Difference between revisions of "2023 AMC 12B Problems/Problem 7"
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==Solution (Solution 1 for dummies)== | ==Solution (Solution 1 for dummies)== | ||
− | Notice log(n^2) can be written as | + | Notice <math>\log(n^2)</math> can be written as <math>2\log(n)</math>. Setting <math>a=\log(n)</math>, the equation becomes <math>\sqrt{\frac{2a-a^2}{a-3}}</math> which can be written as <math>\sqrt{\frac{a(2-a)}{a-3}}</math> |
− | Case 1: a> | + | Case 1: <math>a \ge 3</math> |
− | The expression is undefined when a=3. For a>3, it is trivial to see that the denominator is | + | The expression is undefined when <math>a=3</math>. For <math>a>3</math>, it is trivial to see that the denominator is positive and the numerator is negative, thus resulting in no real solutions. |
− | Case 2: 2 | + | Case 2: <math>2 \le a<3</math> |
− | 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. | + | For <math>a=2</math>, the numerator is zero, giving us a valid solution. When <math>a>2</math>, 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 <math>10^2</math> and <math>10^3-1</math>. There are 900 solutions here. |
− | Case 3: 0<a<2 | + | Case 3: <math>0<a<2</math> |
− | The numerator will be | + | The numerator will be positive but the denominator is negative, no real solutions exist. |
− | Case 4: a=0 | + | Case 4: <math>a=0</math> |
− | The expression evaluates to zero, 1 valid solution exists. | + | The expression evaluates to zero, <math>1</math> valid solution exists. |
− | Case 5: a<0 | + | Case 5: <math>a<0</math> |
− | All values for a<0 requires 0<n<1, no integer solutions exist. | + | All values for <math>a<0</math> requires <math>0<n<1</math>, no integer solutions exist. |
Adding up the cases: | Adding up the cases: | ||
− | 900+1= | + | <math>900+1=\boxed{\textbf{(E) 901}}</math> |
+ | |||
+ | ~woeIsMe | ||
+ | typesetting: paras | ||
+ | |||
+ | ==Video Solution== | ||
+ | |||
+ | https://youtu.be/GGdJJzzbivM | ||
+ | |||
+ | |||
+ | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
− | |||
==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 18:06, 9 December 2023
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
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.