Difference between revisions of "1992 AIME Problems/Problem 6"
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== Problem == | == Problem == | ||
| − | For how many pairs of consecutive integers in <math>\{1000,1001,1002 | + | For how many pairs of consecutive integers in <math>\{1000,1001,1002,\ldots,2000\}</math> is no carrying required when the two integers are added? |
== Solution == | == Solution == | ||
| − | {{solution} | + | 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. |
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| + | 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>. | ||
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{{AIME box|year=1992|num-b=5|num-a=7}} | {{AIME box|year=1992|num-b=5|num-a=7}} | ||
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| + | [[Category:Intermediate Number Theory Problems]] | ||
Revision as of 23:31, 11 November 2007
Problem
For how many pairs of consecutive integers in 
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 
and 
, then 
or 
or 
. 6. Consider 
. 
has no carry if 
. This gives 
possible solutions.
With 
. There obviously must be a carry. Consider 
. have no carry. This gives 
possible solutions. Considering 
, 
have no carry. Thus, the solution is 
.
| 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 | ||
