2024 AMC 10A Problems/Problem 19

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 $7+6+8=\boxed{\textbf{(E)}21}$ .

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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

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
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2024 AMC 10A Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See also