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


First hint

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

Second Hint

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

Final Step

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

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ISI MStat 2018 PSA Problem 11 | Sequence & it's subsequence

This is a problem from ISI MStat 2018 PSA Problem 11 based on Sequence and subsequence.

Sequence & it's subsequence - ISI MStat Year 2018 PSA Problem 11


Let \( {a_{n}}_{n \geq 1}\) be a sequence such that \( a_{1} \leq a_{2} \leq \cdots \leq a_{n} \leq \cdots\)
Suppose the subsequence \( {a_{2 n}}_{n \geq 1}\) is bounded. Then

  • (A) \( \{a_{2 n}\}_{n \geq 1}\) is always convergent but \( \{a_{2 n+1}\}_{n \geq 1} \) need not be convergent.
  • (B) both \( \{a_{2 n}\}_{n \geq 1}\) and \( \{a_{2 n+1}\}_{n \geq 1}\) are always convergent and have the same limit.
  • (C) \( \{a_{3 n}\}_{n \geq 1}\) is not necessarily convergent.
  • (D) both \( \{a_{2 n}\}_{n \geq 1}\) and \( \{a_{2 n+1}\}_{n \geq 1}\) are always convergent but may have different limits.

Key Concepts


Sequence

Subsequence

Check the Answer


Answer: is (B)

ISI MStat 2018 PSA Problem 11

Introduction to Real Analysis by Bertle Sherbert

Try with Hints


Given that \( a_{2n} \) is bounded . Again we have \( a_{1} \leq a_{2} \leq \cdots \leq a_{2n} \leq a_{2n+1} \leq a_{2n+2} \leq \cdots\) which shows that if \( a_{2n} \) is bounded then \( a_{2n+1} \) is also bounded .

Again both \( a_{2n} \) and \( a_{2n+1} \) both are monotonic sequence . Hence both converges .

Now we have to see whether they converges to same limit or not ?

As both \( a_{2n} \) and \( a_{2n+1} \) are bounded hence \( a_{n} \) is bounded and it's already given that it is monotonic . Hence \( a_{n} \) converges . So, it's subsequences must converges to same limit . Hence option (B) is correct .

ISI MStat 2018 PSA Problem 11
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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ISI MStat 2018 PSA Problem 12 | Sequence of positive numbers

This is a problem from ISI MStat 2018 PSA Problem 12 based on Sequence of positive numbers

Sequence of positive numbers - ISI MStat Year 2018 PSA Question 12


Let \(a_n \) ,\( n \ge 1\) be a sequence of positive numbers such that \(a_{n+1} \leq a_{n}\) for all n, and \(\lim {n \rightarrow \infty} a{n}=a .\) Let \(p_{n}(x)\) be the polynomial \( p_{n}(x)=x^{2}+a_{n} x+1\) and suppose \(p_{n}(x)\) has no real roots for every n . Let \(\alpha\) and \(\beta\) be the roots of the polynomial \(p(x)=x^{2}+a x+1 .\) What can you say about \( (\alpha, \beta) \)?

  • (A) \( \alpha=\beta, \alpha\) and \(\beta\) are not real
  • (B) \( \alpha=\beta, \alpha\) and \(\beta\) are real.
  • (C) \(\alpha \neq \beta, \alpha\) and \(\beta\) are real.
  • (D) \(\alpha \neq \beta, \alpha\) and \(\beta\) are not real

Key Concepts


Sequence

Quadratic equation

Discriminant

Check the Answer


Answer: is (D)

ISI MStat 2018 PSA Problem 12

Introduction to Real Analysis by Bertle Sherbert

Try with Hints


Write the discriminant. Use the properties of the sequence \( a_n \) .

Note that as $P_n$ has no real root so discriminant is $(a_n)^2-4<0$ so $|a_n|<2$ and $a_n$'s are positive and decreasing so $0\leq a_n<2$ . So , what can we say about a ?

Therefore we can say that \( 0 \le a < 2 \) hence discriminant of P  , \(a^2-4 \) must be strictly negative so option D.

ISI MStat 2018 PSA Problem 12
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Telescopic Continuity | ISI MStat 2015 PSB Problem 1

This problem is a simple application of the sequential definition of continuity from ISI MStat 2015 PSB Problem 1 based on Telescopic Continuity.

Problem- Telescopic Continuity

Let \(f: R \rightarrow R\) be a function which is continuous at 0 and \(f(0)=1\)
Also assume that \(f\) satisfies the following relation for all \(x\) :
$$
f(x)-f(\frac{x}{2})=\frac{3 x^{2}}{4}+x
$$ Find \(f(3)\).

Prerequisites

Solution

$$
f(3)-f(\frac{3}{2})=\frac{3 \times 3^{2}}{4}+3
$$

$$
f(\frac{3}{2})-f(\frac{\frac{3}{2}}{2})=\frac{3 \times \frac{3}{2}^{2}}{4}+\frac{3}{2}
$$

$$
f(\frac{3}{2^2})-f(\frac{\frac{3}{2^2}}{2})=\frac{3 \times \frac{3}{2^2}^{2}}{4}+\frac{3}{2^2}
$$

\( \cdots \)

$$
f(\frac{3}{2^n})-f(\frac{\frac{3}{2^n}}{2})=\frac{3 \times \frac{3}{2^n}^{2}}{4}+\frac{3}{2^n}
$$

Add them all up. That's the telescopic elegance.

$$
f(3)-f(\frac{3}{2^{n+1}})= \frac{3 \times 3^{2}}{4} \times \sum_{k = 0}^{n} \frac{1}{2^{2k}} + 3 \times \sum_{k = 0}^{n} \frac{1}{2^k} \rightarrow [*]
$$

Observe that \( a_n \to 0 \Rightarrow f(a_n) \to f(0) = 1\) since, \(f(x)\) is continuous at \(x=0\).

