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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ISI MSTAT PSB 2011 Problem 4 | Digging deep into Multivariate Normal

This is an interesting problem from ISI MSTAT PSB 2011 Problem 4 that tests the student's knowledge of how he visualizes the normal distribution in higher dimensions.

The Problem: ISI MSTAT PSB 2011 Problem 4

Suppose that \( X_1,X_2,... \) are independent and identically distributed \(d\) dimensional normal random vectors. Consider a fixed \( x_0 \in \mathbb{R}^d \) and for \(i=1,2,...,\) define \(D_i = \| X_i - x_0 \| \), the Euclidean distance between \( X_i \) and \(x_0\). Show that for every \( \epsilon > 0 \), \(P[\min_{1 \le i \le n} D_i > \epsilon] \rightarrow 0 \) as \( n \rightarrow \infty \)

Prerequisites:

  1. Finding the distribution of the minimum order statistic
  2. Multivariate Gaussian properties

Solution:

First of all, see that \( P(\min_{1 \le i \le n} D_i > \epsilon)=P(D_i > \epsilon)^n \) (Verify yourself!)

But, apparently we are more interested in the event \( \{D_i < \epsilon \} \).

Let me elaborate why this makes sense!

Let \( \phi \) denote the \( d \) dimensional Gaussian density, and let \( B(x_0, \epsilon) \) be the Euclidean ball around \( x_0 \) of radius \( \epsilon \) . Note that \( \{D_i < \epsilon\} \) is the event that the gaussian \( X_i \) will land in this Euclidean ball.

So, if we can show that this event has positive probability for any given $x_0, \epsilon$ pair, we will be done, since then in the limit, we will be exponentiating a number strictly less than 1 by a quantity that is growing larger and larger.

In particular, we have that : \( P(D_i < \epsilon)= \int_{B(x_0, \epsilon)} \phi(x) dx \geq |B(x_0, \epsilon)| \inf_{x \in B(x_0, \epsilon)} \phi(x) \) , and we know that by rotational symmetry and as Gaussians decay as we move away from the centre, this infimum exists and is given by \( \phi(x_0 + \epsilon \frac{x_0}{||x_0||}) \) . (To see that this is indeed a lower bound, note that \( B(x_0, \epsilon) \subset B(0, \epsilon + ||x_0||) \).

So, basically what we have shown here is that exists a \( \delta > 0 \) such that \( P(D_i < \epsilon )>\delta \).

As, \( \delta \) is a lower bound of a probability , hence it a fraction strictly below 1.

Thus, we have \( \lim_{n \rightarrow \infty} P(D_i > \epsilon)^n \leq \lim_{n \rightarrow \infty} (1-\delta)^n = 0 \).

Hence we are done.

Food for thought:

There is a fantastic amount of statistical literature on the equi-density contours of a multivariate Gaussian distribution .

Try to visualize them for non singular and a singular Gaussian distribution separately. They are covered extensively in the books of Kotz and Anderson. Do give it a read!

Some Useful Problems:

Problem from Inequality | PRMO-2018 | Problem 23

Try this beautiful Algebra problem from PRMO, 2018 based on Inequality.

Problem from Inequality | PRMO | Problem-23


What is the largest positive integer n such that \(\frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\)\(\geq n(a+b+c)\).

  • $20$
  • $24$
  • $13$

Key Concepts


algebra

Inequality

Check the Answer


Answer:\(14\)

PRMO-2018, Problem 23

Pre College Mathematics

Try with Hints


we have to find out the largest value of \(n\).......

we know that ,

\(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z} \geq \frac{(a+b+c)^2}{x+y+z}\) .we may use this form to getting the largest positive integer \(n\)

Can you now finish the problem ..........

Therefore,

\(\frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\)\(\geq \frac{(a+b+c)^2}{a(\frac{1}{29}+\frac{1}{31}) +b(\frac{1}{29}+\frac{1}{31})+c(\frac{1}{29}+\frac{1}{31})}\)

\(\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\) \(\geq \frac{(a+b+c)}{(\frac{1}{29} +\frac{1}{31})}\)

\(\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\)\(\geq \frac{a+b+c}{\frac{60}{29 \times 31}}\)

\(\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\)\(\geq \frac{29\times 31}{60}(a+b+c)\)

\(\Rightarrow \frac{a^2}{\frac{b}{29} +\frac{c}{31}} +\frac{b^2}{\frac{c}{29}+\frac{a}{31}} +\frac{c^2}{\frac{a}{29}+\frac{b}{31}}\)\(\geq 14.98(a+b+c)\)

Therefore \(n=14\)

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Ratio and Inequalities | AIME I, 1992 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Ratio and Inequalities.

