Difference between revisions of "2024 AMC 10A Problems/Problem 19"
Line 32: | Line 32: | ||
~lisztepos | ~lisztepos | ||
− | ==Video Solution | + | |
+ | == 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 | https://www.youtube.com/watch?v=JoF4-fo7W1w&pp=ygULYW1jIDEwIDIwMjQ%3D | ||
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.
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 .
~yuvag
Solution 4
Let the common ratio of the geometric sequence be , with . This means that and must both be integers, therefore and are both factors of . We would achieve the smallest ratio if and are consecutive, so by listing out the factors of , we find that , , , , , , , , , are the only consecutive factors (any factors larger than these would result in the ratio simplifying, making it larger than just using consecutive integers). and are the largest, so we find the common ratio to be , making giving us . The sum of its digits is .
~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
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.