Difference between revisions of "2016 AIME II Problems/Problem 1"
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Initially Alex, Betty, and Charlie had a total of <math>444</math> peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats <math>5</math> of his peanuts, Betty eats <math>9</math> of her peanuts, and Charlie eats <math>25</math> of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | Initially Alex, Betty, and Charlie had a total of <math>444</math> peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats <math>5</math> of his peanuts, Betty eats <math>9</math> of her peanuts, and Charlie eats <math>25</math> of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially. | ||
− | ==Solution== | + | ==Solution 1== |
Let <math>r</math> be the common ratio, where <math>r>1</math>. We then have <math>ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16</math>. We now have, letting, subtracting the 2 equations, <math>ar^{2}+-2ar+a=12</math>, so we have <math>3ar=432,</math> or <math>ar=144</math>, which is how much Betty had. Now we have <math>144+\dfrac{144}{r}+144r=444</math>, or <math>144(r+\dfrac{1}{r})=300</math>, or <math>r+\dfrac{1}{r}=\dfrac{25}{12}</math>, which solving for <math>r</math> gives <math>r=\dfrac{4}{3}</math>, since <math>r>1</math>, so Alex had <math>\dfrac{3}{4} \cdot 144=\boxed{108}</math> peanuts. | Let <math>r</math> be the common ratio, where <math>r>1</math>. We then have <math>ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16</math>. We now have, letting, subtracting the 2 equations, <math>ar^{2}+-2ar+a=12</math>, so we have <math>3ar=432,</math> or <math>ar=144</math>, which is how much Betty had. Now we have <math>144+\dfrac{144}{r}+144r=444</math>, or <math>144(r+\dfrac{1}{r})=300</math>, or <math>r+\dfrac{1}{r}=\dfrac{25}{12}</math>, which solving for <math>r</math> gives <math>r=\dfrac{4}{3}</math>, since <math>r>1</math>, so Alex had <math>\dfrac{3}{4} \cdot 144=\boxed{108}</math> peanuts. | ||
Solution by Shaddoll | Solution by Shaddoll | ||
+ | == Solution 2 == | ||
+ | Let the initial numbers of peanuts Alex, Betty and Charlie had be <math>a</math>, <math>b</math>, and <math>c</math> respectively. | ||
+ | Let the final numbers of peanuts, after eating, be <math>a'</math>, <math>b'</math>, and <math>c'</math>. | ||
+ | |||
+ | We are given that <math>a + b + c = 444</math>. Since a total of <math>5 + 9 + 25 = 39</math> peanuts are eaten, we must have <math>a' + b' + c' = 444 - 39 = 405</math>. | ||
+ | Since <math>a'</math>, <math>b'</math>, and <math>c'</math> form an arithmetic progression, we have that <math>a' = b' - x</math> and <math>c' = b' + x</math> for some integer <math>x</math>. | ||
+ | Substituting yields <math>3b' = 405</math> and so <math>b' = 135</math>. Since Betty ate <math>9</math> peanuts, it follows that <math>b = b' + 9 = 144</math>. | ||
+ | |||
+ | Since <math>a</math>, <math>b</math>, and <math>c</math> form a geometric progression, we have that <math>a = \frac{b}{r}</math> and <math>c = br</math>. | ||
+ | Multiplying yields <math>ac = b^2 = 144^2</math>. | ||
+ | Since <math>a + c = 444 - b = 300</math>, it follows that <math>a = 150 - \lambda</math> and <math>c = 150 + \lambda</math> for some integer <math>\lambda</math>. | ||
+ | Substituting yields <math>(150-\lambda)(150+\lambda) = 144^2</math>, which expands and rearranges to <math>\lambda^2 = 150^2-144^2 = 42^2</math>. | ||
+ | Since <math>\lambda > 0</math>, we must have <math>\lambda = 42</math>, and so <math>a = 150 - \lambda = \boxed{108}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2016|n=II|before=First Problem|num-a=2}} | {{AIME box|year=2016|n=II|before=First Problem|num-a=2}} |
Revision as of 18:23, 26 May 2017
Initially Alex, Betty, and Charlie had a total of peanuts. Charlie had the most peanuts, and Alex had the least. The three numbers of peanuts that each person had formed a geometric progression. Alex eats of his peanuts, Betty eats of her peanuts, and Charlie eats of his peanuts. Now the three numbers of peanuts each person has forms an arithmetic progression. Find the number of peanuts Alex had initially.
Solution 1
Let be the common ratio, where . We then have . We now have, letting, subtracting the 2 equations, , so we have or , which is how much Betty had. Now we have , or , or , which solving for gives , since , so Alex had peanuts.
Solution by Shaddoll
Solution 2
Let the initial numbers of peanuts Alex, Betty and Charlie had be , , and respectively. Let the final numbers of peanuts, after eating, be , , and .
We are given that . Since a total of peanuts are eaten, we must have . Since , , and form an arithmetic progression, we have that and for some integer . Substituting yields and so . Since Betty ate peanuts, it follows that .
Since , , and form a geometric progression, we have that and . Multiplying yields . Since , it follows that and for some integer . Substituting yields , which expands and rearranges to . Since , we must have , and so .
See also
2016 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 |