Difference between revisions of "2024 AMC 10A Problems/Problem 7"
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| − | The | + | {{duplicate|[[2024 AMC 10A Problems/Problem 7|2024 AMC 10A #7]] and [[2024 AMC 12A Problems/Problem 6|2024 AMC 12A #6]]}} |
| + | |||
| + | ==Problem== | ||
| + | The product of three integers is <math>60</math>. What is the least possible positive sum of the | ||
| + | three integers? | ||
| + | |||
| + | <math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13</math> | ||
| + | |||
| + | ==Solution 1== | ||
| + | We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split <math>60</math> into three factors and choose negativity. We notice that <math>10\cdot6\cdot1=10\cdot(-6)\cdot(-1)=60</math>, and trying other combinations does not yield lesser results so the answer is <math>10-6-1=\boxed{\textbf{(B) }3}</math>. | ||
| + | |||
| + | ~eevee9406 | ||
| + | |||
| + | == Solution 2== | ||
| + | We have <math>abc = 60</math>. Let <math>a</math> be positive, and let <math>b</math> and <math>c</math> be negative. Then we need <math>a > |b + c|</math>. If <math>a = 6</math>, then <math>|b + c|</math> is at least <math>7</math>, so this doesn't work. If <math>a = 10</math>, then <math>(b,c) = (-6,-1)</math> works, giving <math>10 - 7 = \boxed{\textbf{(B) }3}</math> | ||
| + | |||
| + | ~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]] | ||
| + | |||
| + | == Solution 3 == | ||
| + | We can see that the most optimal solution would be <math>1</math> positive integer and <math>2</math> negative ones (as seen in solution 1). Let the three integers be <math>x</math>, <math>y</math>, and <math>z</math>, and let <math>x</math> be positive and <math>y</math> and <math>z</math> be negative. If we want the optimal solution, we want the negative numbers to be as large as possible. so the answer should be <math>-60 \cdot -1 \cdot 1</math>, where <math>-60 - 1 + 1 = -60</math>... right? | ||
| + | |||
| + | No! Our sum must be a positive number, so that would be invalid! We see that -<math>30, -20, -15</math>, and <math>-10</math> are too far negative to allow the sum to be positive. For example, <math>-15 = z. -15xy=60</math>, so <math>xy = -4</math>. For <math>xy</math> to be the most positive, we will have <math>4</math> and <math>-1</math>. Yet, <math>-15+4-1</math> is still less than <math>0</math>. After <math>-10</math>, the next factor of <math>60</math> would be <math>6</math>. if <math>z</math> = <math>-6</math>, <math>xy = -10</math>. This might be positive! Now, if we have <math>z = -6, y = -1</math>, and <math>x = 10, x + y + z = 3</math>. It cannot be smaller because <math>x = 5</math> and <math>y = -2</math> would result in <math>x + y + z</math> being negative. Therefore, our answer would be <math>\boxed{\textbf{(B) }3}</math>. | ||
| + | |||
| + | ~Moonwatcher22 | ||
| + | |||
| + | == Video Solution by Pi Academy == | ||
| + | |||
| + | https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv | ||
| + | |||
| + | == Video Solution 1 by Power Solve == | ||
| + | https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806 | ||
| + | |||
| + | == Video Solution by Daily Dose of Math == | ||
| + | |||
| + | https://youtu.be/e8eL1l5os30 | ||
| + | |||
| + | ~Thesmartgreekmathdude | ||
| + | |||
| + | ==See also== | ||
| + | {{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}} | ||
| + | {{AMC12 box|year=2024|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 20:18, 9 November 2024
- The following problem is from both the 2024 AMC 10A #7 and 2024 AMC 12A #6, so both problems redirect to this page.
Contents
Problem
The product of three integers is 
. What is the least possible positive sum of the
three integers?
Solution 1
We notice that the optimal solution involves two negative numbers and a positive number. Thus we may split into three factors and choose negativity. We notice that 
, and trying other combinations does not yield lesser results so the answer is 
.
~eevee9406
Solution 2
We have 
. Let 
be positive, and let 
and 
be negative. Then we need 
. If 
, then 
is at least 
, so this doesn't work. If 
, then 
works, giving 
Solution 3
We can see that the most optimal solution would be 
positive integer and 
negative ones (as seen in solution 1). Let the three integers be 
, 
, and 
, and let be positive and and be negative. If we want the optimal solution, we want the negative numbers to be as large as possible. so the answer should be 
, where 
... right?
No! Our sum must be a positive number, so that would be invalid! We see that -
, and 
are too far negative to allow the sum to be positive. For example, 
, so 
. For 
to be the most positive, we will have 
and 
. Yet, 
is still less than 
. After , the next factor of would be 
. if = 
, 
. This might be positive! Now, if we have 
, and 
. It cannot be smaller because 
and 
would result in 
being negative. Therefore, our answer would be 
.
~Moonwatcher22
Video Solution by Pi Academy
https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
Video Solution 1 by Power Solve
https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
Video Solution by Daily Dose of Math
~Thesmartgreekmathdude
See also
| 2024 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 6 |
Followed by Problem 8 | |
| 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 | ||
| 2024 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 5 |
Followed by Problem 7 |
| 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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
