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AIME II Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and permutations | AIME II, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME II, 2015 based on Sequence and permutations.

Sequence and permutations – AIME II, 2015


Call a permutation \(a_1,a_2,….,a_n\) of the integers 1,2,…,n quasi increasing if \(a_k \leq a_{k+1} +2\) for each \(1 \leq k \leq n-1\), find the number of quasi increasing permutations of the integers 1,2,….,7.

  • is 107
  • is 486
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Permutations

Integers

Check the Answer


Answer: is 486.

AIME II, 2015, Question 10

Elementary Number Theory by David Burton

Try with Hints


While inserting n into a string with n-1 integers, integer n has 3 spots where it can be placed before n-1, before n-2, and at the end

Number of permutations with n elements is three times the number of permutations with n-1 elements

or, number of permutations for n elements=3 \(\times\) number of permutations of (n-1) elements

or, number of permutations for n elements=\(3^{2}\) number of permutations of (n-2) elements

……

or, number of permutations for n elements=\(3^{n-2}\) number of permutations of {n-(n-2)} elements

or, number of permutations for n elements=2 \(\times\) \(3^{n-2}\)

forming recurrence relation as the number of permutations =2 \(\times\) \(3^{n-2}\)

for n=3 all six permutations taken and go up 18, 54, 162, 486

for n=7, here \(2 \times 3^{5} =486.\)

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Numbers of positive integers | AIME I, 2012 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Numbers of positive integers.

Numbers of positive integers – AIME 2012


Find the number of positive integers with three not necessarily distinct digits, \(abc\), with \(a \neq 0\) and \(c \neq 0\) such that both \(abc\) and \(cba\) are multiples of \(4\).

  • is 107
  • is 40
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Number Theory

Algebra

Check the Answer


Answer: is 40.

AIME, 2012, Question 1.

Elementary Number Theory by David Burton .

Try with Hints


Here a number divisible by 4 if a units with tens place digit is divisible by 4

Then case 1 for 10b+a and for 10b+c gives 0(mod4) with a pair of a and c for every b

[ since abc and cba divisible by 4 only when the last two digits is divisible by 4 that is 10b+c and 10b+a is divisible by 4]

and case II 2(mod4) with a pair of a and c for every b

Then combining both cases we get for every b gives a pair of a s and a pair of c s

So for 10 b’s with 2 a’s and 2 c’s for every b gives \(10 \times 2 \times 2\)

Then number of ways \(10 \times 2 \times 2\) = 40 ways.

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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


Answer: is 489.

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


\(10 \choose 3\) sets of 3 points which form triangles,

fourth distinct segment excluding 3 segments of triangles=45-3=42

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

Arithmetic Sequence Problem – AIME 2012


The terms of an arithmetic sequence add to \(715\). The first term of the sequence is increased by \(1\), the second term is increased by \(3\), the third term is increased by \(5\), and in general, the \(k\)th term is increased by the \(k\)th odd positive integer. The terms of the new sequence add to \(836\). Find the sum of the first, last, and middle terms of the original sequence.

  • is 107
  • is 195
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Number Theory

Algebra

Check the Answer


Answer: is 195.

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

Try with Hints


After the adding of the odd numbers, the total of the sequence increases by \(836 – 715 = 121 = 11^2\).

Since the sum of the first \(n\) positive odd numbers is \(n^2\), there must be \(11\) terms in the sequence, so the mean of the sequence is \(\frac{715}{11} = 65\).

Since the first, last, and middle terms are centered around the mean, then \(65 \times 3 = 195\)

Hence option B correct.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Algebra and Positive Integer | AIME I, 1987 | Question 8

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1987 based on Algebra and Positive Integer.

Algebra and Positive Integer – AIME I, 1987


What is the largest positive integer n for which there is a unique integer k such that \(\frac{8}{15} <\frac{n}{n+k}<\frac{7}{13}\)?

  • is 107
  • is 112
  • is 840
  • cannot be determined from the given information

Key Concepts


Digits

Algebra

Numbers

Check the Answer


Answer: is 112.

AIME I, 1987, Question 8

Elementary Number Theory by David Burton

Try with Hints


Simplifying the inequality gives, 104(n+k)<195n<105(n+k)

or, 0<91n-104k<n+k

for 91n-104k<n+k, K>\(\frac{6n}{7}\)

and 0<91n-104k gives k<\(\frac{7n}{8}\)

so, 48n<56k<49n for 96<k<98 and k=97

thus largest value of n=112.

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Distance Time | AIME I, 2012 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2012 based on Distance Time.

Distance Time – AIME 2012


When they meet at the milepost, Sparky has been ridden for n miles total. Assume Butch rides Sparky for a miles, and Sundance rides for n-a miles. Thus, we can set up an equation, given that Sparky takes \(\frac{1}{6}\) hours per mile, Butch takes \(\frac{1}{4}\) hours per mile, and Sundance takes \(\frac{2}{5}\) hours per mile.

  • is 107
  • is 279
  • is 840
  • cannot be determined from the given information

Key Concepts


Time

Distance

Speed

Check the Answer


Answer: is 279.

AIME, 2012, Question 4

Problem Solving Strategies by Arther Engel

Try with Hints


After meeting at milepost, Sparky for n miles. Let Butch with Sparky for a miles Sundance with Sparky for n-a miles.

