Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "Random Problem"

(Created page with "Problem The sum<cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</cmath>can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b...")
 
(→‎Solution 1)
 
(19 intermediate revisions by 2 users not shown)
Line 1: Line 1:
−
Problem
+
== Easy Problem==
 
The sum<cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</cmath>can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?
 
The sum<cmath>\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}</cmath>can be expressed as <math>a-\frac{1}{b!}</math>, where <math>a</math> and <math>b</math> are positive integers. What is <math>a+b</math>?
  
 
<math>\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math>
 
<math>\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024</math>
 +
===Solution===
 +
'''Submitted by BinouTheGuineaPig''' | ''A step-by-step solution''
 +
 +
We see that the general form for each term can be expressed in terms of <math>n</math> as follows.
 +
 +
<math>\frac{n}{(n+1)!} =\frac{(n+1)-1}{(n+1)!}</math>
 +
 +
<math>\qquad\qquad = \frac{n+1}{(n+1)(n!)} - \frac{1}{(n+1)!}</math>
 +
 +
<math>\qquad\qquad = \frac{1}{n!} - \frac{1}{(n+1)!}</math>
 +
 +
Now, to find the entire sum, it can be expressed as follows.
 +
 +
<math>\sum_{n=1}^{2022} \frac{n}{(n+1)!}</math>
 +
 +
<math>=\sum_{n=1}^{2022} \frac{1}{n!} - \frac{1}{(n+1)!}</math>
 +
 +
<math>=(\frac{1}{1!}+\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}) - (\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}+\frac{1}{2023!})</math>
 +
 +
Here, we see that a whole chunk of terms cancel each other out, leaving us with
 +
 +
<math>=1-\frac{1}{2023!} \equiv a-\frac{1}{b!}</math>
 +
 +
This means <math>a+b=1+2023=2024</math>.
 +
 +
Therefore, the answer is <math>\boxed{\textbf{(E)}}</math>.
 +
 +
== Medium Problem ==
 +
Show that there exist no finite decimals <math>a = 0.\overline{a_1a_2a_3\ldots a_n}</math> such that when its digits are rearranged to a different decimal <math>b = 0.\overline{a_{b_1}a_{b_2}a_{b_3}\ldots a_{b_n}}</math>, <math>a + b = 1</math>.
 +
 +
===Solution===
 +
???
 +
 +
== Hard-ish Problem ==
 +
A cylinder is inscribed in a circular cone with base radius of <math>7</math> and height of <math>14</math>. What is the maximum possible volume of this cylinder in the form of <math>\frac{a}{b}\pi</math>?
 +
 +
===Solution===
 +
'''Submitted by BinouTheGuineaPig''' | ''A step-by-step solution''
 +
 +
We create a cross-section of the cylinder in the cone, "slicing down" from the apex of the cone as follows.
 +
 +
https://imgur.com/a/Xag3fSE
 +
 +
Volume of the cylinder
 +
 +
<math>V=\pi{r^2}h</math>
 +
 +
<math>V=\pi{r^2}(14-2r)</math>
 +
 +
<math>V=14\pi{r^2}-2\pi{r^3}</math>
 +
 +
Now, we find all inflection points in the graph of this equation, by finding the values of <math>r</math> where the gradient is <math>0</math>, in other words where <math>\frac{dV}{dr}=0</math>.
 +
 +
<math>\frac{dV}{dr}=28\pi{r}-6\pi{r^2}</math>
 +
 +
<math>28\pi{r}-6\pi{r^2}=0</math>
 +
 +
<math>3r^2-14r=0</math>
 +
 +
<math>r(3r-14)=0</math>
 +
 +
<math>r=0</math> or <math>r=\frac{14}{3}</math>
 +
 +
We will obviously use the larger value of <math>r</math> to find the maximum volume of the cylinder, as follows.
 +
 +
<math>V_{max}=14\pi(\frac{14}{3})^2-2\pi(\frac{14}{3})^3</math>
 +
 +
<math>V_{max}=\frac{2744}{27}\pi \equiv \frac{a}{b}\pi</math>
 +
 +
Therefore, the maximum cylinder volume can be expressed as <math>\boxed{\frac{2744}{27}\pi}</math>.
 +
 +
== Hard Problem ==
 +
A regular <math>48</math>-gon is inscribed in a circle with radius <math>1</math>. Let <math>X</math> be the set of distances (not necessarily distinct) from the center of the circle to each side of the <math>48</math>-gon, and <math>Y</math> be the set of distances (not necessarily distinct) from the center of the circle to each diagonal of the <math>48</math>-gon. Let <math>S</math> be the union of <math>X</math> and <math>Y</math>. What is the sum of the squares of all of the elements in <math>S</math>?
 +
 +
===Solution 1===
 +
'''Credits to Daniel Mathias | Submitted by BinouTheGuineaPig'''
 +
 +
For each of the <math>24</math> main diagonals passing through the center there are <math>46</math> vertices not on the diagonal. For each of those vertices, there are two diagonals (or one diagonal and an edge) connecting the vertex to an endpoint of the main diagonal. The distances from the center to these two diagonals are <math>\sin\theta</math> and <math>\cos\theta</math> for some <math>\theta</math> and the sum of the squared distances is <math>\sin^{2}\theta+\cos^{2}\theta=1</math>, as shown.
 +
 +
https://imgur.com/a/gUYewP1
 +
 +
This would give a total sum of <math>24\times46=1104</math>, but since <math>24\times46\times2=2208</math> counts each of <math>1056</math> diagonals and <math>48</math> edges twice, the total sum desired is
 +
 +
<cmath>24\times23=552</cmath>

