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Difference between revisions of "2017 AIME II Problems/Problem 4"

(→‎Solution)
(→‎Solution)
Line 9: Line 9:
  
 
Note that 2017=220221
 
Note that 2017=220221
 +
 
And 2187=3^7=10000000 base 3
 
And 2187=3^7=10000000 base 3
 +
 
There can be a 1,2,...,7 digit number less than 2187
 
There can be a 1,2,...,7 digit number less than 2187
 +
 
Each digit can be 1 or 2
 
Each digit can be 1 or 2
 +
 
So 2^1 one digit numbers and so on up to 2^7 7 digit
 
So 2^1 one digit numbers and so on up to 2^7 7 digit
 +
  
 
Now we have to subtract out numbers from 2018 to 2187
 
Now we have to subtract out numbers from 2018 to 2187
 +
 
The either the number must begin 221... or 222... with  four more digits at the end
 
The either the number must begin 221... or 222... with  four more digits at the end
 +
 
Using 1s and 2s there are 2^4 options for each so:
 
Using 1s and 2s there are 2^4 options for each so:
  

Revision as of 18:00, 23 March 2017

Problem

Find the number of positive integers less than or equal to $2017$ whose base-three representation contains no digit equal to $0$.

Solution

The base-$3$ representation of $2017_{10}$ is $2202201_3$. Because any $7$-digit base-$3$ number that starts with $22$ and has no digit equal to $0$ must be greater than $2017_{10}$, all $7$-digit numbers that have no digit equal to $0$ must start with $21$ or $1$ in base $3$. Of the base-$3$ numbers that have no digit equal to $0$, there are $2^5$ $7$-digit numbers that start with $21$, $2^6$ $7$-digit numbers that start with $1$, $2^6$ $6$-digit numbers, $2^5$ $5$-digit numbers, $2^4$ $4$-digit numbers, $2^3$ $3$-digit numbers, $2^2$ $2$-digit numbers, and $2^1$ $1$-digit numbers. Summing these up, we find that the answer is $2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}$.


Solution 2 (please add latex)

Note that 2017=220221

And 2187=3^7=10000000 base 3

There can be a 1,2,...,7 digit number less than 2187

Each digit can be 1 or 2

So 2^1 one digit numbers and so on up to 2^7 7 digit


Now we have to subtract out numbers from 2018 to 2187

The either the number must begin 221... or 222... with four more digits at the end

Using 1s and 2s there are 2^4 options for each so:

2+4+8+16+32+64+128-2*16=256-2-32=222

See Also

2017 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 3 		Followed by
Problem 5
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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