Difference between revisions of "2006 AIME II Problems/Problem 7"
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Doubling this yields <math>369\times2= 738</math>. | Doubling this yields <math>369\times2= 738</math>. | ||
| − | === Solution 4 === | + | === Solution 4 (Similar to Solution 2)=== |
We proceed by casework on the number of digits of <math>a.</math> | We proceed by casework on the number of digits of <math>a.</math> | ||
Revision as of 11:33, 7 June 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
