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Difference between revisions of "1992 AIME Problems/Problem 6"

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== Solution ==
 
== Solution ==
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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>1abc + 1abc+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.  
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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>c \in \{0, 1, 2, 3, 4\}</math>. <math>1abc + 1abc+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>c \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>c \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>.  

Revision as of 04:51, 15 May 2008

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 $c \in \{0, 1, 2, 3, 4\}$. $1abc + 1abc+1$ has no carry if $a, b \in \{0, 1, 2, 3, 4\}$. This gives $5^3=125$ possible solutions.

With $c \in \{5, 6, 7, 8\}$, 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
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All AIME Problems and Solutions
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