Difference between revisions of "2006 AIME II Problems/Problem 7"
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− | + | == Problem == | |
+ | 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 == | ||
+ | There are <math>\left\lfloor\frac{999}{10}\right\rfloor = 99</math> numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when <math>a</math> or <math>b</math> have a 0 in the tens digit, and since the equation is [[symmetry|symmetric]], we will just count when <math>a</math> has a 0 in the tens digit and multiply by 2 (notice that the only time both <math>a</math> and <math>b</math> can have a 0 in the tens digit is when they are divisible 100, which falls into the above category, so we do not have to worry about [[Principle of Inclusion-Exclusion|overcounting]]). | ||
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+ | Excluding the numbers divisible by 100, which were counted already, there are <math>9</math> numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling <math>9 \cdot 9 = 81</math> such numbers; considering <math>b</math> also and we have <math>81 \cdot 2 = 162</math>. Therefore, there are <math>999 - (99 + 162) = 738</math> such ordered pairs. | ||
+ | |||
+ | == See also == | ||
+ | {{AIME box|year=2006|n=II|num-b=6|num-a=8}} | ||
+ | |||
+ | [[Category:Intermediate Combinatorics Problems]] |
Revision as of 19:07, 25 September 2007
Problem
Find the number of ordered pairs of positive integers such that
and neither
nor
has a zero digit.
Solution
There are numbers up to 1000 that have 0 as their units digit. All of the other excluded possibilities are when
or
have a 0 in the tens digit, and since the equation is symmetric, we will just count when
has a 0 in the tens digit and multiply by 2 (notice that the only time both
and
can have a 0 in the tens digit is when they are divisible 100, which falls into the above category, so we do not have to worry about overcounting).
Excluding the numbers divisible by 100, which were counted already, there are numbers in every hundred numbers that have a tens digit of 0 (this is true from 100 to 900), totaling
such numbers; considering
also and we have
. Therefore, there are
such ordered pairs.
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
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 |