Difference between revisions of "1952 AHSME Problems/Problem 47"
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\textbf{(E) } 4,9,3 </math> | \textbf{(E) } 4,9,3 </math> | ||
| − | == Solution == | + | == Solution #1 == |
| − | <math>\fbox{}</math> | + | The easiest method, which in this case is not very time consuming, is to guess and check and use process of elimination. |
| + | We can immediately rule out (B) since it does not satisfy the third equation. | ||
| + | The first equation allows us to eliminate (A) and (C) because the bases are different and aren't powers of each other. | ||
| + | Finally, we manually check (D) and (E) and see that our answer is <math>\fbox{D}</math> | ||
| + | |||
| + | == Solution #2 == | ||
| + | Let us actually solve the problem through algebra. We start by simplifying each of the equations given to us. | ||
| + | |||
| + | Equation #1 | ||
| + | <cmath>z^x = y^{2x}</cmath> | ||
| + | <cmath>z = y^2</cmath> | ||
| + | |||
| + | Equation #2 | ||
| + | <cmath>2^z = 2 * 4^x</cmath> | ||
| + | <cmath>2^z = 2^{2x + 1}</cmath> | ||
| + | <cmath>z = 2x + 1</cmath> | ||
| + | |||
| + | Equation #3 | ||
| + | <cmath>x + y + z = 16</cmath> | ||
| + | |||
| + | Equations 1 & 2 tell us that <math>z</math> is an odd perfect square. They also tell us that <math>z</math> is the largest of the three integers. | ||
| + | We see that if we assume <math>z = 9</math>, then <math>y = 3</math> and <math>x = 4</math>, which satisfies all three equations. | ||
| + | |||
| + | Therefore, the answer is <math>\fbox{(D) 4,3,9}</math> | ||
== See also == | == See also == | ||
Latest revision as of 19:48, 22 December 2015
Contents
Problem
In the set of equations 
, the integral roots in the order 
are:
Solution #1
The easiest method, which in this case is not very time consuming, is to guess and check and use process of elimination.
We can immediately rule out (B) since it does not satisfy the third equation.
The first equation allows us to eliminate (A) and (C) because the bases are different and aren't powers of each other.
Finally, we manually check (D) and (E) and see that our answer is 
Solution #2
Let us actually solve the problem through algebra. We start by simplifying each of the equations given to us.
Equation #1
![\[z^x = y^{2x}\]](http://latex.artofproblemsolving.com/8/5/d/85d9ebde324d698585f99cf4b704cc685dd525c5.png)
![\[z = y^2\]](http://latex.artofproblemsolving.com/2/6/4/2649c82df303f92dcb5db34b2aaa2ef4f4f6f50f.png)
Equation #2
![\[2^z = 2 * 4^x\]](http://latex.artofproblemsolving.com/8/e/5/8e5c41bac7f9b44627d79f1c09c102b9792bf609.png)
![\[2^z = 2^{2x + 1}\]](http://latex.artofproblemsolving.com/b/4/b/b4bdd207777280a56bbe418622f2365fc828e13d.png)
![\[z = 2x + 1\]](http://latex.artofproblemsolving.com/e/c/d/ecda4cad1829b190d9f03bf05d4b168317632f88.png)
Equation #3
![\[x + y + z = 16\]](http://latex.artofproblemsolving.com/8/d/4/8d42c36783239bb62023c9c1b5aaa113c6210cd3.png)
Equations 1 & 2 tell us that 
is an odd perfect square. They also tell us that is the largest of the three integers.
We see that if we assume 
, then 
and 
, which satisfies all three equations.
Therefore, the answer is 
See also
| 1952 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 46 |
Followed by Problem 48 | |
| 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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
