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 "1992 AIME Problems/Problem 6"

(solution)
(→‎Solution: grammar)
Line 5: Line 5:
 
Consider what carrying means: If carrying is needed to add two numbers with digits <math>abcd</math> and <math>efgh</math>, then <math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6.  Consider <math>d \in \{0, 1, 2, 3, 4\}</math>. <math>abcd + abcd+1</math> has no carry if <math>a, b \in \{0, 1, 2, 3, 4\}</math>. This gives <math>5^3=125</math> possible solutions.  
 
Consider what carrying means: If carrying is needed to add two numbers with digits <math>abcd</math> and <math>efgh</math>, then <math>h+d\ge 10</math> or <math>c+g\ge 10</math> or <math>b+f\ge 10</math>. 6.  Consider <math>d \in \{0, 1, 2, 3, 4\}</math>. <math>abcd + abcd+1</math> has no carry if <math>a, b \in \{0, 1, 2, 3, 4\}</math>. This gives <math>5^3=125</math> possible solutions.  
  
−
With <math>d \in \{5, 6, 7, 8\}</math>. There obviously must be a carry. Consider <math>c = 9</math>. <math>a, b \in \{0, 1, 2, 3, 4\}</math> have no carry. This gives <math>5^2=25</math> possible solutions. Considering <math>b = 9</math>,  <math>a \in \{0, 1, 2, 3, 4, 9\}</math> have no carry. Thus, the solution is <math>125 + 25 + 6=\boxed{156}</math>.  
+
With <math>d \in \{5, 6, 7, 8\}</math>, there obviously must be a carry. Consider <math>c = 9</math>. <math>a, b \in \{0, 1, 2, 3, 4\}</math> have no carry. This gives <math>5^2=25</math> possible solutions. Considering <math>b = 9</math>,  <math>a \in \{0, 1, 2, 3, 4, 9\}</math> have no carry. Thus, the solution is <math>125 + 25 + 6=\boxed{156}</math>.  
  
 
{{AIME box|year=1992|num-b=5|num-a=7}}
 
{{AIME box|year=1992|num-b=5|num-a=7}}
  
 
[[Category:Intermediate Number Theory Problems]]
 
[[Category:Intermediate Number Theory Problems]]

Revision as of 22:31, 11 November 2007

Problem

For how many pairs of consecutive integers in $\{1000,1001,1002,\ldots,2000\}$ is no carrying required when the two integers are added?

Solution

Consider what carrying means: If carrying is needed to add two numbers with digits $abcd$ and $efgh$, then $h+d\ge 10$ or $c+g\ge 10$ or $b+f\ge 10$. 6. Consider $d \in \{0, 1, 2, 3, 4\}$. $abcd + abcd+1$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$. This gives $5^3=125$ possible solutions.

With $d \in \{5, 6, 7, 8\}$, there obviously must be a carry. Consider $c = 9$. $a, b \in \{0, 1, 2, 3, 4\}$ have no carry. This gives $5^2=25$ possible solutions. Considering $b = 9$, $a \in \{0, 1, 2, 3, 4, 9\}$ have no carry. Thus, the solution is $125 + 25 + 6=\boxed{156}$.

1992 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
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=1992_AIME_Problems/Problem_6&oldid=19389"