Difference between revisions of "1985 AHSME Problems/Problem 21"
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==Problem== | ==Problem== | ||
− | How many integers <math> x </math> satisfy the equation < | + | How many integers <math>x</math> satisfy the equation <cmath>\left(x^2-x-1\right)^{x+2} = 1?</cmath> |
<math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these} </math> | <math> \mathrm{(A)\ } 2 \qquad \mathrm{(B) \ }3 \qquad \mathrm{(C) \ } 4 \qquad \mathrm{(D) \ } 5 \qquad \mathrm{(E) \ }\text{none of these} </math> | ||
==Solution== | ==Solution== | ||
− | + | We recall that for real numbers <math>a</math> and <math>b</math>, there are exactly <math>3</math> ways in which we can have <math>a^b = 1</math>, namely <math>a = 1</math>; <math>b = 0</math> and <math>a \neq 0</math>; or <math>a = -1</math> and <math>b</math> is an even integer. | |
− | < | + | The first case therefore gives <cmath>\begin{align*}x^2-x-1 = 1 &\iff x^2-x-2 = 0 \&\iff (x-2)(x+1) = 0 \&\iff x = 2 \text{ or } x = -1.\end{align*}</cmath> |
− | <math> | + | Similarly, the second case gives <math>x+2 = 0</math>, i.e. <math>x = -2</math>, and this indeed gives <math>x^2-x-1 = 4+2-1 = 5 \neq 0</math>, so <math>x = -2</math> is a further valid solution. |
− | <math> x=2, | + | Lastly, for the third case, we have <cmath> |
− | + | Accordingly, the possible values of <math>x = -2</math>, <math>-1</math>, <math>0</math>, or <math>2</math>, yielding a total of <math>\boxed{\text{(C)} \ 4}</math> solutions. | |
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==See Also== | ==See Also== | ||
{{AHSME box|year=1985|num-b=20|num-a=22}} | {{AHSME box|year=1985|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 22:29, 19 March 2024
Problem
How many integers satisfy the equation
Solution
We recall that for real numbers and , there are exactly ways in which we can have , namely ; and ; or and is an even integer.
The first case therefore gives
Similarly, the second case gives , i.e. , and this indeed gives , so is a further valid solution.
Lastly, for the third case, we have but would give , which is odd, whereas gives , which is even. Therefore, this case gives only one further solution, namely .
Accordingly, the possible values of , , , or , yielding a total of solutions.
See Also
1985 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
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