Difference between revisions of "2016 AIME II Problems/Problem 1"
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==Solution 1== | ==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. | ||
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== Solution 2 (Quadratic Formula)== | == Solution 2 (Quadratic Formula)== | ||
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<cmath>a=150\pm6\sqrt{625-576}=108, 192</cmath> | <cmath>a=150\pm6\sqrt{625-576}=108, 192</cmath> | ||
Since Alex's peanut number was the lowest of the trio, and <math>3*192>444</math>, Alex initially had <math>\boxed{108}</math> peanuts. | Since Alex's peanut number was the lowest of the trio, and <math>3*192>444</math>, Alex initially had <math>\boxed{108}</math> peanuts. | ||
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== Solution 3 == | == Solution 3 == | ||
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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>. | 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>. | Since <math>\lambda > 0</math>, we must have <math>\lambda = 42</math>, and so <math>a = 150 - \lambda = \boxed{108}</math>. | ||
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+ | == Solution 4== | ||
+ | Bashing is not difficult. All we have to consider is the first equation. We can write it as <math>x*(1+r+r^2) = 444</math>. The variable <math>x</math> must be an integer, and after trying all the factors of <math>444</math>, it's clear that <math>r</math> is a fraction smaller than <math>10</math>. When calculating the coefficient of <math>x</math>, we must consider that the fraction produced will very likely have a numerator that divides <math>444</math>. Trying a couple will make it easier to find the fraction, and soon you will find that <math>\frac{4}{3}</math> gives a numerator of <math>37</math>, a rather specific factor of <math>444</math>. Solving for the rest will give you an integer value of <math>\boxed{108}</math>. This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals. | ||
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== 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}} | ||
+ | {{MAA Notice}} |
Latest revision as of 19:00, 18 October 2020
Contents
Problem
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 2 (Quadratic Formula)
Let be Alex's peanuts and the common ratio. Then we have . Adding to both sides and factoring,
For the common difference, . Simplifying, . Factoring,
Substitute in the second equation to get . Expanding and applying the quadratic formula,
Taking out from under the radical leaves
Since Alex's peanut number was the lowest of the trio, and , Alex initially had peanuts.
Solution 3
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 .
Solution 4
Bashing is not difficult. All we have to consider is the first equation. We can write it as . The variable must be an integer, and after trying all the factors of , it's clear that is a fraction smaller than . When calculating the coefficient of , we must consider that the fraction produced will very likely have a numerator that divides . Trying a couple will make it easier to find the fraction, and soon you will find that gives a numerator of , a rather specific factor of . Solving for the rest will give you an integer value of . This is by no means a good solution, but it may be faster in a competition if you don't want to mess with several other equations. This is purely up to different individuals.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.