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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]]}}
 
==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>
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==See also==
 
==See also==
 
{{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}}
 
{{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}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:38, 8 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)

~eevee9406

Solution 2

We have $720 = 2^4 * 3^2 * 5$. We want to find factors $x$ and $y$ where $y>x$ such that $\frac{y}{x}$ is minimized, as $720 * \frac{y}{x}$ 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 $720 * \frac{16}{15} = 768$, 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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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