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AMC-8 Math Olympiad Number Theory

Page number counting |AMC 8- 2010 -|Problem 21

Try this beautiful problem from Algebra about Page number counting

Page number counting | AMC-8, 2010 |Problem 21


Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read \(\frac{1}{5}\) of the pages plus more, and on the second day she read  \(\frac{1}{4}\) of the remaining pages plus 15 pages. On the third day she read \(\frac{1}{3}\) of the remaining pages plus 18 pages. She then realized that there were only  62 pages left to read, which she read the next day. How many pages are in this book?

  • 320
  • 240
  • 200

Key Concepts


Algebra

Arithmetic

multiplication

Check the Answer


Answer:$240$

AMC-8, 2010 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


assume that the number of all pages be \(x\)

Can you now finish the problem ……….

count day by day

can you finish the problem……..

Let x be the number of pages in the book

First day ,Hui Read \(\frac{x}{5} + 12\) pages

After first day Remaining pages=\(\{x-(\frac{x}{5}+12)\}\)=\(\frac{4x}{5} -12\)

Second day ,Hui Read \(\frac{1}{4} (\frac{4x}{5} -12) +15=\frac{x}{5} +12\)

After Second day Remaining pages= \((\frac{4x}{5} -12) -(\frac{x}{5} +12)\)=\(\frac{4x}{5} -\frac{x}{5}-24\)=\(\frac{3x}{5} -24\)

Third day,Hui read \(\frac {1}{3} (\frac{3x}{5} -24) +18\) =\((\frac{x}{5} -8+18)\)=\(\frac{x}{5} +10\)

After Third day Remaining pages = \((\frac{3x}{5} -24) -(\frac{x}{5} +10)\) =\(\frac{2x}{5} – 34\)

Now by the condition, \(\frac{2x}{5} – 34 = 62\)

\(\Rightarrow 2x-170=310\)

\(\Rightarrow 2x=480\)

\(\Rightarrow x=240\)

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College Mathematics

Sequence and Series: IIT JAM 2016 Problem 24

Sequence and series of real numbers


A sequence of real numbers is an one to one mapping from $\mathbb{N}$ (the set of natural numbers) to a subset of $\mathbb{R}$(the set of all real numbers).

Sum of the terms of a real sequence sequence is called a series.

Try the problem


Find the sum of the series $\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{n^2+n-2}$.

$\textbf{(A)}\quad \frac13\ln 2-\frac{5}{18}\quad \textbf{(B)}\quad \frac13\ln 2-\frac{5}{6}\quad \textbf{(C)}\quad \frac23\ln 2-\frac{5}{18}\quad \textbf{(D)}\quad \frac23\ln 2-\frac{5}{6}\quad$

IIT JAM 2016, PROBLEM 24

Sequence and Series of real numbers.

6 out of 10

Introduction to Real Analysis : Robert G. Bartle & Donald R. Sherbert.

Knowledge Graph


Sequence of real numbers-knowledge graph

Use some hints


This problem is a very basic one, this problem can easily be solved by step by step solution. The steps are:

Step 1 : First we will ignore the summation part. We will factorize the denominator, because we are going step by step so our aim is to simplify the given problem first.

Step 2: After factorizing the the denominator we will reach to a position where we have to use partial fraction to go forward.

Step 3: In this step we will take care of the $(-1)^n$ part, like how it will affect the series.

Step 4: After taking care of the $(-1)^n$ we will now expand the summation (breaking it into infinite sum).

Step 5 : So after $4$ steps we are halfway done now just the last simplification is left we will use the value

$\ln 2= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\ldots$

to simplify it further.

Now try to solve the entire problem by following these steps !!!

Let us see how to execute STEP 1 .

$\displaystyle\sum_{n=2}^{\infty} \frac{ (-1)^n }{ n^2+n-2 }$, here we are only concerned with the denominator part so,

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

So we have successfully factorized the denominator. Now you can reach to the part where we have to use the concept of partial fraction!!!

Let us execute STEP 2 now.

After factorizing the denominator in HINT 1 we get

$\displaystyle\sum_{n=2}^{\infty} \frac{(-1)^n}{(n+2)(n-1)}$

Now our aim here is to seperate $(n+2)$ and $(n-1)$ so we can do this by using partial fraction.

