Difference between revisions of "2006 AIME A Problems/Problem 7"

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== Problem ==
 
== Problem ==
Find the number of ordered pairs of positive integers <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a zero digit.
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Find the number of [[ordered pair]]s of [[positive]] [[integer]]s <math> (a,b) </math> such that <math> a+b=1000 </math> and neither <math> a </math> nor <math> b </math> has a [[zero]] [[digit]].
  
 
== Solution ==
 
== Solution ==
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There are <math>\frac{1000}{10} = 100</math> numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities for <math>a</math> are when <math>a</math> or <math>b</math> have a 0 in the tens digit, or if the tens digit of <math>a</math> is 0 or 9. Excluding the hundreds, which were counted above already, there are <math>9 \cdot 2 = 18</math> numbers in every hundred numbers that have a tens digit of 0 or 9, totaling <math>10 \cdot 18 = 180</math> such numbers. However, the numbers from 1 to 9 and 991 to 999 do not have 0s, so we must subtract <math>18</math> from that to get <math>162</math>. Therefore, there are <math>1000 - (100 + 162) = 738</math> such ordered pairs.
  
 
== See also ==
 
== See also ==
*[[2006 AIME II Problems]]
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{{AIME box|year=2006|n=II|num-b=6|num-a=8}}
  
 
[[Category:Intermediate Combinatorics Problems]]
 
[[Category:Intermediate Combinatorics Problems]]

Revision as of 20:27, 21 February 2007

Problem

Find the number of ordered pairs of positive integers $(a,b)$ such that $a+b=1000$ and neither $a$ nor $b$ has a zero digit.

Solution

There are $\frac{1000}{10} = 100$ numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities for $a$ are when $a$ or $b$ have a 0 in the tens digit, or if the tens digit of $a$ is 0 or 9. Excluding the hundreds, which were counted above already, there are $9 \cdot 2 = 18$ numbers in every hundred numbers that have a tens digit of 0 or 9, totaling $10 \cdot 18 = 180$ such numbers. However, the numbers from 1 to 9 and 991 to 999 do not have 0s, so we must subtract $18$ from that to get $162$. Therefore, there are $1000 - (100 + 162) = 738$ such ordered pairs.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions