2024 AMC 10A Problems/Problem 19
- 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]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3 (Similar to previous solutions)
- 5 Solution 4
- 6 Solution 5 (Quick but not much recommended)
- 7 Video Solution by Power Solve
- 8 Video Solution by SF
- 9 Detailed Video Solution by Scholars Foundation
- 10 Video Solution by Pi Academy
- 11 Video Solution by SpreadTheMathLove
- 12 See also
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
Note: Another reason that is because the
(as the middle term in a geometric series is always the geometric mean [the geometric mean is the square root of the product of the first and last terms of the series]) and squaring on both sides results in
.
~ThatPrimePunnyGuy
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
Solution 5 (Quick but not much recommended)
To find the smallest possible value of such that
, we need to find the
factors of
that are closest to each other. After a little bit of light bash we find that those
numbers are
and
. Note that the
numbers don't necessarily have to multiply to
. You could have used prime factorization to find this. Next all we do is find
which is just:
.
-jb2015007
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 |
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