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

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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]]}}
  
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==Problem==
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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>
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math>
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==Solution 1==
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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>.
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(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)
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Note: The reason why <math>ab=720^2</math> is because <math>b/720 = 720/a</math>. Rearranging this gives <math>ab = 720^2</math>
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~eevee9406
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==Solution 2==
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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>.
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~i_am_suk_at_math_2
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== Solution 3 (Similar to previous solutions) ==
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To minimize the value of <math>b</math>, where it has to be an integer, and it has to be greater than 720, we can express the common ratio as <math>\dfrac{n+1}{n}</math>, where the value has to be greater than <math>1</math>, and <math>n</math>, and <math>n+1</math> have to be factors of 720. Since the bigger the denominator gets, the smaller the value of the fraction, we essentially have to find the biggest value for <math>n</math>, where itself and <math>n+1</math> are factors of <math>720</math>. From here, we can check whether <math>n(n+1) = 720</math> yields an integer root, which it doesn't. So, then we check the next biggest factor of <math>720</math>, which is <math>360</math>. <math>n(n+1) = 360</math>, this doesn't have an integer root either. So, then we check the next biggest factor which is <math>240</math>, <math>n(n+1) = 240</math>, which we get <math>15</math> as a root. This means the common ration is <math>\dfrac{15}{16}</math>. We then multiply <math>\dfrac{15}{16}</math> times <math>720</math> and add up the digits getting <math>\boxed{\textbf{(E) } 21}</math>.
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~yuvag
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== Solution 4 ==
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Let the common ratio of the geometric sequence be <math>\frac{x}{y}</math>, with <math>x>y</math>. This means that <math>720(\frac{x}{y})</math> and <math>720(\frac{y}{x})</math> must both be integers, therefore <math>x</math> and <math>y</math> are both factors of <math>720</math>. We would achieve the smallest ratio <math>\frac{x}{y}</math> if <math>x</math> and <math>y</math> are consecutive, so by listing out the factors of <math>720</math>, we find that <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, <math>5</math>, <math>6</math>, <math>8</math>, <math>9</math>, <math>15</math>, <math>16</math> are the only consecutive factors (any factors larger than these would result in the ratio simplifying, making it larger than just using consecutive integers). <math>15</math> and <math>16</math> are the largest, so we find the common ratio to be <math>\frac{16}{15}</math>, making <math>b=720(\frac{16}{15})</math> giving us <math>768</math>. The sum of its digits is <math>\boxed{\textbf{(E) } 21}</math>.
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~lisztepos
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==Video Solution 1 by SF==
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https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D
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==Detailed Video Solution by Scholars Foundation==
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https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D
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==Video Solution by Pi Academy==
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https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE
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==See also==
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{{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}}
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{{AMC12 box|year=2024|ab=A|num-b=11|num-a=13}}
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{{MAA Notice}}

Latest revision as of 21:49, 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

Solution 3 (Similar to previous solutions)

To minimize the value of $b$, where it has to be an integer, and it has to be greater than 720, we can express the common ratio as $\dfrac{n+1}{n}$, where the value has to be greater than $1$, and $n$, and $n+1$ have to be factors of 720. Since the bigger the denominator gets, the smaller the value of the fraction, we essentially have to find the biggest value for $n$, where itself and $n+1$ are factors of $720$. From here, we can check whether $n(n+1) = 720$ yields an integer root, which it doesn't. So, then we check the next biggest factor of $720$, which is $360$. $n(n+1) = 360$, this doesn't have an integer root either. So, then we check the next biggest factor which is $240$, $n(n+1) = 240$, which we get $15$ as a root. This means the common ration is $\dfrac{15}{16}$. We then multiply $\dfrac{15}{16}$ times $720$ and add up the digits getting $\boxed{\textbf{(E) } 21}$.

~yuvag

Solution 4

Let the common ratio of the geometric sequence be $\frac{x}{y}$, with $x>y$. This means that $720(\frac{x}{y})$ and $720(\frac{y}{x})$ must both be integers, therefore $x$ and $y$ are both factors of $720$. We would achieve the smallest ratio $\frac{x}{y}$ if $x$ and $y$ are consecutive, so by listing out the factors of $720$, we find that $1$, $2$, $3$, $4$, $5$, $6$, $8$, $9$, $15$, $16$ are the only consecutive factors (any factors larger than these would result in the ratio simplifying, making it larger than just using consecutive integers). $15$ and $16$ are the largest, so we find the common ratio to be $\frac{16}{15}$, making $b=720(\frac{16}{15})$ giving us $768$. The sum of its digits is $\boxed{\textbf{(E) } 21}$.

~lisztepos

Video Solution 1 by SF

https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D

Detailed Video Solution by Scholars Foundation

https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D

Video Solution by Pi Academy

https://youtu.be/fW7OGWee31c?si=oq7toGPh2QaksLHE

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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