Difference between revisions of "2021 CMC 12A Problems/Problem 4"

Revision as of 20:59, 3 January 2021

The following problem is from both the 2021 CMC 12A #4 and 2021 CMC 10A #5, so both problems redirect to this page.

Problem

There exists a positive integer $N$ such that $\frac{\tfrac{1}{999}+\tfrac{1}{1001}}{2}=\frac{1}{N}.$ What is the sum of the digits of $N$ ?

$\textbf{(A) } 36\qquad\textbf{(B) } 45\qquad\textbf{(C) } 54\qquad\textbf{(D) } 63\qquad\textbf{(E) } 72\qquad$

Solution

Let $a=1000$ . Then $\frac{\tfrac{1}{a-1}+\frac{1}{a+1}}{2}-\frac{1}{a}=\frac{1}{N}$ or $\frac{1}{a^{3}-a}=\frac{1}{N}$ , so $N=a^{3}-a=1000^{3}-1000=999,999,000$ whose digits sum to $\boxed{\textbf{(C) } 54}$ .

Video Solution

https://youtu.be/H4uIpLB-6Jo

~Punxsutawney Phil

2021 CMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 12 Problems and Solutions

2021 CMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All CMC 10 Problems and Solutions

@@ Line 14: / Line 14: @@
 ==See also==
-{{CMC12 box|year=2021|ab=A|num-a=3|num-a=5}}
+{{CMC12 box|year=2021|ab=A|num-b=3|num-a=5}}
-{{CMC10 box|year=2021|ab=A|num-a=4|num-a=6}}
+{{CMC10 box|year=2021|ab=A|num-b=4|num-a=6}}

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Difference between revisions of "2021 CMC 12A Problems/Problem 4"

Revision as of 20:59, 3 January 2021

Contents

Problem

Solution

Video Solution

See also