Latest revision as of 17:07, 24 March 2023

Easy Problem

The sum\[\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\cdots+\frac{2021}{2022!}\]can be expressed as $a-\frac{1}{b!}$, where $a$ and $b$ are positive integers. What is $a+b$?

$\textbf{(A)}\ 2020 \qquad\textbf{(B)}\ 2021 \qquad\textbf{(C)}\ 2022 \qquad\textbf{(D)}\ 2023 \qquad\textbf{(E)}\ 2024$

Solution

Submitted by BinouTheGuineaPig | A step-by-step solution

We see that the general form for each term can be expressed in terms of $n$ as follows.

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

$\qquad\qquad = \frac{n+1}{(n+1)(n!)} - \frac{1}{(n+1)!}$

$\qquad\qquad = \frac{1}{n!} - \frac{1}{(n+1)!}$

Now, to find the entire sum, it can be expressed as follows.

$\sum_{n=1}^{2022} \frac{n}{(n+1)!}$

$=\sum_{n=1}^{2022} \frac{1}{n!} - \frac{1}{(n+1)!}$

$=(\frac{1}{1!}+\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}) - (\cancel{\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\cdots+\frac{1}{2022!}}+\frac{1}{2023!})$

Here, we see that a whole chunk of terms cancel each other out, leaving us with

$=1-\frac{1}{2023!} \equiv a-\frac{1}{b!}$

This means $a+b=1+2023=2024$.

Therefore, the answer is $\boxed{\textbf{(E)}}$.

Medium Problem

Show that there exist no finite decimals $a = 0.\overline{a_1a_2a_3\ldots a_n}$ such that when its digits are rearranged to a different decimal $b = 0.\overline{a_{b_1}a_{b_2}a_{b_3}\ldots a_{b_n}}$, $a + b = 1$.

Solution

???

Hard-ish Problem

A cylinder is inscribed in a circular cone with base radius of $7$ and height of $14$. What is the maximum possible volume of this cylinder in the form of $\frac{a}{b}\pi$?

Solution

Submitted by BinouTheGuineaPig | A step-by-step solution

We create a cross-section of the cylinder in the cone, "slicing down" from the apex of the cone as follows.

https://imgur.com/a/Xag3fSE

Volume of the cylinder

$V=\pi{r^2}h$

$V=\pi{r^2}(14-2r)$

$V=14\pi{r^2}-2\pi{r^3}$

Now, we find all inflection points in the graph of this equation, by finding the values of $r$ where the gradient is $0$, in other words where $\frac{dV}{dr}=0$.

$\frac{dV}{dr}=28\pi{r}-6\pi{r^2}$

$28\pi{r}-6\pi{r^2}=0$

$3r^2-14r=0$

$r(3r-14)=0$

$r=0$ or $r=\frac{14}{3}$

We will obviously use the larger value of $r$ to find the maximum volume of the cylinder, as follows.

$V_{max}=14\pi(\frac{14}{3})^2-2\pi(\frac{14}{3})^3$

$V_{max}=\frac{2744}{27}\pi \equiv \frac{a}{b}\pi$

Therefore, the maximum cylinder volume can be expressed as $\boxed{\frac{2744}{27}\pi}$.

Hard Problem

A regular $48$-gon is inscribed in a circle with radius $1$. Let $X$ be the set of distances (not necessarily distinct) from the center of the circle to each side of the $48$-gon, and $Y$ be the set of distances (not necessarily distinct) from the center of the circle to each diagonal of the $48$-gon. Let $S$ be the union of $X$ and $Y$. What is the sum of the squares of all of the elements in $S$?

Solution 1

Credits to Daniel Mathias | Submitted by BinouTheGuineaPig

For each of the $24$ main diagonals passing through the center there are $46$ vertices not on the diagonal. For each of those vertices, there are two diagonals (or one diagonal and an edge) connecting the vertex to an endpoint of the main diagonal. The distances from the center to these two diagonals are $\sin\theta$ and $\cos\theta$ for some $\theta$ and the sum of the squared distances is $\sin^{2}\theta+\cos^{2}\theta=1$, as shown.

https://imgur.com/a/gUYewP1

This would give a total sum of $24\times46=1104$, but since $24\times46\times2=2208$ counts each of $1056$ diagonals and $48$ edges twice, the total sum desired is

\[24\times23=552\]

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Random_Problem&oldid=191285"