Difference between revisions of "1993 AHSME Problems/Problem 3"

(Created page with "== Problem == <math>\frac{15^{30}}{45^{15}} =</math> <math>\text{(A)} \left(\frac{1}{3}\right)^{15}\quad \text{(B)} \left(\frac{1}{3}\right)^{2}\quad \text{(C)} 1\quad \text{(D)...")
 
m (Solution)
 
(One intermediate revision by one other user not shown)
Line 2: Line 2:
 
<math>\frac{15^{30}}{45^{15}} =</math>
 
<math>\frac{15^{30}}{45^{15}} =</math>
  
<math>\text{(A)} \left(\frac{1}{3}\right)^{15}\quad
+
<math>\text{(A) } \left(\frac{1}{3}\right)^{15}\quad
\text{(B)} \left(\frac{1}{3}\right)^{2}\quad
+
\text{(B) } \left(\frac{1}{3}\right)^{2}\quad
\text{(C)} 1\quad
+
\text{(C) } 1\quad
\text{(D)} 3^{15}\quad
+
\text{(D) } 3^{15}\quad
\text{(E)} 5^{15}</math>
+
\text{(E) } 5^{15}</math>
  
 
== Solution ==
 
== Solution ==
 +
 +
First we must convert these to the same bases. We can rewrite <math>45^{15}</math> as <math>15^{15} \cdot 3^{15}</math> Now
 +
 +
<math>\frac{15^{30}}{15^{15} \cdot 3^{15}}</math>
 +
 +
Canceling....
 +
 +
<math>\frac{15^{15}}{3^{15}} \Rightarrow (\frac{15}{3})^{15} \Rightarrow 5^{15}</math>
 +
 
<math>\fbox{E}</math>
 
<math>\fbox{E}</math>
  

Latest revision as of 23:34, 29 November 2016

Problem

$\frac{15^{30}}{45^{15}} =$

$\text{(A) } \left(\frac{1}{3}\right)^{15}\quad \text{(B) } \left(\frac{1}{3}\right)^{2}\quad \text{(C) } 1\quad \text{(D) } 3^{15}\quad \text{(E) } 5^{15}$

Solution

First we must convert these to the same bases. We can rewrite $45^{15}$ as $15^{15} \cdot 3^{15}$ Now

$\frac{15^{30}}{15^{15} \cdot 3^{15}}$

Canceling....

$\frac{15^{15}}{3^{15}} \Rightarrow (\frac{15}{3})^{15} \Rightarrow 5^{15}$

$\fbox{E}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 2
Followed by
Problem 4
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png