2017 AIME II Problems/Problem 4

Revision as of 12:19, 11 August 2017 by Eswa2000 (talk | contribs) (Solution 2)

Problem

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

Solution

Solution 1

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

Note that $2017=2202201_{3}$, and $2187=3^7=10000000_{3}$. There can be a $1,2,...,7$ digit number less than $2187$, and each digit can either 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 $2187$

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

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

$2+4+8+16+32+64+128-2*16=256-2-32=\boxed{222}$

See Also

2017 AIME II (ProblemsAnswer KeyResources)
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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