Difference between revisions of "2005 AIME II Problems/Problem 5"
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There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11= \boxed{054}</math>. | There are <math> 44-2+1=43 </math> possibilities for the square case and <math> 12-2+1=11 </math> possibilities for the cube case. Thus, the answer is <math> 43+11= \boxed{054}</math>. | ||
− | Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>. | + | Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs <math>(a,b)</math>, and not for the number of possible values of <math>b</math>. Were the problem to ask for the number of possible values of <math>b</math>, the values of <math>b^6</math> under <math>2005</math> would have to be subtracted, which would just be <math>2</math> values: <math>2^6</math> and <math>3^6</math>. However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).) |
==Solution II == | ==Solution II == |
Revision as of 20:05, 7 August 2018
Contents
Problem
Determine the number of ordered pairs of integers such that and
Solution I
The equation can be rewritten as Multiplying through by and factoring yields . Therefore, or , so either or .
- For the case , note that and . Thus, all values of from to will work.
- For the case , note that while . Therefore, for this case, all values of from to work.
There are possibilities for the square case and possibilities for the cube case. Thus, the answer is .
Note that Inclusion-Exclusion does not need to be used, as the problem is asking for ordered pairs , and not for the number of possible values of . Were the problem to ask for the number of possible values of , the values of under would have to be subtracted, which would just be values: and . However, the ordered pairs where b is to the sixth power are distinct, so they are not redundant. (For example, the pairs (4, 64) and (8, 64).)
Solution II
Let . Then our equation becomes . Multiplying through by and solving the quadratic gives us or . Hence or .
For the first case , can range from 2 to 44, a total of 43 values. For the second case , can range from 2 to 12, a total of 11 values.
Thus the total number of possible values is .
See also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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