2017 AIME II Problems/Problem 4
Find the number of positive integers less than or equal to whose base-three representation contains no digit equal to .
The base- representation of is . Because any -digit base- number that starts with and has no digit equal to must be greater than , all -digit numbers that have no digit equal to must start with or in base . Of the base- numbers that have no digit equal to , there are -digit numbers that start with , -digit numbers that start with , -digit numbers, -digit numbers, -digit numbers, -digit numbers, -digit numbers, and -digit numbers. Summing these up, we find that the answer is .
Note that , and . There can be a digit number less than , and each digit can either be or . So one digit numbers and so on up to digit.
Now we have to subtract out numbers from to
Then either the number must begin or with four more digits at the end
Using s and s there are options for each so:
Solution 3 (Casework)
Since the greatest power of that can be used is , we can do these cases.
Coefficient of : Then if the number has only , it has 2 choices (1 or 2). Likewise if the number has both a and term, there are 4 choices, and so on until , so the sum is .
Coefficient of : Any combination of or for the remaining coefficients works, so . Why is this less than ? Because the first time around, leading zeroes didn't count. But this time, all coefficients of need 1 and 2.
Coefficient of : Look at coefficient. If 1, all of them work because . That's 32 cases. Now of this coefficient is 2, then at the coefficient of is at least 1. However, , so our answer is .
|2017 AIME II (Problems • Answer Key • Resources)|
|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.