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?

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, $\frac{2x}{5} - 34 = 62$

$\Rightarrow 2x-170=310$

$\Rightarrow 2x=480$

$\Rightarrow x=240$

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

IIT JAM 2016, PROBLEM 24

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

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

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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.

Books

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

Series Convergence: IIT JAM 2018 Problem 12

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 $$\sum_{n=3}^{\infty} \frac{a^{n}}{n^{b}\left(\log _{e} n\right)^{c}} ?$$ (A) $|a|<1, b \in \mathbb{R}, c \in \mathbb{R}$ (B) $a=1, b>1, c \in \mathbb{R}$ (C) $a=1, b \geq 0, c<1$ (D) $a=-1, b \geq 0, c>0$

Start with hints

[Hint 1] 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.] [Hint 2] $\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 convergent if $l<1 $ , $\sum u_n $ is divergent if $ l>1 $. [Hint 3]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 $x 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. [Hint 4]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. [Hint 5]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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Series Convergence: IIT JAM 2018 Problem 12

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Warm Yourself With An Mcq

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Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27.4" text_font="Raleway||||||||" background_color="#f4f4f4" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" box_shadow_style="preset2"]Let a, b, c $latex \in \mathbb{R} $ Which of the following values of a ,b, c do NOT result in the convergence of the series $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_en)^{c}} $ (a) $latex |a|<1 , b \in \mathbb{R} , c \in \mathbb{R} $ (b) $latex a=1 , b>1 , c \in \mathbb{R} $ (c) $latex a=1 , b \leq 1 , c<1 $ (d) $latex a=-1 , b \geq , c>0 $

[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="4.1" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="4.1"]IIT JAM 2018 Problem 12 [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="4.1" open="off"]Convergence of a seris [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="4.1" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="4.1" open="off"]Real Analysis By S.K Mapa[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#8300e9" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="4.1" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="4.1"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="4.1"] 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 $latex {u_n}$ be a monotone decreasing sequence of positive real numbers and lim $latex u_n = 0$ , then the alternating series $latex u_1 - u_2 + u_3 - u_4 + ...........$ is convergent.][/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="4.1"] $latex \sum_{n=3}^{\infty} \frac{a^{n}}{n^{b} (log_e^{n})^{c}} $ Let us talk about option D first $latex a=-1, b \geq 0 , c<0 $ $latex \sum_{n=3}^{\infty} \frac{(-1)^{n}}{n^{b}(ln^{n})^{c}} $ where $latex \frac{1}{n^{b}(ln n)^{c}} \longrightarrow 0 $ as $latex 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 $latex f(n) $ of non-negative real numbers, the series $latex \sum_{n=1}^{\infty} f(n) $ converges if and only if the "condensed series" $latex \sum_{n=0}^{\infty} 2^{n} f(2^{n})$ 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 $latex \sum u_n $ be a series of positive real numbers and let $latex lim \frac{u_n +1}{u_n} = l $ Then $latex \sum u_n $ is convegent if $latex l<1 $ , $latex \sum u_n $ is divergent if $latex l>1 $.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="4.1"]Moving on to option c $latex a=1 , b \geq 0 , c<1 $ $latex \sum_{n=3}^{\infty} \frac{1}{n^{b} (log_{e}^{n})^{c}} := S $(say) Observe if we have $latex c=b=\frac{1}{3} $ thus $latex 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 $latex \sum u_n $ and $latex \sum v_n $ be two series of positive real numbers and there is a natural number m such that $latex u_n \leq kv_n $ for all $latex n \geq m,k $ being a fixed positive number. Then (i) $latex \sum u_n $ is convergent if $latex \sum v_n $ is convergent. we have S is divergent. (ii) $latex \sum u_n $ is divergent if $latex \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 $latex b>1 $ say $latex b=2 $ and make $latex c $ smaller Let $latex c= \frac{1}{2} $ Thus $latex S= \sum_{n=3}^{\infty} \frac{1}{n^{2} (log_{e}^{n})^{\frac{1}{2}}} $ Here $latex \sum_{n=3}^{\infty} \frac{1}{n^{2} (log n)^{\frac{1}{2}}} < \sum \frac{1}{n^{2}} < \infty $ So, $latex S $ is convergent and c is one correct answer. Look for the others.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="4.1"]Option b $latex a=1,b>1,c \in \mathbb{R}$ $latex S=\sum_{n=3}^{\infty} \frac{1}{n^{b}(ln n)^{c}}$ If $latex c=2 $ clearly by comparison test $latex S $ will be convergent Now the question is that, can we find one example such that the series will be divergent? Observe that, if $latex c \geq 0 $ then as $latex b>1 $ we will get that the series is convergent. What will happen if $latex c<0 $ Here Cauchy Condensation test comes into play Consider $latex a=2>1 $ thus $latex \sum_{n=3}^{\infty} \frac{2^{n}}{(2^{n})^{b} (ln 2^{n})^{c}} =\sum_{n=3}^{\infty} \frac{{2^{n}}^{1-b}} {n^{c} (ln 2)^{c}} =\frac{1}{(ln 2)^{c}} \sum_{n=3}^{\infty} \frac{1}{(2^{b-1})^{n} n^{c}}$ Now we have to use D’ Alembert’s Ratio test : Consider $latex a_n = \frac{1}{(2^{b-1})^{n} n^{c}} $ Thus $latex \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 $latex c $ and here the series is convergent always and that is why option b is not correct.[/et_pb_tab][et_pb_tab title="Hint 5" _builder_version="4.1"]option a) $latex |a| < 1, b \in \mathbb{R} , c \in \mathbb{R} $ Here if we consider $latex 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 $latex a,b,c $ such that the series will be divergent. Consider $latex a_n = \frac{a^{n}}{n^{b} (\log_e n )^{c}} $ $latex \frac{a_{n+1}}{a_n} = \frac{a}{(1+ \frac{1}{n})^{b} (\frac{(log{n+1}}{log{n})}^{c}}$ Now we know that $latex \frac{n+1}{n} \longrightarrow 1 $ We have to think about $latex \frac{\log(n+1)}{\log n}$ Let us consider $latex \lim_{x \to \infty} \frac{\log(x+1)}{\log(x)}=\lim_{x \to \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to \infty} \frac{x}{x+1}=1$[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 $latex {\lim}_{x \to c} f(x) = {\lim}_{x \to c} g(x) = 0 $ or $latex -\infty , +\infty , g'(x) \neq 0 $ for all x in I with $latex x \neq c$ and $latex {\lim}_{x \to c} \frac{f'(x)}{g'(x)}$ exist then $latex {\lim}_{x \to c} \frac{f(x)}{g(x)} = {\lim} \frac{f'(x)}{g'(x)}$ ] So, $latex \frac{\log(n+1)}{\log n} \to 1$ And hence $latex \lim{n \to \infty} |\frac{a_{n+1}}{a_n}|=|a|<1$. So,the series is convergent $latex \forall |a|<1,b,c \in \mathbb{R} $Hence c. is the only correct answer.[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="4.1" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#e06100" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" hover_enabled="0" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3"]

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Sum Of Series: IIT JAM 2018 Problem 13

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) $1+e^{-1}$

Start with hints

Do you really need a hint? Try it first!

[Hint 1] 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. [Hint 2]$ \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? [Hint 3] $\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.

Let's have a look into the graphs

Connected Program at Cheenta

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