Difference between revisions of "2006 AIME II Problems/Problem 7"
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== Solution == | == Solution == | ||
=== Solution 1 === | === Solution 1 === | ||
− | 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 | + | 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 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 by 100, which falls into the above category, so we do not have to worry about [[Principle of Inclusion-Exclusion|overcounting]]). |
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) = \boxed{738}</math> such ordered pairs. | 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) = \boxed{738}</math> such ordered pairs. |
Revision as of 06:12, 4 July 2009
Problem
Find the number of ordered pairs of positive integers such that
and neither
nor
has a zero digit.
Contents
[hide]Solution
Solution 1
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 by 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.
Solution 2
Let and
be 3 digit numbers:
cde +fgh ---- 1000
and
must add up to
,
and
must add up to
, and
and
must add up to
. Since none of the digits can be 0, there are
possibilites if both numbers are three digits.
There are two other scenarios. and
can be a three digit number and a two digit number, or a three digit number and a one digit number. For the first scenario, there are
possibilities (the two accounting for whether
or
has three digits) and for the second case there are
possibilities. Thus, thus total possibilities for
is
.
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 |