Difference between revisions of "2003 AMC 10B Problems/Problem 14"

Revision as of 10:06, 10 June 2011

Given that $3^8\cdot5^2=a^b,$ where both $a$ and $b$ are positive integers, find the smallest possible value for $a+b$ .

$\textbf{(A) } 25 \qquad\textbf{(B) } 34 \qquad\textbf{(C) } 351 \qquad\textbf{(D) } 407 \qquad\textbf{(E) } 900$

$3^8\cdot5^2 = (3^4)^2\cdot5^2 = (3^4\cdot5)^2 = 405^2$

$405$ is not a perfect power, so the smallest possible value of $a+b$ is $405+2=\boxed{\mathrm{(D) \ } 407}$ .

2003 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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
All AMC 10 Problems and Solutions

@@ Line 7: / Line 7: @@
 ==Solution==
-<math>3^8\cdot5^2</math>
+<cmath>3^8\cdot5^2 = (3^4)^2\cdot5^2 = (3^4\cdot5)^2 = 405^2</cmath>
-<math>=3^2\cdot3^2\cdot3^2\cdot3^2\cdot5^2</math>
+<math>405</math> is not a perfect power, so the smallest possible value of <math>a+b</math> is <math>405+2=\boxed{\mathrm{(D) \ } 407}</math>.
-<math>=(3\cdot3\cdot3\cdot3\cdot5)^2</math>
-<math>=405^2</math>
-<math>405</math> is not a perfect power, so the smallest possible value of <math>a+b</math> is <math>405+2=\boxed{407\textbf{ (D)}}</math>.
 ==See Also==
 {{AMC10 box|year=2003|ab=B|num-b=13|num-a=15}}