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Difference between revisions of "2023 AIME II Problems/Problem 1"
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The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is <math>990.</math> Find the greatest number of apples growing on any of the six trees.
 
The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is <math>990.</math> Find the greatest number of apples growing on any of the six trees.
  
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==Solution==
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==Solution 1==
  
 
In the arithmetic sequence, let <math>a</math> be the first term and <math>d</math> be the common difference, where <math>d>0.</math> The sum of the first six terms is <cmath>a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.</cmath>
 
In the arithmetic sequence, let <math>a</math> be the first term and <math>d</math> be the common difference, where <math>d>0.</math> The sum of the first six terms is <cmath>a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.</cmath>
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~MRENTHUSIASM
 
~MRENTHUSIASM
  
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==Solution 2==
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Let the terms in the sequence be defined as <math>a_1, a_2, ..., a_6</math>.
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Since the sequence is arithmetic, <math>a_1+a_6=a_2+a_5=a_3+a_4</math>. So, <math>\sum_{i=1}^6 a_i=3(a_1+a_6)=990</math>. Hence, <math>(a_1+a_6)=330</math>, and since <math>a_6=2a_1</math>, <math>3a_1=330\implies a_1=110</math> and <math>a_6=\boxed{220}</math>.
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~Kiran
 
==See also==
 
==See also==
 
{{AIME box|year=2023|before=First Problem|num-a=2|n=II}}
 
{{AIME box|year=2023|before=First Problem|num-a=2|n=II}}
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:00, 16 February 2023

Problem

The numbers of apples growing on each of six apple trees form an arithmetic sequence where the greatest number of apples growing on any of the six trees is double the least number of apples growing on any of the six trees. The total number of apples growing on all six trees is $990.$ Find the greatest number of apples growing on any of the six trees.

Solution 1

In the arithmetic sequence, let $a$ be the first term and $d$ be the common difference, where $d>0.$ The sum of the first six terms is \[a+(a+d)+(a+2d)+(a+3d)+(a+4d)+(a+5d) = 6a+15d.\] We are given that \begin{align*} 6a+15d &= 990, \\ 2a &= a+5d. \end{align*} The second equation implies that $a=5d.$ Substituting this into the first equation, we get \begin{align*} 6(5d)+15d &=990, \\ 45d &= 990 \\ d &= 22. \end{align*} It follows that $a=110.$ Therefore, the greatest number of apples growing on any of the six trees is $a+5d=\boxed{220}.$

~MRENTHUSIASM

Solution 2

Let the terms in the sequence be defined as $a_1, a_2, ..., a_6$.

Since the sequence is arithmetic, $a_1+a_6=a_2+a_5=a_3+a_4$. So, $\sum_{i=1}^6 a_i=3(a_1+a_6)=990$. Hence, $(a_1+a_6)=330$, and since $a_6=2a_1$, $3a_1=330\implies a_1=110$ and $a_6=\boxed{220}$.

~Kiran

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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