Difference between revisions of "2017 AIME II Problems/Problem 4"
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− | + | ==Problem== | |
Find the number of positive integers less than or equal to <math>2017</math> whose base-three representation contains no digit equal to <math>0</math>. | Find the number of positive integers less than or equal to <math>2017</math> whose base-three representation contains no digit equal to <math>0</math>. | ||
− | + | ==Solution== | |
The base-<math>3</math> representation of <math>2017_{10}</math> is <math>2202201_3</math>. Because any <math>7</math>-digit base-<math>3</math> number that starts with <math>22</math> and has no digit equal to <math>0</math> must be greater than <math>2017_{10}</math>, all <math>7</math>-digit numbers that have no digit equal to <math>0</math> must start with <math>21</math> or <math>1</math> in base <math>3</math>. Of the base-<math>3</math> numbers that have no digit equal to <math>0</math>, there are <math>2^5</math> <math>7</math>-digit numbers that start with <math>21</math>, <math>2^6</math> <math>7</math>-digit numbers that start with <math>1</math>, <math>2^6</math> <math>6</math>-digit numbers, <math>2^5</math> <math>5</math>-digit numbers, <math>2^4</math> <math>4</math>-digit numbers, <math>2^3</math> <math>3</math>-digit numbers, <math>2^2</math> <math>2</math>-digit numbers, and <math>2^1</math> <math>1</math>-digit numbers. Summing these up, the answer is <math>2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}</math>. | The base-<math>3</math> representation of <math>2017_{10}</math> is <math>2202201_3</math>. Because any <math>7</math>-digit base-<math>3</math> number that starts with <math>22</math> and has no digit equal to <math>0</math> must be greater than <math>2017_{10}</math>, all <math>7</math>-digit numbers that have no digit equal to <math>0</math> must start with <math>21</math> or <math>1</math> in base <math>3</math>. Of the base-<math>3</math> numbers that have no digit equal to <math>0</math>, there are <math>2^5</math> <math>7</math>-digit numbers that start with <math>21</math>, <math>2^6</math> <math>7</math>-digit numbers that start with <math>1</math>, <math>2^6</math> <math>6</math>-digit numbers, <math>2^5</math> <math>5</math>-digit numbers, <math>2^4</math> <math>4</math>-digit numbers, <math>2^3</math> <math>3</math>-digit numbers, <math>2^2</math> <math>2</math>-digit numbers, and <math>2^1</math> <math>1</math>-digit numbers. Summing these up, the answer is <math>2^5+2^6+2^6+2^5+2^4+2^3+2^2+2^1=\boxed{222}</math>. | ||
+ | =See Also= | ||
+ | {{AIME box|year=2017|n=II|num-b=3|num-a=5}} | ||
+ | {{MAA Notice}} |
Revision as of 12:51, 23 March 2017
Problem
Find the number of positive integers less than or equal to whose base-three representation contains no digit equal to
.
Solution
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, the answer is
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.