Difference between revisions of "2006 AIME II Problems/Problem 7"
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Revision as of 15:17, 15 September 2019
Problem
Find the number of ordered pairs of positive integers such that
and neither
nor
has a zero digit.
Contents
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 .
Solution 4 (Similar to Solution 2)
We proceed by casework on the number of digits of
Case 1: Both and
have three digits
We now use constructive counting. For the hundreds digit of we see that there are
options - the numbers
through
(If
that means that
will be a two digit number, and if
will have two digits). Similarly, the tens digit can be
as well because a tens digit of
is obviously prohibited and a tens digit of
will lead to a tens digit of
in the other number. The units digit can be anything from
Hence, there are
possible values in this case.
Case 2: (or
) has two digits
If has two digits, the only restrictions are that the units digit must not be
and the tens digit must not be
(because then that would lead to
beginning with
). There thus are
possibilities for
and we have to multiply by
because there are the same number of possibilities for
Thus, there are
possible values in this case.
Case 3: (or
) has one digit
This is easy -- can be anything from
to
for a total of
possible values. We multiply this by
to account for the single digit
values, so we have
possible values for this case.
Adding them all up, we get and we're done.
Solution by Ilikeapos
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 |
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