$\frac{1}{(n+2)(n-1)}=\frac{A}{n+2}+\frac{B}{(n-1)}$

$\Rightarrow A(n-1)+B(n+2)=1$

Now taking $\underline{n=1}$ we get

$A(1-1)+B(1+2)=1$

$B=\frac{1}{3}$

Again taking $\underline{n=-2}$, we get:

$A(-2-1)+B(-2+2)=1$

therefore, $A=-\frac{1}{3}$

So we get, $\frac{1}{(n+2)(n-1)}=\frac{1}{3}[\frac{1}{n-1}-\frac{1}{n+2}]$

So our aim here is successful we have separated $(n+2)$ and $(n-1)$ . Now can you proceed further ???

Now after using partial fraction we get.

$\displaystyle\sum_{n=2}^{\infty} (-1)^n\frac{1}{3}[\frac{1}{n-1}+\frac{1}{n+2}]$

Now we will execute STEP 4 i.e., we will take care of the $(-1)^n$ and at the same time we will execute STEP 4 i.e, we will split it into the infinite sum.

$\frac{1}{3}\displaystyle\sum_{n=2}^{\infty} [\frac{1}{n-1}+\frac{1}{n+2}](-1)^n$

$\Rightarrow \frac{1}{3}[(1-\frac{1}{4})-(\frac{1}{2}-\frac{1}{5})+(\frac13-\frac16)\ldots]$

$\Rightarrow \frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$

Now after this can you applying the formula of ‘$\ln 2$’ to finish the problem!!

Now only STEP 5 is left to execute. So we will use the infinite series of $\ln 2$.

$\frac13[(1-\frac12+\frac13-\frac14\ldots)+(-\frac14+\frac15-\frac16\ldots)]$

$=\frac13[\ln 2+\ln 2-(1-\frac12+\frac13)]$

$=\frac13(2\ln 2-\frac56)$

$=\frac23 \ln 2 – \frac{5}{18}$

Hence the anser is C.

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College Mathematics

Series Convergence: IIT JAM 2018 Problem 12

Warm Yourself With An Mcq

[h5p id="13"]

Understand the problem

Let a, b, c \in \mathbb{R}  Which of the following values of a ,b, c do NOT result in the convergence of the series  \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}} (a) |a|<1 , b \in \mathbb{R} , c \in \mathbb{R} (b) a=1 , b>1 , c \in \mathbb{R} (c) a=1 , b \leq 1 , c<1 (d) a=-1 , b \geq , c>0  

Source of the problem
IIT JAM 2018 Problem 12 
Topic
Convergence of a seris 
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

Do you really need a hint? Try it first!

 One disclaimer: In this question you will  see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to  give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz’s test states that if {u_n} be a monotone decreasing sequence of positive real numbers and lim u_n = 0 , then the alternating series u_1 - u_2 + u_3 - u_4 + ........... is convergent.]
 \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} Let us talk about option D first a=-1, b \geq 0 , c<0 \sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}}   where \frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 as n \longrightarrow \infty Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence f(n) of non-negative real numbers, the series \sum_{n=1}^{\infty} f(n) converges if and only if the “condensed series” \sum_{n=0}^{\infty} 2^{n} f(2^{n}) converges.Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let \sum u_n be a series of positive real numbers and let lim \frac{u_n +1}{u_n} = l Then \sum u_n is convegent if l<1 , \sum u_n is divergent if l>1 .
Moving on to option c a=1 , b \geq 0 , c<1 \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S (say) Observe if we have c=b=\frac{1}{3} thus S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty So, by comparison test(Let \sum u_n and \sum v_n be two series of positive real numbers and there is a natural number m such that u_n \leq kv_n for all n \geq m,k being a fixed positive number. Then (i)   \sum u_n is convergent if \sum v_n is convergent.  we have S is divergent. (ii) \sum u_n is divergent if \sum u_n is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making b>1 say b=2 and make c smaller  Let c= \frac{1}{2} Thus S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} Here \sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty So, S is convergent and c is one correct answer. Look for the others.
Option b a=1,b>1,c \in \mathbb{R} S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}} If c=2 clearly by comparison test S will be convergent  Now the question is that, can we find one example such that the series will be divergent? Observe that, if c \geq 0 then as b>1 we will get that the series is convergent. What will happen if c<0   Here Cauchy Condensation test comes into play  Consider a=2>1 thus \sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}}  =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}}  =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}} Now we have to use D’ Alembert’s Ratio test : Consider a_n = \frac{1}{(2^{b-1})^{n} n^{c}}  Thus \frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 Hence the series is convergent and so the series is convergent for any value of c and here the series is convergent always and that is why option b is not correct.
option a) |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} Here if we consider a<0 , b<0 , c<0 The series is convergent by Leibnitz test . So, the question is whether we can find out some values of a,b,c such that the series will be divergent. Consider a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}  Now we know that  \frac{n+1}{n} \longrightarrow 1 We have to think about \frac{\log(n+1)}{\log n} Let us consider \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1[using L’Hopital’s rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if  {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 or -\infty , +\infty , g'(x) \neq 0 for all x in I with x \neq c and {\lim}_{x \to c} \frac{f'(x)}{g'(x)} exist then {\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}  ] So, \frac{\log(n+1)}{\log n} \to 1 And hence \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1. So,the series is convergent \forall |a|<1,b,c \in \mathbb{R} Hence c. is the only correct answer.

