2024 AMC 10A Problems/Problem 19

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 by Power Solve

https://www.youtube.com/watch?v=B0JOMiiCtAo

Video Solution 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

Video Solution by SpreadTheMathLove

https://www.youtube.com/watch?v=6SQ74nt3ynw

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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