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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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Math Olympiad Math Olympiad Videos USA Math Olympiad

Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

Geometric Sequence Problem – AIME 2009


Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

  • is 500
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Sequence

Series

Real Analysis

Check the Answer


Answer: is 840.

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

Try with Hints


3-digit sequence a, ar, \(ar^{2}\). The largest geometric number must have a<=9.

ar \(ar^{2}\) less than 9 r fraction less than 1 For a=9 is \(\frac{2}{3}\) then number 964.

a>=1 ar and \(ar^{2}\) greater than 1 r is 2 and number is 124. Then difference 964-124=840.

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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 Math Olympiad USA Math Olympiad

GCD and Sequence | AIME I, 1985 | Question 13

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1985 based on GCD and Sequence.

GCD and Sequence – AIME I, 1985


The numbers in the sequence 101, 104,109,116,…..are of the form \(a_n=100+n^{2}\) where n=1,2,3,——-, for each n, let \(d_n\) be the greatest common divisor of \(a_n\) and \(a_{n+1}\), find the maximum value of \(d_n\) as n ranges through the positive integers.

  • is 107
  • is 401
  • is 840
  • cannot be determined from the given information

Key Concepts


GCD

Sequence

Integers

Check the Answer


Answer: is 401.

AIME I, 1985, Question 13

Elementary Number Theory by David Burton

Try with Hints


\(a_n=100+n^{2}\) \(a_{n+1}=100+(n+1)^{2}=100 + n^{2} +2n +1\) and \(a_{n+1}-a_{n}=2n +1\)

\(d_{n}|(2n+1)\) and \(d_{n}|(100 +n^{2})\) then \(d_{n}|[(100+n^{2})-100(2n+1)]\) then \(d_{n}|(n^{2}-200n)\)

here \(n^{2} -200n=0\) then n=200 then \(d_{n}\)=2n+1=2(200)+1=401.

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AMC 10 USA Math Olympiad

Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.

Problem on Series and Sequences (SMO Test)


For each positive integer \(n \geq 1\) , we define the recursive relation given by \(a_{n+1} = \frac {1}{1+a_{n}} \).

Suppose that \(a_{1} = a_{2012}\).Find the sum of the squares of all

possible values of \(a_{1}\).

  • 2
  • 3
  • 6
  • 12

Key Concepts


Series and Sequence

Functional Equation

Recursive Relation

Check the Answer


Answer: 3

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


If you got stuck start from here :

At first we have to understand the sequence it is following

Given that : \(a_{n+1} = \frac {1}{1+a_{n}} \)

Let \(a_{1} = a\)

so , \(a_{2} = \frac {1}{1 + a_{1}}\) = \(\frac {1}{1 + a}\)

Again, \(a_{3} = \frac {1}{1+a_{2}} = \frac {1+a}{2+a} \)

For , \(a_{4} = \frac {1}{1+a_{3}} = \frac {2+a}{3+2a} \)

And , \(a_{5} = \frac {1}{1+a_{4}} = \frac {3+2a}{5+3a} \)

and so on……..

Try to do the rest …………………………….

Looking at the previous hint ………………

In general we can say ……………..

\(a_{n} = \frac {F_{n} + F_{n-1}a}{F_{n+1} +F_{n}a}\)

Where \(F_{1} = 0 , F_{ 2} = 1\) and \(F_{n+1} = F_{n} \) for all value of \(n\geq 1\)

Try to do the rest …….

Here is the rest of the solution,

If \(a_{2012} = \frac {F_{2012}+F_{2011}a}{F_{2013} + F{2012}a} = a \)

Then \((a^2+a-1 )F_{2012} = 0\)

Since \(F_{2012}>0\) we have \(a^2 +a -1 = 0\) ……………………….(1)

Assume x and y are the two roots of the \(eq^n (1)\), then

\(x^2 + y^2 = (x+y)^2 -2xy = (-1)^2 – 2(-1) = 3\) (Answer)

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Trigonometry Problem | AIME I, 2015 | Question 13

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

Trigonometry Problem – AIME 2015


With all angles measured in degrees, the product (\prod_{k=1}^{45} \csc^2(2k-1)^\circ=m^n), where (m) and (n) are integers greater than 1. Find (m+n).

  • is 107
  • is 91
  • is 840
  • cannot be determined from the given information

Key Concepts


Trigonometry

Sequence

Algebra

Check the Answer


Answer: is 91.

AIME, 2015, Question 13.

Plane Trigonometry by Loney .

Try with Hints


Let (x = \cos 1^\circ + i \sin 1^\circ). Then from the identity[\sin 1 = \frac{x – \frac{1}{x}}{2i} = \frac{x^2 – 1}{2 i x},]we deduce that (taking absolute values and noticing (|x| = 1))[|2\sin 1| = |x^2 – 1|.]

But because \(\csc\) is the reciprocal of \(\sin\) and because \(\sin z = \sin (180^\circ – z)\), if we let our product be \(M\) then[\frac{1}{M} = \sin 1^\circ \sin 3^\circ \sin 5^\circ \dots \sin 177^\circ \sin 179^\circ][= \frac{1}{2^{90}} |x^2 – 1| |x^6 – 1| |x^{10} – 1| \dots |x^{354} – 1| |x^{358} – 1|]because \(\sin\) is positive in the first and second quadrants.

Now, notice that \(x^2, x^6, x^{10}, \dots, x^{358}\) are the roots of \(z^{90} + 1 = 0.\) Hence, we can write \((z – x^2)(z – x^6)\dots (z – x^{358}) = z^{90} + 1\), and so[\frac{1}{M} = \dfrac{1}{2^{90}}|1 – x^2| |1 – x^6| \dots |1 – x^{358}| = \dfrac{1}{2^{90}} |1^{90} + 1| = \dfrac{1}{2^{89}}.]It is easy to see that \(M = 2^{89}\) and that our answer is \(2+89=91\).

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