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"
(→‎Video Solutions)
 
(4 intermediate revisions by 2 users not shown)
Line 10: Line 10:
 
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{040} </math> numbers.
 
A number is divisible by four if its last two digits are divisible by 4. Thus, we require that <math> 10b + a </math> and <math> 10b + c</math> are both divisible by <math> 4 </math>. If <math> b </math> is odd, then <math> a </math> and <math> c </math> must both be <math> 2 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 2 </math> or <math> 6 </math>. If <math> b </math> is even, then <math> a </math> and <math> c </math> must be <math> 0 \pmod 4 </math> meaning that <math> a </math> and <math> c </math> are <math> 4 </math> or <math> 8 </math>. For each choice of <math> b </math> there are <math> 2 </math> choices for <math> a </math> and <math> 2 </math> for <math> c </math> for a total of <math> 10 \cdot 2 \cdot 2 = \boxed{040} </math> numbers.
  
−
== Video Solution by Richard Rusczyk ==
+
== Solution 3 ==
  
−
https://artofproblemsolving.com/videos/amc/2012aimei/289
+
For this number to fit the requirements <math>bc</math> and <math>ba</math> must be divisible by 4. So <math>bc = 00, 04, 08, 12, 16, ... , 92, 96</math> and so must <math>ba</math> for each two digits of <math>bc</math>. There are two possibilities for <math>ba</math> if <math>b</math> is odd and three possibilities if <math>b</math> is even. So there are <math>2^{2} \cdot 5+3^{2} \cdot 4 = 65</math> possibilities but this overcounts when <math>a</math> or <math>c = 0</math>. So when <math>bc = 00, 20, 40, 60, 80</math> and the corresponding <math>ba</math> should be removed, so <math>65 - 5 \cdot 3 = 50</math>. But we are still overcounting when <math>b</math> is even because then <math>a</math> can be 0. So the answer is <math>50 - 10 \cdot 1 = \boxed{040}</math> 
  
−
~ dolphin7
+
~LuisFonseca123
  
−
== Video Solution ==
+
== Video Solution by OmegaLearn ==
 
https://youtu.be/ZhAZ1oPe5Ds?t=3235
 
https://youtu.be/ZhAZ1oPe5Ds?t=3235
  
 
~ pi_is_3.14
 
~ pi_is_3.14
  
−
==Video Solution==
+
== Video Solutions ==
 +
 
 +
https://artofproblemsolving.com/videos/amc/2012aimei/289
  
 
https://www.youtube.com/watch?v=T8Ox412AkZc
 
https://www.youtube.com/watch?v=T8Ox412AkZc
−
~Shreyas S
 
  
 
== See also ==
 
== See also ==

Latest revision as of 02:59, 21 January 2023

Problem

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 $4$.

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.

Solution 3

For this number to fit the requirements $bc$ and $ba$ must be divisible by 4. So $bc = 00, 04, 08, 12, 16, ... , 92, 96$ and so must $ba$ for each two digits of $bc$. There are two possibilities for $ba$ if $b$ is odd and three possibilities if $b$ is even. So there are $2^{2} \cdot 5+3^{2} \cdot 4 = 65$ possibilities but this overcounts when $a$ or $c = 0$. So when $bc = 00, 20, 40, 60, 80$ and the corresponding $ba$ should be removed, so $65 - 5 \cdot 3 = 50$. But we are still overcounting when $b$ is even because then $a$ can be 0. So the answer is $50 - 10 \cdot 1 = \boxed{040}$

~LuisFonseca123

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=3235

~ pi_is_3.14

Video Solutions

https://artofproblemsolving.com/videos/amc/2012aimei/289

https://www.youtube.com/watch?v=T8Ox412AkZc

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2012_AIME_I_Problems/Problem_1&oldid=187190"