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

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Series Convergence: IIT JAM 2018 Problem 12

Warm Yourself With An Mcq

[h5p id="13"]

Understand the problem

Let a, b, c \in \mathbb{R}  Which of the following values of a ,b, c do NOT result in the convergence of the series  \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}} (a) |a|<1 , b \in \mathbb{R} , c \in \mathbb{R} (b) a=1 , b>1 , c \in \mathbb{R} (c) a=1 , b \leq 1 , c<1 (d) a=-1 , b \geq , c>0  

Source of the problem
IIT JAM 2018 Problem 12 
Topic
Convergence of a seris 
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

Do you really need a hint? Try it first!

 One disclaimer: In this question you will  see that for some option the series is clearly convergent and for some option it might be convergent and might not be. So the question wordings are not very clear. Now having that disclaimer, what we have to find is the options where we have the series might be or might not be convergent. I want to end this hint here to  give you a bit more room to search. Look for Leibnitz rule for alternating series.[ In mathematics Leibnitz’s test states that if {u_n} be a monotone decreasing sequence of positive real numbers and lim u_n = 0 , then the alternating series u_1 - u_2 + u_3 - u_4 + ........... is convergent.]
 \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} Let us talk about option D first a=-1, b \geq 0 , c<0 \sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}}   where \frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 as n \longrightarrow \infty Hence the series is convergence in this case. So option D is rejected. Now look for the other option and see Cauchy condensation test. (For a non increasing sequence f(n) of non-negative real numbers, the series \sum_{n=1}^{\infty} f(n) converges if and only if the “condensed series” \sum_{n=0}^{\infty} 2^{n} f(2^{n}) converges.Moreover if they converge,the sum of the condensed series is no more than twice as large of the sum as original) and D’ Alembert’s test( Let \sum u_n be a series of positive real numbers and let lim \frac{u_n +1}{u_n} = l Then \sum u_n is convegent if l<1 , \sum u_n is divergent if l>1 .
Moving on to option c a=1 , b \geq 0 , c<1 \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S (say) Observe if we have c=b=\frac{1}{3} thus S = \sum_{n=3}^{\infty} \frac{1}{n^{\frac{1}{3}} (log_{e}^{n})^{\frac{1}{3}}} > \sum_{n=3}^{\infty} \frac{1}{n^{\frac{2}{3}}} \longrightarrow \infty So, by comparison test(Let \sum u_n and \sum v_n be two series of positive real numbers and there is a natural number m such that u_n \leq kv_n for all n \geq m,k being a fixed positive number. Then (i)   \sum u_n is convergent if \sum v_n is convergent.  we have S is divergent. (ii) \sum u_n is divergent if \sum u_n is divergent.) Now the question is: Can we get some point where the series is convergent? The first bet would be making b>1 say b=2 and make c smaller  Let c= \frac{1}{2} Thus S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} Here \sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty So, S is convergent and c is one correct answer. Look for the others.
Option b a=1,b>1,c \in \mathbb{R} S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}} If c=2 clearly by comparison test S will be convergent  Now the question is that, can we find one example such that the series will be divergent? Observe that, if c \geq 0 then as b>1 we will get that the series is convergent. What will happen if c<0   Here Cauchy Condensation test comes into play  Consider a=2>1 thus \sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}}  =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}}  =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}} Now we have to use D’ Alembert’s Ratio test : Consider a_n = \frac{1}{(2^{b-1})^{n} n^{c}}  Thus \frac{a_{n+1}}{a_n} = \frac{(2^{b-1})^{n} n^{c}}{(2^{b-1})^{n+1} (n+1)^{c}} \longrightarrow \frac{1}{2^{b-1}} < 1 Hence the series is convergent and so the series is convergent for any value of c and here the series is convergent always and that is why option b is not correct.
option a) |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} Here if we consider a<0 , b<0 , c<0 The series is convergent by Leibnitz test . So, the question is whether we can find out some values of a,b,c such that the series will be divergent. Consider a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}  Now we know that  \frac{n+1}{n} \longrightarrow 1 We have to think about \frac{\log(n+1)}{\log n} Let us consider \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1[using L’Hopital’s rule which states that for function f and g which are differentiable on an open interval I except possibly at a point c contained in I if  {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 or -\infty , +\infty , g'(x) \neq 0 for all x in I with x \neq c and {\lim}_{x \to c} \frac{f'(x)}{g'(x)} exist then {\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}  ] So, \frac{\log(n+1)}{\log n} \to 1 And hence \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1. So,the series is convergent \forall |a|<1,b,c \in \mathbb{R} Hence c. is the only correct answer.

