Difference between revisions of "2006 AIME II Problems/Problem 7"
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We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result. | We first must notice that we can find all the possible values of <math>a</math> between <math>1</math> and <math>500</math> and then double that result. | ||
− | When <math>1< | + | When <math>1 < a < 100</math> there are <math>9\times9 = 81</math> possible solution for <math>a</math> so that neither <math>a</math> nor <math>b</math> has a zero in it, counting <math>1</math> through <math>9</math>, <math>11</math> through <math>19</math>, ..., <math>81</math> through <math>89</math>. |
− | When <math>100< | + | When <math>100 < a < 200</math> there are <math>9\times8 =72</math> possible solution for a so that neither a nor b has a zero in it, counting <math>111</math> through <math>119</math>, <math>121</math> through <math>129</math>, ..., <math>181</math> through <math>189</math>. |
− | This can clearly be extended to <math>100k< | + | This can clearly be extended to <math>100k < a < 100(k+1)</math> where <math>k</math> is an integer and <math>0 <k < 9</math>. |
− | Thus for <math>100 | + | Thus for <math>100 < a < 500</math> there are <math>72\times4</math> = <math>288</math> possible values of <math>a</math>. |
− | Thus when <math>1< | + | Thus when <math>1 < a < 500</math> there are <math>288 + 81 =369</math> possible values of <math>a</math> and <math>b</math>. |
− | Doubling this yields <math>369 | + | Doubling this yields <math>369\times2= 738</math>. |
== See also == | == See also == |
Revision as of 20:03, 13 March 2015
Problem
Find the number of ordered pairs of positive integers such that
and neither
nor
has a zero digit.
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
.
Solution 3
We first must notice that we can find all the possible values of between
and
and then double that result.
When there are
possible solution for
so that neither
nor
has a zero in it, counting
through
,
through
, ...,
through
.
When
there are
possible solution for a so that neither a nor b has a zero in it, counting
through
,
through
, ...,
through
.
This can clearly be extended to
where
is an integer and
.
Thus for
there are
=
possible values of
.
Thus when there are
possible values of
and
.
Doubling this yields .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.