2018 AMC 10A Problems/Problem 18
Contents
Problem
How many nonnegative integers can be written in the form where for ?
Solution 1
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 for the balanced ternary 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 non-negative integers, so subtracting from the total gives . (RegularHexagon, KLBBC minor changes)
Solution 3
Note that the number of total possibilities (ignoring the conditions set by the problem) is . So, E is clearly unrealistic.
Note that if is 1, then it's impossible for to be negative. Therefore, if is 1, there are possibilities. (We also must convince ourselves that these different sets of coefficients must necessarily yield different integer results.)
As A, B, and C are all less than 2187, the answer must be
Solution 4
Note that we can do some simple casework: If , then we can choose anything for the other 7 variables, so this give us . If and , then we can choose anything for the other 6 variables, giving us . If , , and , then we have . Continuing in this vein, we have ways to choose the variables' values, except we have to add 1 because we haven't counted the case where all variables are 0. So our total sum is . Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.
Solution 5
The key is to realize that this question is basically taking place in if each value of was increased by , essentially making it into base . Then the range would be from to , yielding different values. Since the distribution for all the question originally gave is symmetrical, we retain the positive integers and one but discard the negative integers. Thus, we are left with the answer, . --anna0kear
Solution 6
First, set for all . The range would be the integers for which . If for all , our set expands to include all integers such that . Similarly, when we get , and when the range is . The pattern continues until we reach , where . Because we are only looking for positive integers, we filter out all , leaving us with all integers between , inclusive. The answer becomes . --anna0kear
Solution 7
To get the number of integers, we can get the highest positive integer that can be represented using where for .
Note that the least nonnegative integer that can be represented is , when all . The highest number will be the number when all . That will be
Therefore, there are positive integers and nonnegative integers (while including ) that can be represented. Our answer is
~OlutosinNGA
Solution 8
Notice that there are options for since each can take the value of or or . Now we want to find how many of them are positive and then we can add one in the end to account for (they are asking for non-negative).
By symmetry(look out for these on the contest), we see that exactly half of them are positive. So So now we will add because of the to account for the non-negative solutions.
So our final answer is which is .
Video Solution
~IceMatrix
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 |
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