Difference between revisions of "2024 AMC 10A Problems/Problem 19"
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(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) | ||
− | + | 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> | |
~eevee9406 | ~eevee9406 |
Revision as of 18:57, 8 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
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.