Difference between revisions of "2016 AIME II Problems/Problem 1"

Revision as of 21:20, 7 September 2019

Problem

Initially Alex, Betty, and Charlie had a total of $444$ 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 $5$ of his peanuts, Betty eats $9$ of her peanuts, and Charlie eats $25$ 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 $r$ be the common ratio, where $r>1$ . We then have $ar-9-(a-5)=a(r-1)-4=ar^{2}-25-(ar-9)=ar(r-1)-16$ . We now have, letting, subtracting the 2 equations, $ar^{2}+-2ar+a=12$ , so we have $3ar=432,$ or $ar=144$ , which is how much Betty had. Now we have $144+\dfrac{144}{r}+144r=444$ , or $144(r+\dfrac{1}{r})=300$ , or $r+\dfrac{1}{r}=\dfrac{25}{12}$ , which solving for $r$ gives $r=\dfrac{4}{3}$ , since $r>1$ , so Alex had $\dfrac{3}{4} \cdot 144=\boxed{108}$ peanuts.

Solution by Shaddoll

Solution 2 (Quadratic Formula)

Let $a$ be Alex's peanuts and $k$ the common ratio. Then we have $a(k^2+k+1)=444$ . Adding $k$ to both sides and factoring,

$\frac{444}{a}+k=(k+1)^2$

For the common difference, $ak=5-(a-5)=ak^2-25-(ak-9)$ . Simplifying, $k^2-2k+1=\frac{12}{a}$ . Factoring, $(k-1)^2=\frac{12}{a}$

$(k+1)^2-(k-1)^2=4k$ , so $4k=\frac{444-20}{a}$ . Then, $k=\frac{432}{3a}=\frac{432}{a}=\frac{144}{a}$ . Substitute k in $(k-1)^2=\frac{12}{a}$ to get $(\frac{144-a}{a})^2=\frac{12}{a}$ . Simplifying, expanding, and applying the quadratic formula,

\[a=150\pm\frac{\sqrt{300^2-4(144^2)}{2}\] (Error compiling LaTeX. Unknown error_msg)

Taking out $4^2\cdot3^2$ from under the radical leaves $a=150\pm6\sqrt{625-576}=108, 192$ Since Alex's peanut number was the lowest of the trio, and $3*192>444$ , Alex initially had \boxed{108} peanutes.

(Solution by BJHHar)

Solution 3

Let the initial numbers of peanuts Alex, Betty and Charlie had be $a$ , $b$ , and $c$ respectively. Let the final numbers of peanuts, after eating, be $a'$ , $b'$ , and $c'$ .

We are given that $a + b + c = 444$ . Since a total of $5 + 9 + 25 = 39$ peanuts are eaten, we must have $a' + b' + c' = 444 - 39 = 405$ . Since $a'$ , $b'$ , and $c'$ form an arithmetic progression, we have that $a' = b' - x$ and $c' = b' + x$ for some integer $x$ . Substituting yields $3b' = 405$ and so $b' = 135$ . Since Betty ate $9$ peanuts, it follows that $b = b' + 9 = 144$ .

Since $a$ , $b$ , and $c$ form a geometric progression, we have that $a = \frac{b}{r}$ and $c = br$ . Multiplying yields $ac = b^2 = 144^2$ . Since $a + c = 444 - b = 300$ , it follows that $a = 150 - \lambda$ and $c = 150 + \lambda$ for some integer $\lambda$ . Substituting yields $(150-\lambda)(150+\lambda) = 144^2$ , which expands and rearranges to $\lambda^2 = 150^2-144^2 = 42^2$ . Since $\lambda > 0$ , we must have $\lambda = 42$ , and so $a = 150 - \lambda = \boxed{108}$ .

@@ Line 7: / Line 7: @@
 Solution by Shaddoll
-== Solution 2 ==
+== Solution 2 (Quadratic Formula)==
+Let <math>a</math> be Alex's peanuts and <math>k</math> the common ratio. Then we have <math>a(k^2+k+1)=444</math>. Adding <math>k</math> to both sides and factoring,
+<cmath>\frac{444}{a}+k=(k+1)^2</cmath>
+For the common difference, <math>ak=5-(a-5)=ak^2-25-(ak-9)</math>. Simplifying, <math>k^2-2k+1=\frac{12}{a}</math>. Factoring,
+<cmath>(k-1)^2=\frac{12}{a}</cmath>
+<math>(k+1)^2-(k-1)^2=4k</math>, so <math>4k=\frac{444-20}{a}</math>. Then, <math>k=\frac{432}{3a}=\frac{432}{a}=\frac{144}{a}</math>. Substitute k in <math>(k-1)^2=\frac{12}{a}</math> to get <math>(\frac{144-a}{a})^2=\frac{12}{a}</math>. Simplifying, expanding, and applying the quadratic formula,
+<cmath>a=150\pm\frac{\sqrt{300^2-4(144^2)}{2}</cmath>
+Taking out <math>4^2\cdot3^2</math> from under the radical leaves
+<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 \boxed{108} peanutes.
+(Solution by BJHHar)
+== Solution 3 ==
 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>.

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

Page

Toolbox

Search