Revision as of 19:22, 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 $2017=220221_{3}$ , 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:

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

@@ Line 17: / Line 17: @@
 Using <math>1</math>s and <math>2</math>s there are <math>2^4</math> options for each so:
-<math>2+4+8+16+32+64+128-2*16=256-2-32=222</math>
+<math>2+4+8+16+32+64+128-2*16=256-2-32=\boxed{222}</math>
 =See Also=
 {{AIME box|year=2017|n=II|num-b=3|num-a=5}}
 {{MAA Notice}}

2017 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AIME II Problems/Problem 4"

Revision as of 19:22, 23 March 2017

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