Difference between revisions of "2024 AMC 10A Problems/Problem 19"
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== Solution 3 (Similar to previous solutions) == | == 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>\ | + | 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>. |
Somebody pls fix my terrible <math>\LaTeX</math> | Somebody pls fix my terrible <math>\LaTeX</math> |
Revision as of 10:22, 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
~eevee9406
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
Solution 3 (Similar to previous solutions)
To minimize the value of , where it has to be an integer, and it has to be greater than 720, we can express the common ratio as , where the value has to be greater than , and , and 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 , where itself and are factors of . From here, we can check whether yields an integer root, which it doesn't. So, then we check the next biggest factor of , which is . , this doesn't have an integer root either. So, then we check the next biggest factor which is , , which we get as a root. This means the common ration is . We then multiply times and add up the digits getting .
Somebody pls fix my terrible
~yuvag
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.