Then
\(\frac{a}{6} + \frac{n-a}{4}\) = \(\frac{n-a}{6} + \frac{2a}{5}\) implies that \(a = \frac{5}{19}n\)

Then integral value of n is 19 and a = 5 and \(t = \frac{13}{3}\) hours that is 260 minutes. Then \(19 + 260 = {279}\).

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Algebra Arithmetic Math Olympiad USA Math Olympiad

Arithmetic Mean | AIME I, 2015 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Arithmetic Mean.

Arithmetic Mean of Number Theory – AIME 2015


Consider all 1000-element subsets of the set {1, 2, 3, … , 2015}. From each such subset choose the least element. The arithmetic mean of all of these least elements is \(\frac{p}{q}\), where \(p\) and \(q\) are relatively prime positive integers. Find \(p + q\).

  • is 107
  • is 431
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequalities

Algebra

Number Theory

Check the Answer


Answer: is 431.

AIME, 2015, Question 12

Elementary Number Theory by David Burton

Try with Hints


Each 1000-element subset \({ a_1, a_2,a_3,…,a_{1000}}\) of \({1,2,3,…,2015}\) with \(a_1<a_2<a_3<…<a_{1000}\) contributes \(a_1\) to sum of least element of each subset and set \({a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\). \(a_1\) ways to choose a positive integer \(k\) such that \(k<a_1+1<a_2+1,a_3+1<…<a_{1000}+1\) (\(k\) can be anything from \(1\) to \(a_1\) inclusive

Thus, the number of ways to choose the set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is equal to the sum. But choosing a set \({k,a_1+1,a_2+1,a_3+1,…,a_{1000}+1}\) is same as choosing a 1001-element subset from \({1,2,3,…,2016}\)!

average =\(\frac{2016}{1001}\)=\(\frac{288}{143}\). Then \(p+q=288+143={431}\)

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Arithmetic Geometry Math Olympiad USA Math Olympiad

Arithmetic and geometric mean | AIME I, 2000 Question 6

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2000 based on Arithmetic and geometric mean with Algebra.

Arithmetic and geometric mean with Algebra – AIME 2000


Find the number of ordered pairs (x,y) of integers is it true that \(0 \lt y \lt 10^{6}\) and that the arithmetic mean of x and y is exactly 2 more than the geometric mean of x and y.

  • is 107
  • is 997
  • is 840
  • cannot be determined from the given information

Key Concepts


Algebra

Equations

Ordered pair

Check the Answer


Answer: is 997.

AIME, 2000, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


 given that \(\frac{x+y}{2}=2+({xy})^\frac{1}{2}\) then solving we have \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and-2

given that \(y \gt x\) then \(y^\frac{1}{2}\)-\(x^\frac{1}{2}\)=+2 and here maximum integer value of \(y^\frac{1}{2}\)=\(10^{3}-1\)=999 whose corresponding \(x^\frac{1}{2}\)=997 and decreases upto \(y^\frac{1}{2}\)=3 whose corresponding \(x^\frac{1}{2}\)=1

then number of pairs (\(x^\frac{1}{2}\),\(y^\frac{1}{2}\))=number of pairs of (x,y)=997.

.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and fraction | AIME I, 2000 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2000 based on Sequence and fraction.

Sequence and fraction – AIME I, 2000


A sequence of numbers \(x_1,x_2,….,x_{100}\) has the property that, for every integer k between 1 and 100, inclusive, the number \(x_k\) is k less than the sum of the other 99 numbers, given that \(x_{50}=\frac{m}{n}\), where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 173
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 173.

AIME I, 2000, Question 10

Elementary Number Theory by Sierpinsky

Try with Hints


Let S be the sum of the sequence \(x_k\)

given that \(x_k=S-x_k-k\) for any k

taking k=1,2,….,100 and adding

\(100S-2(x_1+x_2+….+x_{100})=1+2+….+100\)

\(\Rightarrow 100S-2S=\frac{100 \times 101}{2}=5050\)

\(\Rightarrow S=\frac{2525}{49}\)

for \(k=50, 2x_{50}=\frac{2525}{49}-50=\frac{75}{49}\)

\(\Rightarrow x_{50}=\frac{75}{98}\)

\(\Rightarrow m+n\)=75+98

=173.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Finding smallest positive Integer | AIME I, 1996 Problem 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME I, 1996 based on Finding the smallest positive Integer.

Finding smallest positive Integer – AIME I, 1996


Find the smallest positive integer solution to \(tan19x=\frac{cos96+sin96}{cos96-sin96}\).

  • is 107
  • is 159
  • is 840
  • cannot be determined from the given information

Key Concepts


Functions

Trigonometry

Integers

Check the Answer


Answer: is 159.

AIME I, 1996, Question 10

Plane Trigonometry by Loney

Try with Hints


\(\frac{cos96+sin96}{cos96-sin96}\)

=\(\frac{sin(90+96)+sin96}{sin(90+96)-sin96}\)

=\(\frac{sin186+sin96}{sin186-sin96}\)

=\(\frac{sin(141+45)+sin(141-45)}{sin(141+45)-sin(141-45)}\)

=\(\frac{2sin141cos45}{2cos141sin45}\)

=tan141

here \(tan(180+\theta)\)=\(tan\theta\)

\(\Rightarrow 19x=141+180n\) for some integer n is first equation

multiplying equation with 19 gives

\(x \equiv 141\times 19 \equiv 2679 \equiv 159(mod180)\) [since 2679 divided by 180 gives remainder 159]

\(\Rightarrow x=159\).

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