Difference between revisions of "2018 AMC 10A Problems/Problem 18"
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| + | == Problem == | ||
How many nonnegative integers can be written in the form <cmath>a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,</cmath> | How many nonnegative integers can be written in the form <cmath>a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,</cmath> | ||
where <math>a_i\in \{-1,0,1\}</math> for <math>0\le i \le 7</math>? | where <math>a_i\in \{-1,0,1\}</math> for <math>0\le i \le 7</math>? | ||
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~RegularHexagon | ~RegularHexagon | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}} | ||
| + | {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}} | ||
| + | {{MAA Notice}} | ||
Revision as of 17:33, 8 February 2018
Contents
Problem
How many nonnegative integers can be written in the form ![\[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\]](http://latex.artofproblemsolving.com/1/2/e/12e09c391eb94ff988e6dcfc490ddb6c72e79342.png)
where 
for 
?
Solution
This looks like balanced ternary, in which all the integers with absolute values less than 
are represented in 
digits. There are 8 digits. Plugging in 8 into the formula gives a maximum bound of 
, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are 
integers or 
.
Solution 2
Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all 
. The total number of ways to pick 
from 
is 
. 
gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives 
.
~RegularHexagon
See Also
| 2018 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2018 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
