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]]}} | {{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== | ==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> | ||
<math>\textbf{(A) } 9 \qquad \textbf{(B) } 12 \qquad \textbf{(C) } 16 \qquad \textbf{(D) } 18 \qquad \textbf{(E) } 21</math> | <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== | ==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>. | + | 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>. |
(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) | (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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==Solution 2== | ==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>. | + | 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 | ~i_am_suk_at_math_2 |
Revision as of 00:29, 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.
Contents
[hide]Problem
The first three terms of a geometric sequence are the integers and where What is the sum of the digits of the least possible value of
Solution 1
For a geometric sequence, we have , and we can test values for . We find that and works, and we can test multiples of in between the two values. Finding that none of the multiples of 5 divide besides itself, we know that the answer is .
(Note: To find the value of without bashing, we can observe that , and that multiplying it by gives us , which is really close to . ~ YTH)
Note: The reason why is because . Rearranging this gives
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Solution 2
We have . We want to find factors and where such that is minimized, as will then be the least possible value of . After experimenting, we see this is achieved when and , which means our value of is , so our sum is .
~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.