Difference between revisions of "2023 AMC 12B Problems/Problem 7"
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for <math>\frac{a(2-a)}{a-3} >0 </math> , | for <math>\frac{a(2-a)}{a-3} >0 </math> , |
Latest revision as of 22:08, 19 August 2024
Contents
Problem
For how many integers does the expression
represent a real number, where log denotes the base
logarithm?
Solution
We have
Because is an integer and
is well defined,
must be a positive integer.
Case 1: or
.
The above expression is 0. So these are valid solutions.
Case 2: .
Thus, and
.
To make the above expression real, we must have
.
Thus,
.
Thus,
.
Hence, the number of solutions in this case is 899.
Putting all cases together, the total number of solutions is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Solution (Solution 1 for dummies)
Notice can be written as
. Setting
, the equation becomes
which can be written as
Case 1:
The expression is undefined when
. For
, it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions.
Case 2:
For
, the numerator is zero, giving us a valid solution. When
, both the denominator and numerator are negative so all real values of a in this interval is a solution to the equation. All integers of n that makes this true are between
and
. There are 900 solutions here.
Case 3:
The numerator will be positive but the denominator is negative, no real solutions exist.
Case 4:
The expression evaluates to zero,
valid solution exists.
Case 5:
All values for
requires
, no integer solutions exist.
Adding up the cases:
~woeIsMe typesetting: paras
Solution 3 (3 degree polynominal graph)
for ,
transform it into a(a-2)(a-3) <0 ,
use the following graph to quickly confirm
1) a < 0 or
2) 2 < a <3
Video Solution
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See Also
2023 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 6 |
Followed by Problem 8 |
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.