Hence take limit \( n \to \infty \) on \([*]\), and we get \( f(3) - f(0) = \frac{3 \times 3^{2}}{4} \times \frac{4}{3} + 3 \times 2 = 15 \).

Food for Thought

Combination of Sequence | B.Stat Objective | TOMATO 79

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Combination of Sequence.

Logic and combination of sequence (B.Stat Objective)


The two sequence of numbers {1,4,16,64,.....} and {3,12,48,192,.....} are mixed as follows {1,3,4,12,16,48,64,192,....}. One of the numbers in the mixed series is 1048576. Then the number immediately preceeding it is

  • 262144
  • 786432
  • 814572
  • 786516

Key Concepts


Logic

Sequence

Integers

Check the Answer


Answer: 786432.

B.Stat Objective Question 79

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

The first series is of form \(4^{r}\) for \(r \geq 0\) \(r \in\) set of natural numbers the second series is of form \(3 \times 4^{r}\) for \(r \geq 0\) \(r \in\) set of natural numbers and the third series is of \(4^{r}\),\(3 \times 4^{r}\) in alternate element form for \(r \geq 0\) \(r \in\) set of natural numbers

Second Hint

given that 1048576=\(4^{r}\)=\(4^{10}\)

Final Step

then preceeding term \(3 \times 4^{9}\)=(3)(262144)=786432.

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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


First hint

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

Second Hint

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

Final Step

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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Composite number Problem | B.Stat Objective | TOMATO 75

Try this TOMATO problem from I.S.I. B.Stat Entrance Objective Problem based on Sequence and composite number.

Composite number Problem (B.Stat Objective)


Consider the sequence \(a_1\)=101, \(a_2\)=10101,\(a_3\)=1010101 and so on. Then \(a_k\) is a composite number ( that is not a prime number)

  • if and only if \(k \geq 2\) and \(11|(10^{k+1}+1)\)
  • if and only if \(k \geq 2\) and k-2 is divisible by 3
  • if and only if \(k \geq 2\) and \(11|(10^{k+1}-1)\)
  • if and only if \(k \geq 2\)

Key Concepts


Logic

Sequence

Composite number

Check the Answer


Answer: if and only if \(k \geq 2\) and k-2 is divisible by 3

B.Stat Objective Question 75

Challenges and Thrills of Pre-College Mathematics by University Press

Try with Hints


First hint

for \(a_k\) \(k \geq 2\) may be prime also then not considering this here

Second Hint

for \(a_{8}\) \(10^{9}-1\) and \(10^{9}+1\) not divisible by 11

Final Step

8-2 is divisible by 3 and \(a_{8}\) is composite number then \(a_{k}\) is composite if and only if \(k \geq 2\) and k-2 is divisible by 3.

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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


First hint

\(a_{n+1}a_{n-1}\)=\(a_{n}^{2}+2007\) 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

Second Hint

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

Final Step

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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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


First hint

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

Second Hint

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

Final Step

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

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Problem on Series | SMO, 2009 | Problem No. 25

Try this beautiful problem from Singapore Mathematics Olympiad, SMO, 2009 based on Problem on Series.

Problem on Series | SMO Test


Given that \(x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n\) where each \( a_r\) is an integer , \(r = 0,1,2,...,n\)

Find the value of n such that \(a_0 +a_1 +a_2+a_3 +...........+a_{n-2}+a_{n-1} = 60 -\frac{n(n-1)}{2}\)?

  • 2
  • 5
  • 6
  • 0

Key Concepts


Series Problem

Algebra

Check the Answer


Answer : 5

Singapore Mathematics Olympiad, 2009

Challenges and Thrills - Pre - college Mathematics

Try with Hints


If you got stuck in this sum then we can try by understanding the pattern we are using here.If we assume x = 1 thenn the expression will be

\(x+(1+x)^2+(1+x)^3+.........(1+x)^n = a_0 + a_1 x+a_2 x^2+..+a_n x^n\)

The right hand side of this equation will be

\(a_0 +a_1 + a_2 +a_3+..................+ a_n\)

and the left hand side of the given equation be like

\( 1 + 2^2 + 2^ 3 + ......+2^n\)

\( 1 + 2^2 + 2^ 3 + ......+2^n\) = \(a_0 +a_1 + a_2 +a_3+..................+ a_n\)

Try the rest of the sum..............

In the next hint we continue from the previous hint:

so the expression , \( 1 + 2^2 + 2^ 3 + ......+2^n\) = \(2^{n+1} - 3\)

Again , \(a_1 = 1+2+3+..........+n = \frac {n(n+1)}{2}\)

so \(a_n = 1\)

Now we have almost reach the final step.I am not showing it now.Try first.........

Now coming back to the last step :

\( 60 - \frac {n(n+1)}{2} + \frac {n(n+1}{2} + 1 = (2^{n+1} - 3\)

n = 5 (Answer)

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