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 Alabama ARML TST Problems/Problem 4"

(→‎Solution: divide latex)
(I can't seem to follow this; but I believe the answer is 17)
 
Line 1: Line 1:
 
==Problem==
 
==Problem==
 
For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of <math>12</math>?
 
For how many [[ordered pair]]s of [[digit]]s <math>(A,B)</math> is <math>2AB8</math> a [[multiple]] of <math>12</math>?
 +
 
==Solution==
 
==Solution==
 +
A number is divisible by <math>12</math> if it is divisible by <math>3</math> and <math>4</math>. A number is divisible by <math>4</math> if its last two digits are divisible by <math>4</math>, so <math>4|\overline{B8} \Longrightarrow B = 0,2,4,6,8</math>. A number is divisible by <math>3</math> if the sum of its digits is <math>3</math>, so
  
−
We wish for <math>2000+100A+10B+8 \equiv 0 \pmod {12}\Longleftrightarrow 4A+10B</math><math>\equiv 8 \pmod {12} \Longleftrightarrow 2A+5B\equiv 4 \pmod 6</math>. Thus <math>B\equiv 0 \pmod 2</math>. Let <math>B=2C\rightarrow A+2C\equiv 2 \pmod 3</math>; <math>C<5</math>,<math>A<10</math>, one of the eqns. must be true:
+
*<math>3| s(2A08) = 10 + A \Longrightarrow A = 2,5,8</math> - 3 solutions
−
 
+
*<math>3| s(2A28) = 12 + A \Longrightarrow A = 0,3,6,9</math> - 4 solutions
−
<math>A+2C=2\rightarrow</math> 2 ways
+
*<math>3| s(2A48) = 14 + A \Longrightarrow A = 1,4,7</math> - 3 solutions
−
 
+
*<math>3| s(2A68) = 16 + A \Longrightarrow A = 2,5,8</math> - 3 solutions
−
<math>A+2C=5\rightarrow</math> 3 ways
+
*<math>3| s(2A88) = 18 + A \Longrightarrow A = 0,3,6,9</math> - 4 solutions
−
 
 
−
<math>A+2C=2\rightarrow</math> 4 ways
 
−
 
 
−
<math>A+2C=2\rightarrow</math> 4 ways
 
−
 
 
−
<math>A+2C=2\rightarrow</math> 3 ways
 
−
 
 
−
<math>A+2C=2\rightarrow</math> 2 ways
 
  
−
We have a total of <math>18</math> ways.
+
These sum to <math>17</math>.
  
 
==See also==
 
==See also==
Line 23: Line 17:
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]
−
 
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Latest revision as of 19:34, 5 January 2008

Problem

For how many ordered pairs of digits $(A,B)$ is $2AB8$ a multiple of $12$?

Solution

A number is divisible by $12$ if it is divisible by $3$ and $4$. A number is divisible by $4$ if its last two digits are divisible by $4$, so $4|\overline{B8} \Longrightarrow B = 0,2,4,6,8$. A number is divisible by $3$ if the sum of its digits is $3$, so

  • $3| s(2A08) = 10 + A \Longrightarrow A = 2,5,8$ - 3 solutions
  • $3| s(2A28) = 12 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions
  • $3| s(2A48) = 14 + A \Longrightarrow A = 1,4,7$ - 3 solutions
  • $3| s(2A68) = 16 + A \Longrightarrow A = 2,5,8$ - 3 solutions
  • $3| s(2A88) = 18 + A \Longrightarrow A = 0,3,6,9$ - 4 solutions

These sum to $17$.

See also

2005 Alabama ARML TST (Problems)
Preceded by:
Problem 3 		Followed by:
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_Alabama_ARML_TST_Problems/Problem_4&oldid=21955"