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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 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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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 Calculus Math Olympiad USA Math Olympiad

Sequence and greatest integer | AIME I, 2000 | Question 11

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

Sequence and greatest integer – AIME I, 2000


Let S be the sum of all numbers of the form \(\frac{a}{b}\),where a and b are relatively prime positive divisors of 1000, find greatest integer that does not exceed \(\frac{S}{10}\).

  • is 107
  • is 248
  • is 840
  • cannot be determined from the given information

Key Concepts


Equation

Algebra

Integers

Check the Answer


Answer: is 248.

AIME I, 2000, Question 11

Elementary Number Theory by Sierpinsky

Try with Hints


We have 1000=(2)(2)(2)(5)(5)(5) and \(\frac{a}{b}=2^{x}5^{y} where -3 \leq x,y \leq 3\)

sum of all numbers of form \(\frac{a}{b}\) such that a and b are relatively prime positive divisors of 1000

=\((2^{-3}+2^{-2}+2^{-1}+2^{0}+2^{1}+2^{2}+2^{3})(5^{-3}+5^{-2}+5^{-1}+5^{0}+5^{1}+5^{2}+5^{3})\)

\(\Rightarrow S= \frac{(2^{-3})(2^{7}-1)}{2-1} \times\) \(\frac{(5^{-3})(5^{7}-1)}{5-1}\)

=2480 + \(\frac{437}{1000}\)

\(\Rightarrow [\frac{s}{10}]\)=248.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Series and sum | AIME I, 1999 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Series and sum.

Series and sum – AIME I, 1999


given that \(\displaystyle\sum_{k=1}^{35}sin5k=tan\frac{m}{n}\) where angles are measured in degrees, m and n are relatively prime positive integer that satisfy \(\frac{m}{n} \lt 90\), find m+n.

  • is 107
  • is 177
  • is 840
  • cannot be determined from the given information

Key Concepts


Angles

Triangles

Side Length

Check the Answer


Answer: is 177.

AIME I, 2009, Question 5

Plane Trigonometry by Loney

Try with Hints


s=\(\displaystyle\sum_{k=1}^{35}sin5k\)

s(sin5)=\(\displaystyle\sum_{k=1}^{35}sin5ksin5=\displaystyle\sum_{k=1}^{35}(0.5)[cos(5k-5)-cos(5k+5)]\)=\(\frac{1+cos5}{sin5}\)

\(=\frac{1-cos(175)}{sin175}\)=\(tan\frac{175}{2}\) then m+n=175+2=177.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Problem on Fibonacci sequence | AIME I, 1988 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1988 based on Fibonacci sequence.

Fibonacci sequence Problem – AIME I, 1988


Find a if a and b are integers such that \(x^{2}-x-1\) is a factor of \(ax^{17}+bx^{16}+1\).

  • is 107
  • is 987
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Digits

Sets

Check the Answer


Answer: is 987.

AIME I, 1988, Question 13

Elementary Number Theory by David Burton

Try with Hints


Let F(x)=\(ax^{17}+bx^{16}+1\)

Let P(x) be polynomial such that

\(P(x)(x^{2}-x-1)=F(x)\)

constant term of P(x) =(-1)

now \((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\) where \(c_{i}\)=coefficient

comparing the coefficients of x we get the terms

since F(x) has no x term, then \(c_{15}\)=1

getting \(c_{14}\)

\((x^{2}-x-1)(c_1x^{15}+c_2x^{14}+….+c_{15}x-1)\)

=terms +\(0x^{2}\) +terms

or, \(c_{14}=-2\)

proceeding in the same way \(c_{13}=3\), \(c_{12}=-5\), \(c_{11}=8\) gives a pattern of Fibonacci sequence

or, coefficients of P(x) are Fibonacci sequence with alternating signs

or, a=\(c_1=F_{16}\) where \(F_{16}\) is 16th Fibonacci number

or, a=987.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Sequence and Integers | AIME I, 2007 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2007 based on Sequence and Integers.

Sequence and Integers – AIME I, 2007


A sequence is defined over non negetive integral indexes in the following way \(a_0=a_1=3\), \( a_{n+1}a_{n-1}=a_n^{2}+2007\), find the greatest integer that does not exceed \(\frac{a_{2006}^{2}+a_{2007}^{2}}{a_{2006}a_{2007}}\)

  • is 107
  • is 224
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Inequalities

Integers

Check the Answer


Answer: is 224.

AIME I, 2007, Question 14

Elementary Number Theory by David Burton

Try with Hints


\(a_{n+1}a_{n-1}\)=\(a_{n}^{2}+2007\) then \(a_{n-1}^{2} +2007 =a_{n}a_{n-2}\) adding these \(\frac{a_{n-1}+a_{n+1}}{a_{n}}\)=\(\frac{a_{n}+a_{n-2}}{a_{n-1}}\), let \(b_{j}\)=\(\frac{a_{j}}{a_{j-1}}\) then \(b_{n+1} + \frac{1}{b_{n}}\)=\(b_{n}+\frac{1}{b_{n-1}}\) then \(b_{2007} + \frac{1}{b_{2006}}\)=\(b_{3}+\frac{1}{b_{2}}\)=225

here \(\frac{a_{2007}a_{2005}}{a_{2006}a_{2005}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\) then \(b_{2007}\)=\(\frac{a_{2007}}{a_{2006}}\)=\(\frac{a_{2006}^{2}+2007}{a_{2006}a_{2005}}\)\( \gt \)\(\frac{a_{2006}}{a_{2005}}\)=\(b_{2006}\)

then \(b_{2007}+\frac{1}{b_{2007}} \lt b_{2007}+\frac{1}{b_{2006}}\)=225 which is small less such that all \(b_{j}\) s are greater than 1 then \(\frac{a_{2006}^{2}+ a_{2007}^{2}}{a_{2006}a_{2007}}\)=\(b_{2007}+\frac{1}{b_{2007}}\)=224.

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

Number and Series | Number Theory | AIME I, 2015

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.

Number Theory and Series – AIME 2015


The expressions A = \((1 \times 2)+(3 \times 4)+….+(35 \times 36)+37\) and B = \(1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)\) are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.

  • is 107
  • is 648
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Number Theory

Check the Answer


Answer: is 648.

AIME, 2015, Question 1

Elementary Number Theory by David Burton

Try with Hints


B-A=\(-36+(2 \times 3)+….+(2 \times 36)\)

=\(-36+4 \times (1+2+3+….+18)\)

=\(-36+(4 \times \frac{18 \times 19}{2})\)=648.

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