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| − | {{duplicate|[[2024 AMC 10A Problems/Problem 7|2024 AMC 10A #7]] and [[2024 AMC 12A Problems/Problem 6|2024 AMC 12A #6]]}}
| + | #redirect [[2024 AMC 12A Problems/Problem 5]] |
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| − | ==Problem==
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| − | The product of three integers is <math>60</math>. What is the least possible positive sum of the
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| − | three integers?
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| − | <math>\textbf{(A) }2\qquad\textbf{(B) }3\qquad\textbf{(C) }5\qquad\textbf{(D) }6\qquad\textbf{(E) }13</math>
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| − | ==Solution 1==
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| − | 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>.
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| − | ~eevee9406
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| − | == Solution 2==
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| − | 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>
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| − | ~[[User:pog|pog]],~[[User:Mathkiddus|mathkiddus]]
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| − | == Solution 3 ==
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| − | 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?
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| − | 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>.
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| − | ~Moonwatcher22
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| − | == Video Solution by Math from my desk ==
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| − | https://www.youtube.com/watch?v=ptw2hYKoAIs&t=2s
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| − | == Video Solution (🚀 2 min solve 🚀) ==
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| − | https://youtu.be/u42_QzyO9zA
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| − | <i>~Education, the Study of Everything</i>
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| − | == Video Solution by Pi Academy ==
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| − | https://youtu.be/6qYaJsgqkbs?si=K2Ebwqg-Ro8Yqoiv
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| − | == Video Solution 1 by Power Solve ==
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| − | https://youtu.be/j-37jvqzhrg?si=aggRgbnyZ3QjYwZF&t=806
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| − | == Video Solution by Daily Dose of Math ==
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| − | https://youtu.be/e8eL1l5os30
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| − | ~Thesmartgreekmathdude
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| − | ==Video Solution by SpreadTheMathLove==
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| − | https://www.youtube.com/watch?v=_o5zagJVe1U
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| − | ==See also==
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| − | {{AMC10 box|year=2024|ab=A|num-b=6|num-a=8}}
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| − | {{AMC12 box|year=2024|ab=A|num-b=5|num-a=7}}
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| − | {{MAA Notice}}
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