Ratio and Inequalities - AIME I, 1992


A tennis player computes her win ratio by dividing the number of matches she has won by the total number of matches she has played. At the start of a weekend, her win ratio is exactly 0.500. During the weekend, she plays four matches, winning three and losing one. At the end of the weekend, her win ratio is greater than 0.503. Find the largest number of matches she could have won before the weekend began.

  • is 107
  • is 164
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Ratios

Inequalities

Check the Answer


Answer: is 164.

AIME I, 1992, Question 3

Elementary Algebra by Hall and Knight

Try with Hints


First hint

Let x be number of matches she has played and won then \(\frac{x}{2x}=\frac{1}{2}\)

Second Hint

and \(\frac{x+3}{2x+4}>\frac{503}{1000}\)

Final Step

\(\Rightarrow 1000x+3000 > 1006x+2012\)

\(\Rightarrow x<\frac{988}{6}\)

\(\Rightarrow\) x=164.

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Length and Inequalities | AIME I, 1994 | Question 12

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1994 based on Length and Inequalities.

Length and Inequalities - AIME I, 1994


A fenced, rectangular field measures 24 meters by 52 meters.An agricultural researcher has 1994 meters of fence that can be used for internal fencing to partition the field into congruent, square test plots. The entire field must be partitioned, and the sides of the squares must be parallel to the edges of the field, find the largest number of square test plots into which the field can be partitioned using all or some of the 1994 meters of fence.

  • is 107
  • is 702
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Inequalities

Length

Check the Answer


Answer: is 702.

AIME I, 1994, Question 12

Inequalities (Little Mathematical Library) by Korovkin

Try with Hints


First hint

Number of squares in a row=\(\frac{52n}{24}\)=\(\frac{13n}{6}\) squares in every row

Second Hint

each vertical fence lengths 24 for \(\frac{13n}{6}-1\) vertical fences

each horizontal fence lengths 52 for n-1 such fences

Final Step

total length of internal fencing 24 (\(\frac{13n}{6}-1\))+52(n-1)=104n-76 \( \leq 1994\)

\(\Rightarrow n \leq \frac{1035}{52}\)

\(\Rightarrow n \leq 19\)

the largest multiple of 6 that is \( \leq 19 \)

\(\Rightarrow n=18\)

required number =\(\frac{13n^{2}}{6}\)=702.

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Problem on Inequality | ISI - MSQMS - B, 2018 | Problem 2a

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 2a


(a) Prove that if $x>0, y>0$ and $x+y=1,$ then $\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right) \geq 9$

Key Concepts


Algebra

Inequality

Numbers

Check The Answer


But Try the Problem First...

Answer: $xy \leq \frac{1}{4}$

ISI - MSQMS - B, 2018, Problem 2A

"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"

Try with Hints


We have to show that ,

$(1+\frac{1}{x})(1+\frac{1}{y}) \geq 9$

i.e $1+ \frac{1}{x} + \frac{1}{y} +\frac{1}{xy} \geq 9$

Since $x+y =1$

Therefore the above equation becomes $\frac{2}{xy} \geq 8$

ie $xy \leq \frac{1}{4}$

Now with this reduced form of the equation why don't you give it a try yourself,I am sure you can do it.

Applying AM $\geq$ GM on $x,y$

So you are just one step away from solving your problem,go on.............

Therefore, $\frac{x+y}{2} \geq (xy)^\frac{1}{2}$

$\Rightarrow \frac{1}{2} \geq (xy)^\frac{1}{2}$

Squaring both sides we get, $xy \leq \frac{1}{4}$

Hence the result follows.

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Problem on Integral Inequality | ISI - MSQMS - B, 2015

Try this problem from ISI-MSQMS 2015 which involves the concept of Integral Inequality.

INTEGRAL INEQUALITY | ISI 2015 | MSQMS | PART B | PROBLEM 7b


Show that $1<\int_{0}^{1} e^{x^{2}} d x<e$

Key Concepts


Real Analysis

Inequality

Numbers

Check The Answer


But Try the Problem First...

ISI - MSQMS - B, 2015, Problem 7b

"INEQUALITIES: AN APPROACH THROUGH PROBLEMS BY BJ VENKATACHALA"

Try with Hints


We have to show that ,

$1<\int_{0}^{1} e^{x^{2}} d x<e$

$ 0< x <1$

It implies, $0 < x^2 <1$

Now with this reduced form of the equation why don't you give it a try yourself, I am sure you can do it.

Thus, $ e^0 < e^{x^2} <e^1 $

i.e $1 < e^{x^2} <e $

So you are just one step away from solving your problem, go on.............

