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Difference between revisions of "2024 AMC 10A Problems/Problem 19"

m (→‎Solution 2)
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{{duplicate|[[2024 AMC 10A Problems/Problem 19|2024 AMC 10A #19]] and [[2024 AMC 12A Problems/Problem 12|2024 AMC 12A #12]]}}
 
{{duplicate|[[2024 AMC 10A Problems/Problem 19|2024 AMC 10A #19]] and [[2024 AMC 12A Problems/Problem 12|2024 AMC 12A #12]]}}
 +
 
==Problem==
 
==Problem==
 
The first three terms of a geometric sequence are the integers <math>a,\,720,</math> and <math>b,</math> where <math>a<720<b.</math> What is the sum of the digits of the least possible value of <math>b?</math>
 
The first three terms of a geometric sequence are the integers <math>a,\,720,</math> and <math>b,</math> where <math>a<720<b.</math> What is the sum of the digits of the least possible value of <math>b?</math>
  
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math>
 
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math>
 +
 
==Solution 1==
 
==Solution 1==
−
For a geometric sequence, we have <math>ab=720^2=2^8 3^4 5^2</math>, and we can test values for <math>b</math>. We find that <math>b=768</math> and <math>a=675</math> works, and we can test multiples of <math>5</math> in between the two values. Finding that none of the multiples of 5 divide <math>720^2</math> besides <math>720</math> itself, we know that the answer is <math>7+6+8=\boxed{\textbf{(E)}21}</math>.
+
For a geometric sequence, we have <math>ab=720^2=2^8 3^4 5^2</math>, and we can test values for <math>b</math>. We find that <math>b=768</math> and <math>a=675</math> works, and we can test multiples of <math>5</math> in between the two values. Finding that none of the multiples of 5 divide <math>720^2</math> besides <math>720</math> itself, we know that the answer is <math>7+6+8=\boxed{\textbf{(E) } 21}</math>.
  
 
(Note: To find the value of <math>b</math> without bashing, we can observe that <math>2^8=256</math>, and that multiplying it by <math>3</math> gives us <math>768</math>, which is really close to <math>720</math>. ~ YTH)
 
(Note: To find the value of <math>b</math> without bashing, we can observe that <math>2^8=256</math>, and that multiplying it by <math>3</math> gives us <math>768</math>, which is really close to <math>720</math>. ~ YTH)
Line 14: Line 16:
  
 
==Solution 2==
 
==Solution 2==
−
We have <math>720 = 2^4 \cdot 3^2 \cdot 5</math>. We want to find factors <math>x</math> and <math>y</math> where <math>y>x</math> such that <math>\frac{y}{x}</math> is minimized, as <math>720 \cdot \frac{y}{x}</math> will then be the least possible value of <math>b</math>. After experimenting, we see this is achieved when <math>y=16</math> and <math>x=15</math>, which means our value of <math>b</math> is <math>720 \cdot \frac{16}{15} = 768</math>, so our sum is <math>7+6+8=\boxed{\textbf{(E)}21}</math>.
+
We have <math>720 = 2^4 \cdot 3^2 \cdot 5</math>. We want to find factors <math>x</math> and <math>y</math> where <math>y>x</math> such that <math>\frac{y}{x}</math> is minimized, as <math>720 \cdot \frac{y}{x}</math> will then be the least possible value of <math>b</math>. After experimenting, we see this is achieved when <math>y=16</math> and <math>x=15</math>, which means our value of <math>b</math> is <math>720 \cdot \frac{16}{15} = 768</math>, so our sum is <math>7+6+8=\boxed{\textbf{(E) } 21}</math>.
  
 
~i_am_suk_at_math_2
 
~i_am_suk_at_math_2

Revision as of 00:29, 9 November 2024

The following problem is from both the 2024 AMC 10A #19 and 2024 AMC 12A #12, so both problems redirect to this page.

Problem

The first three terms of a geometric sequence are the integers $a,\,720,$ and $b,$ where $a<720<b.$ What is the sum of the digits of the least possible value of $b?$

$\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21$

Solution 1

For a geometric sequence, we have $ab=720^2=2^8 3^4 5^2$, and we can test values for $b$. We find that $b=768$ and $a=675$ works, and we can test multiples of $5$ in between the two values. Finding that none of the multiples of 5 divide $720^2$ besides $720$ itself, we know that the answer is $7+6+8=\boxed{\textbf{(E) } 21}$.

(Note: To find the value of $b$ without bashing, we can observe that $2^8=256$, and that multiplying it by $3$ gives us $768$, which is really close to $720$. ~ YTH)

Note: The reason why $ab=720^2$ is because $b/720 = 720/a$. Rearranging this gives $ab = 720^2$

~eevee9406

Solution 2

We have $720 = 2^4 \cdot 3^2 \cdot 5$. We want to find factors $x$ and $y$ where $y>x$ such that $\frac{y}{x}$ is minimized, as $720 \cdot \frac{y}{x}$ will then be the least possible value of $b$. After experimenting, we see this is achieved when $y=16$ and $x=15$, which means our value of $b$ is $720 \cdot \frac{16}{15} = 768$, so our sum is $7+6+8=\boxed{\textbf{(E) } 21}$.

~i_am_suk_at_math_2

See also

2024 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions
2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 11 		Followed by
Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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