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 "2012 AIME I Problems/Problem 1"

m (→‎Solutions: -Leading zeroes are important for AIME.)
m (→‎Solution 1)
Line 7: Line 7:
  
 
A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 \cdot 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 \cdot 10 = \boxed{040}</math>.
 
A positive integer is divisible by <math>4</math> if and only if its last two digits are divisible by <math>4.</math> For any value of <math>b</math>, there are two possible values for <math>a</math> and <math>c</math>, since we find that if <math>b</math> is even, <math>a</math> and <math>c</math> must be either <math>4</math> or <math>8</math>, and if <math>b</math> is odd, <math>a</math> and <math>c</math> must be either <math>2</math> or <math>6</math>. There are thus <math>2 \cdot 2 = 4</math> ways to choose <math>a</math> and <math>c</math> for each <math>b,</math> and <math>10</math> ways to choose <math>b</math> since <math>b</math> can be any digit. The final answer is then <math>4 \cdot 10 = \boxed{040}</math>.
−
.
+
 
 
=== Solution 2 ===
 
=== Solution 2 ===
  

Revision as of 11:31, 29 March 2013

Problem 1

Find the number of positive integers with three not necessarily distinct digits, $abc$, with $a \neq 0$ and $c \neq 0$ such that both $abc$ and $cba$ are multiples of $4$.

Solutions

Solution 1

A positive integer is divisible by $4$ if and only if its last two digits are divisible by $4.$ For any value of $b$, there are two possible values for $a$ and $c$, since we find that if $b$ is even, $a$ and $c$ must be either $4$ or $8$, and if $b$ is odd, $a$ and $c$ must be either $2$ or $6$. There are thus $2 \cdot 2 = 4$ ways to choose $a$ and $c$ for each $b,$ and $10$ ways to choose $b$ since $b$ can be any digit. The final answer is then $4 \cdot 10 = \boxed{040}$.

Solution 2

A number is divisible by four if its last two digits are divisible by 4. Thus, we require that $10b + a$ and $10b + c$ are both divisible by $4$. If $b$ is odd, then $a$ and $c$ must both be $2 \pmod 4$ meaning that $a$ and $c$ are $2$ or $6$. If $b$ is even, then $a$ and $c$ must be $0 \pmod 4$ meaning that $a$ and $c$ are $4$ or $8$. For each choice of $b$ there are $2$ choices for $a$ and $2$ for $c$ for a total of $10 \cdot 2 \cdot 2 = \boxed{040}$ numbers.

See also

2012 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_1&oldid=51958"