Therefore, Integrating the inequality with limits $0$ to $1$ we get, $\int\limits_0^1 \mathrm dx < \int\limits_0^1 e^{x^2} \mathrm dx < \int\limits_0^1e \mathrm dx$

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Inequality Problem From ISI - MSQMS - B, 2017 | Problem 3a

Try this problem from ISI-MSQMS 2017 which involves the concept of Inequality.

INEQUALITY | ISI 2017| MSQMS | PART B | PROBLEM 3a


(a) Prove that $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a),$ for all positive distinct values of $a, b, c$

Key Concepts


Algebra

Inequality

Numbers


Try with Hints


We know if we have $n$ numbers say $a_1,a_2,.....,a_n$ then AM $\geq$ GM implies

$\frac{a_1+a_2+....+a_n}{n} \geq (a_1.a_2........a_n)^\frac{1}{n}$

I assure you that the sum the can be done just by using this simple inequality,why don't you just give it a try?

Applying AM $\geq$ GM on $a^5,a^5,a^5,b^5,c^5$ we get

$3a^5+b^5+c^5 \geq 5a^3bc$

Similarly, $a^5+3b^5+c^5 \geq 5ab^3c$

$a^5+b^5+3c^5 \geq 5abc^3$

Adding the above three equations we get $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2)$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

Now we have to show that $a^2+b^2+c^2 > ab+bc+ca$

Applying AM $\geq$ GM on $a^2,b^2$ we get,

$a^2+b^2 \geq 2ab$

Similarly,$b^2+c^2 \geq 2bc$

$a^2+c^2 \geq 2ca$

Adding the above three equations we get $a^2+b^2+c^2 \geq ab+bc+ca$

We are almost there ,so just try the last step yourself.

Therefore, $a^5+b^5+c^5 \geq abc(a^2+b^2+c^2) \geq abc(a^2+b^2+c^2)$

i.e, $a^{5}+b^{5}+c^{5}>a b c(a b+b c+c a)$

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Perfect square Problem | AIME I, 1999 | Question 3

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Perfect square and Integers.

Perfect square Problem - AIME I, 1999


Find the sum of all positive integers n for which \(n^{2}-19n+99\) is a perfect square.

  • is 107
  • is 38
  • is 840
  • cannot be determined from the given information

Key Concepts


Perfect Square

Integers

Inequalities

Check the Answer


Answer: is 38.

AIME I, 1999, Question 3

Elementary Number Theory by David Burton

Try with Hints


First hint

\((n-10)^{2}\) \(\lt\) \(n^{2}-19n+99\) \(\lt\) \((n-8)^{2}\) and \(n^{2}-19n+99\) is perfect square then \(n^{2}-19n+99\)=\((n-9)^{2}\) that is n=18

Second Hint

and \(n^{2}-19n+99\) also perfect square for n=1,9,10

Final Step

then adding 1+9+10+18=38.

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Problem on Inequality | ISI - MSQMS - B, 2018 | Problem 4b

Try this problem from ISI-MSQMS 2018 which involves the concept of Inequality.

INEQUALITY | ISI 2018| MSQMS | PART B | PROBLEM 4b


Let $a>0$ and $n \in \mathbb{N} .$ Show that
"
$$
\frac{a^{n}}{1+a+a^{2}+\ldots+a^{2 n}}<\frac{1}{2 n}
$$

Key Concepts


Algebra

Inequality

Numbers


Try with Hints


We know if $\frac{a}{b} \geq c$

Then $\frac{b}{a} \leq c$

$\frac{a^n}{1+a+a^2+.......+a^{2n}}$ < $\frac{1}{2n}$

$\frac{1+a+a^2+.....+a^{2n}}{a^n}$ > $2n$

Now why don't you just give it a try yourself,try to conclude something from the previous line,I know you can.

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+.............+(a^n+\frac{1}{a^n})> 2n$

So you have all the pieces of the jigsaw puzzle with you,and the puzzle is about to be completed,just try to place the remaining few pieces in its correct position

$a+\frac{1}{a} \geq 2$

$a^2+\frac{1}{a^2} \geq 2$

.

.

.

$a^n+\frac{1}{a^n} \geq 2$

Adding the above inequalities we get,$(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+..........+(a^n+\frac{1}{a^n}) \geq 2n$

We are almost there ,so just try the last step yourself.

Therefore, $1+(a+\frac{1}{a})+(a^2+\frac{1}{a^2})+..........+(a^n+\frac{1}{a^n}) \geq 2n+1 $ > $2n$

Thus, $\frac{a^n}{1+a+a^2+......+a^n}$<$\frac{1}{2n}$

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