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| − | == Problem ==
| + | #REDIRECT [[2006 AIME I Problems/Problem 7]] |
| − | 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]].
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| − | == 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.
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| − | == See also ==
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| − | {{AIME box|year=2006|n=II|num-b=6|num-a=8}}
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| − | [[Category:Intermediate Combinatorics Problems]]
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