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Sum Of Series: IIT JAM 2018 Problem 13

What about a small warm up MCQ!!!!!

[h5p id="14"]

Understand the problem

Let a_n = n+\frac{1}{n} , n \in \mathbb{N} . Then the sum of the series \sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} is (A) e^{-1}-1 (B) e^{-1} (C) 1-e^{-1} (D) 1+e^{-1}
Source of the problem
IIT JAM 2018 Problem 13
Topic
Series
Difficulty Level
Easy
Suggested Book
Real Analysis By S.K Mapa

Start with hints

Do you really need a hint? Try it first!

Consider a_n = n + \frac{1}{n} , n \in \mathbb{N} We have to use e^{x} = 1 + \frac{x}{1!} + \frac{x^{2}}{2!}+..... Specifically e^{1} = 1+ \frac{1}{1!} + \frac{1}{2!}+.... And e^{-1} = 1 - \frac{1}{1!} + \frac{1}{2!} - \frac{1}{3!} + ....... Do you want to play with it
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{n+1 + \frac{1}{n+1}}{n!} = \sum_{n=1}^{\infty}[ (-1)^{n+1} \frac{1}{(n-1)!} + (-1)^{n+1} \frac{1}{n!} + (-1)^{n+1} \frac{1}{(n+1)!}]   Now we will be breaking it term by term for the ease of calculation. Can you do it from here?
\sum_{n=1}^{\infty} (-1)^{n+1} \frac{a_{n+1}}{n!} = [1-\frac{1}{1!} + \frac{1}{2!} - ......] + [\frac{1}{1!} - \frac{1}{2!} + \frac{1}{3!} - .....] + [\frac{1}{2!} - \frac{1}{3!} + \frac{1}{4!} -.......] = e^{-1} + [1-e^{-1}] + [e^{-1} + 1 - 1] = e^{-1} + 1  So option (D) is our required answer.

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ISI MStat PSB 2009 Problem 4 | Polarized to Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 4. It is based on the idea of Polar Transformations, but need a good deal of observation o realize that. Give it a Try it !
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ISI MStat PSB 2009 Problem 6 | abNormal MLE of Normal

This is a very beautiful sample problem from ISI MStat PSB 2009 Problem 6. It is based on the idea of Restricted Maximum Likelihood Estimators, and Mean Squared Errors. Give it a Try it !
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ISI MStat PSB 2009 Problem 3 | Gamma is not abNormal

This is a very simple but beautiful sample problem from ISI MStat PSB 2009 Problem 3. It is based on recognizing density function and then using CLT. Try it !
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ISI MStat PSB 2009 Problem 1 | Nilpotent Matrices

This is a very simple sample problem from ISI MStat PSB 2009 Problem 1. It is based on basic properties of Nilpotent Matrices and Skew-symmetric Matrices. Try it !
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ISI MStat PSB 2006 Problem 2 | Cauchy & Schwarz come to rescue

This is a very subtle sample problem from ISI MStat PSB 2006 Problem 2. After seeing this problem, one may think of using Lagrange Multipliers, but one can just find easier and beautiful way, if one is really keen to find one. Can you!

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

Try this problem from ISI MSQMS 2018 which involves the concept of Inequality. You can use the sequential hints provided to solve the problem.

Data, Determinant and Simplex

This problem is a beautiful problem connecting linear algebra, geometry and data. Go ahead and dwelve into the glorious connection.

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

Try this problem from ISI MSQMS 2015 which involves the concept of Integral Inequality and real analysis. You can use the sequential hints provided to solve the problem.