Difference between revisions of "2016 AIME I Problems/Problem 7"
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We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely <math>ab<-2016</math> and <math>-2015\approx -2016</math>. | We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely <math>ab<-2016</math> and <math>-2015\approx -2016</math>. | ||
− | The factors of <math>(\text{-})2015</math> are <math>(\text{-})5\cdot 13\cdot 31</math>, so the <math>(a+1,b+1)=\{(-1, | + | The factors of <math>(\text{-})2015</math> are <math>(\text{-})5\cdot 13\cdot 31</math>, so the <math>(a+1,b+1)=\{(-1,2015),(-5,403), (-13,155),(-31,65)\}</math> and the sets flipped. |
Similarly from the second case of <math>(a-1,b-1)</math> we also have <math>8</math> solutions. | Similarly from the second case of <math>(a-1,b-1)</math> we also have <math>8</math> solutions. | ||
− | Thus, <math>(a,b),(b,a)=\{(\mp 2,\pm | + | Thus, <math>(a,b),(b,a)=\{(\mp 2,\pm 2014),(\mp 6,\pm 402),(\mp 14,\pm 154),(\mp 32,\pm 64)\}</math>. Surely, all of their products, <math>ab=-4028,-2412,-2156,-2048<2016</math>. |
So there are <math>87+16=\boxed{103}</math> solutions. | So there are <math>87+16=\boxed{103}</math> solutions. |
Revision as of 20:00, 26 June 2016
Contents
Problem
For integers and consider the complex number
Find the number of ordered pairs of integers such that this complex number is a real number.
Solution
We consider two cases:
Case 1: .
In this case, if then and . Thus so . Thus , yielding values. However since , we have . Thus there are allowed tuples in this case.
Case 2: .
In this case, we want Squaring, we have the equations (which always holds in this case) and Then if and , let . If , Note that for every one of these solutions. If , then Again, for every one of the above solutions. This yields solutions. Similarly, if and , there are solutions. Thus, there are a total of solutions in this case.
Thus, the answer is .
(Solution by gundraja)
Solution 2
Similar to Solution 1, but concise:
First, we set the imaginary expression to , so that or , of which there are possibilities. But because the denominator would be . So this gives solutions.
Then we try to cancel the imaginary part with the square root of the real part, which must be negative. So . by Simon's Factoring Trick.
We must have the negative part lesser in magnitude than the positive part, because an increase in magnitude of a lesser number is MORE than a decrease in the magnitude of a positive number, so the product will net to be more magnitude, namely and .
The factors of are , so the and the sets flipped.
Similarly from the second case of we also have solutions.
Thus, . Surely, all of their products, .
So there are solutions.
See also
2016 AIME I (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 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.