 11a + . This is equal to a+b, and solving that equation gives = = r.
Clearly, B is divisible by 1 doesn't make sense, so these are the only possibilities. Summing these fractions give , so the answer is 169 + 550 = 719.
~Anonymus
Problem 6
Consider the L-shaped region formed by three unit squares joined at their sides, as shown below. Two points and are chosen independently and uniformly at random from inside the region. The probability that the midpoint of also lies inside this L-shaped region can be expressed as where and are relatively prime positive integers. Find
Problem 7
Each vertex of a regular dodecagon (-gon) is to be colored either red or blue, and thus there are possible colorings. Find the number of these colorings with the property that no four vertices colored the same color are the four vertices of a rectangle.
Problem 8
Let where Find the value of the product
Problem 9
Circles and intersect at two points and and their common tangent line closer to intersects and at points and respectively. The line parallel to that passes through intersects and for the second time at points and respectively. Suppose and Then the area of trapezoid is where and are positive integers and is not divisible by the square of any prime. Find
Problem 10
Let be the number of ways to place the integers through in the cells of a grid so that for any two cells sharing a side, the difference between the numbers in those cells is not divisible by One way to do this is shown below. Find the number of positive integer divisors of
Problem 11
Find the number of collections of distinct subsets of with the property that for any two subsets and in the collection,
Problem 12
In with side lengths and let be the midpoint of Let be the point on the circumcircle of such that is on There exists a unique point on segment such that Then can be written as where and are relatively prime positive integers. Find
Problem 13
Let be an acute angle such that Find the number of positive integers less than or equal to such that is a positive integer whose units digit is
Problem 14
A cube-shaped container has vertices and where and are parallel edges of the cube, and and are diagonals of faces of the cube, as shown. Vertex of the cube is set on a horizontal plane so that the plane of the rectangle is perpendicular to vertex is meters above vertex is meters above and vertex is meters above The cube contains water whose surface is parallel to at a height of meters above The volume of water is cubic meters, where and are relatively prime positive intgers. Find
Problem 15
For each positive integer let be the least positive integer multiple of such that Find the number of positive integers less than or equal to that satisfy
See also
2023 AIME II (Problems • Answer Key • Resources) | ||
Preceded by 2023 AIME I |
Followed by 2024 AIME I | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
- American Invitational Mathematics Examination
- AIME Problems and Solutions
- Mathematics competition resources
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.