Difference between revisions of "2024 AMC 10A Problems/Problem 19"
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| + | ==Solution 2== | ||
| + | We have <math>720 = 2^4 * 3^2 * 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 * \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 * \frac{16}{15} = 768</math>, so our sum is <math>7+6+8=\boxed{\textbf{(E)}21}</math>. | ||
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| + | ~i_am_suk_at_math_2 | ||
==See also== | ==See also== | ||
{{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}} | {{AMC10 box|year=2024|ab=A|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:30, 8 November 2024
Contents
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 
.
~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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
