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Difference between revisions of "2000 AMC 12 Problems/Problem 1"

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In the year <math>2001</math>, the United States will host the International Mathematical Olympiad. Let <math> \displaystyle I,M,</math> and <math>\displaystyle O</math> be distinct positive integers such that the product <math>I \cdot M \cdot O = 2001 </math>. What is the largest possible value of the sum <math>\displaystyle I + M + O</math>?
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{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}}
  
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<math> \mathrm{(A) \ 23 } \qquad \mathrm{(B) \ 55 } \qquad \mathrm{(C) \ 99 } \qquad \mathrm{(D) \ 111 } \qquad \mathrm{(E) \ 671 }  </math>
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==Problem==
  
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In the year <math>2001</math>, the United States will host the [[International Mathematical Olympiad]].  Let <math>I,M,</math> and <math>O</math> be distinct [[positive integer]]s such that the product <math>I \cdot M \cdot O = 2001 </math>.  What is the largest possible value of the sum <math>I + M + O</math>?
  
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== Solution ==
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<math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math>
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The sum is the highest if two factors are the lowest!
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So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671</math>, (E).
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== Solution 1 (Verifying the Statement)==
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First, we need to recognize that a number is going to be largest only if, of the <math>3</math> [[factor]]s, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like <math>30</math>. It becomes much more clear that this is true, and in this situation, the value of <math>I + M + O</math> would be <math>18</math>. Now, we use this process on <math>2001</math> to get <math>667 * 3 * 1</math> as our <math>3</math> factors.
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Hence, we have <math>667 + 3 + 1 = \boxed{\text{(E) 671}}</math>.
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~armang32324
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== Solution 2==
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The sum is the highest if two [[factor]]s are the lowest.
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So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E) 671}}</math>.
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== Solution 3 (Answer Choices) ==
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We see since <math>2 + 0 + 0 + 1</math> is divisible by <math>3</math>, we can eliminate all of the first <math>4</math> answer choices because they are way too small and get <math>\boxed{\text{(E) 671}}</math> as our final answer.
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== Solution 4 (Faster Way) ==
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Notice <math>2001 = 3 * 29 * 23</math>, so we can just maximize this with <math>667 * 3 * 1</math>, which has a sum of <math>671</math>. Our answer is <math>\boxed{\text{(E) 671}}</math>.
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-itsj
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==Video Solution by Daily Dose of Math==
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https://www.youtube.com/watch?v=aSzsStkkYeA
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~Thesmartgreekmathdude
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==See Also==
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{{AMC12 box|year=2000|before=First<br />Question|num-a=2}}
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{{AMC10 box|year=2000|before=First<br />Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]
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[[Category:Introductory Number Theory Problems]]
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{{MAA Notice}}

Latest revision as of 23:53, 17 November 2024

The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.

Problem

In the year $2001$, the United States will host the International Mathematical Olympiad. Let $I,M,$ and $O$ be distinct positive integers such that the product $I \cdot M \cdot O = 2001$. What is the largest possible value of the sum $I + M + O$?

$\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671$

Solution 1 (Verifying the Statement)

First, we need to recognize that a number is going to be largest only if, of the $3$ factors, two of them are small. If we want to make sure that this is correct, we could test with a smaller number, like $30$. It becomes much more clear that this is true, and in this situation, the value of $I + M + O$ would be $18$. Now, we use this process on $2001$ to get $667 * 3 * 1$ as our $3$ factors. Hence, we have $667 + 3 + 1 = \boxed{\text{(E) 671}}$.

~armang32324

Solution 2

The sum is the highest if two factors are the lowest.

So, $1 \cdot 3 \cdot 667 = 2001$ and $1+3+667=671 \Longrightarrow \boxed{\text{(E) 671}}$.

Solution 3 (Answer Choices)

We see since $2 + 0 + 0 + 1$ is divisible by $3$, we can eliminate all of the first $4$ answer choices because they are way too small and get $\boxed{\text{(E) 671}}$ as our final answer.

Solution 4 (Faster Way)

Notice $2001 = 3 * 29 * 23$, so we can just maximize this with $667 * 3 * 1$, which has a sum of $671$. Our answer is $\boxed{\text{(E) 671}}$.

-itsj

Video Solution by Daily Dose of Math

https://www.youtube.com/watch?v=aSzsStkkYeA

~Thesmartgreekmathdude

See Also

2000 AMC 12 (Problems • Answer Key • Resources)
Preceded by
First
Question 		Followed by
Problem 2
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 AMC 12 Problems and Solutions
2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
First
Question 		Followed by
Problem 2
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 AMC 10 Problems and Solutions

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