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

Latest revision as of 09:19, 14 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 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

@@ Line 1: / Line 1: @@
-The test isn't available yet, but here is a puzzle for you.
+{{duplicate|[[2024 AMC 10A Problems/Problem 19|2024 AMC 10A #19]] and [[2024 AMC 12A Problems/Problem 12|2024 AMC 12A #12]]}}
-A bus left from the Scarlet Devil Mansion; <math>3</math> people boarded at the start. At Hakugyokurou, <math>1</math> person got off and <math>\frac{1}{2}</math> a person boarded. At Yukari's house, <math>1</math> person left the bus. How many people are on the bus at the end?
+==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>
-<math>\mathrm{(A) \ } 0 \qquad \mathrm{(B) \ } 1 \qquad \mathrm{(C) \ } \frac{3}{2}\qquad \mathrm{(D) \ } 2 \qquad \mathrm{(E) \ } 9</math>
+<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math>
-play touhou
+==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>.
-https://en.touhouwiki.net/wiki/Touhou_Wiki
+(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)
-https://en.touhouwiki.net/wiki/Lyrics:_%E3%83%81%E3%83%AB%E3%83%8E%E3%81%AE%E3%83%91%E3%83%BC%E3%83%95%E3%82%A7%E3%82%AF%E3%83%88%E3%81%95%E3%82%93%E3%81%99%E3%81%86%E6%95%99%E5%AE%A4
-==Video Solution==
+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>
-https://m.youtube.com/watch?v=GNeyqFfXf7k&pp=ygUaY2lybm8ncyBwZXJmZWN0IG1hdGggY2xhc3M%3D&t=49
+~eevee9406
+==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>.
+~i_am_suk_at_math_2
+== Solution 3 (Similar to previous solutions) ==
+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>.
+~yuvag
+== Solution 4 ==
+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>.
+~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
+==See also==
+{{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}}
+{{AMC12 box|year=2024|ab=A|num-b=11|num-a=13}}
+{{MAA Notice}}

2024 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
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All AMC 12 Problems and Solutions

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