Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2005 AMC 12A Problems/Problem 21"

(solution)
 
m (Improved formatting of problem statement)
 
(One intermediate revision by one other user not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
How many ordered triples of [[integer]]s <math>(a,b,c)</math>, with <math>a \ge 2</math>, <math>b\ge 1</math>, and <math>c \ge 0</math>, satisfy both <math>\log_a b = c^{2005}</math> and <math>a + b + c = 2005</math>?
+
How many ordered triples of integers <math>(a,b,c)</math>, with <math>a \geq 2</math>, <math>b \geq 1</math>, and <math>c \geq 0</math>, satisfy both <math>\log_{a}b = c^{2005}</math> and <math>a + b + c = 2005</math>?
  
 
<math>\mathrm{(A)} \ 0 \qquad \mathrm{(B)} \ 1 \qquad \mathrm{(C)} \ 2 \qquad \mathrm{(D)} \ 3 \qquad \mathrm{(E)} \ 4</math>
 
<math>\mathrm{(A)} \ 0 \qquad \mathrm{(B)} \ 1 \qquad \mathrm{(C)} \ 2 \qquad \mathrm{(D)} \ 3 \qquad \mathrm{(E)} \ 4</math>
Line 17: Line 17:
  
 
[[Category:Introductory Algebra Problems]]
 
[[Category:Introductory Algebra Problems]]
 +
{{MAA Notice}}

Latest revision as of 15:03, 1 July 2025

Problem

How many ordered triples of integers $(a,b,c)$, with $a \geq 2$, $b \geq 1$, and $c \geq 0$, satisfy both $\log_{a}b = c^{2005}$ and $a + b + c = 2005$?

$\mathrm{(A)} \ 0 \qquad \mathrm{(B)} \ 1 \qquad \mathrm{(C)} \ 2 \qquad \mathrm{(D)} \ 3 \qquad \mathrm{(E)} \ 4$

Solution

$a^{c^{2005}} = b$

Casework upon $c$:

  • $c = 0$: Then $a^0 = b \Longrightarrow b = 1$. Thus we get $(2004,1,0)$.
  • $c = 1$: Then $a^1 = b \Longrightarrow a = b$. Thus we get $(1002,1002,1)$.
  • $c \ge 2$: Then the exponent of $a$ becomes huge, and since $a \ge 2$ there is no way we can satisfy the second condition. Hence we have two ordered triples $\mathrm{(C)}$.

See also

2005 AMC 12A (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
All AMC 12 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_21&oldid=253033"