Difference between revisions of "2023 AMC 12B Problems/Problem 7"
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+ | ~woeIsMe | ||
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==See Also== | ==See Also== | ||
{{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}} | {{AMC12 box|year=2023|ab=B|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:54, 15 November 2023
Problem
For how many integers does the expressionrepresent 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 log(n^2) can be written as 2log(n). Setting a=log(n), the equation becomes sqrt((2a-a^2)/(a-3)) which can be written as sqrt((a(a-2))/(a-3))
Case 1: a>=3 The expression is undefined when a=3. For a>3, it is trivial to see that the denominator is negative and the numerator is positive, thus resulting in no real solutions.
Case 2: 2<=a<3 For a=2, the numerator is zero, giving us a valid solution. When a>2, 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 10^2 and 10^3-1. There are 900 solutions here.
Case 3: 0<a<2 The numerator will be negative but the denominator is positive, no real solutions exist.
Case 4: a=0 The expression evaluates to zero, 1 valid solution exists.
Case 5: a<0 All values for a<0 requires 0<n<1, no integer solutions exist.
Adding up the cases: 900+1=
Side Note: Someone please fix this to make it readable. I don't know anything about latex.
~woeIsMe
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.