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

Latest revision as of 05:25, 9 December 2023

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 lowest 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 $667 + 3 + 1 = \boxed{\text{(E) 671}}$

Solution By: armang32324

Solution 2

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

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

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 $\boxed{\text{E}}$ as our final answer.

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 1: / Line 1: @@
-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 integer]]s 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>?
+{{duplicate|[[2000 AMC 12 Problems|2000 AMC 12 #1]] and [[2000 AMC 10 Problems|2000 AMC 10 #1]]}}
-<math> \mathrm{(A) \ 23 } \qquad \mathrm{(B) \ 55 } \qquad \mathrm{(C) \ 99 } \qquad \mathrm{(D) \ 111 } \qquad \mathrm{(E) \ 671 }  </math>
+==Problem==
+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>?
-== Solution ==
+<math>\textbf{(A)}\ 23 \qquad \textbf{(B)}\ 55 \qquad \textbf{(C)}\ 99 \qquad \textbf{(D)}\ 111 \qquad \textbf{(E)}\ 671</math>
-The sum is the highest if two [[factor]]s are the lowest!
-So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \mathrm{(E)}</math>.
+== Solution 1 (Verifying the Statement)==
+First, we need to recognize that a number is going to be lowest 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.
+Hence, we have <math>667 + 3 + 1 = \boxed{\text{(E) 671}}</math>
+Solution By: armang32324
+== Solution 2==
+The sum is the highest if two [[factor]]s are the lowest.
+So, <math>1 \cdot 3 \cdot 667 = 2001</math> and <math>1+3+667=671 \Longrightarrow \boxed{\text{(E)}}</math>.
+== Solution 3 (Answer Choices) ==
+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}}</math> as our final answer.
 ==See Also==
-* [[2000 AMC 12]]
-* [[2000 AMC 12/Problem 2 | Next problem]]
+{{AMC12 box|year=2000|before=First<br />Question|num-a=2}}
+{{AMC10 box|year=2000|before=First<br />Question|num-a=2}}
 [[Category:Introductory Algebra Problems]]
+[[Category:Introductory Number Theory Problems]]
+{{MAA Notice}}

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