Difference between revisions of "Random Problem"

Revision as of 14:37, 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

???

Hardish 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 is $\frac{a}{b}\pi$ ?

Solution

???

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

???

@@ Line 4: / Line 4: @@
 <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 ==

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Difference between revisions of "Random Problem"

Revision as of 14:37, 24 March 2023

Contents

Easy Problem

Solution

Medium Problem

Solution

Hardish Problem

Solution

Hard Problem

Solution