2006 AIME II Problems/Problem 7
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 
.
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 | ||
