2018 AMC 10A Problems/Problem 18

Revision as of 16:33, 30 December 2020 by Krisha12 (talk | contribs)

Problem

How many nonnegative integers can be written in the form \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] where $a_i\in \{-1,0,1\}$ for $0\le i \le 7$?

$\textbf{(A) } 512 \qquad  \textbf{(B) } 729 \qquad  \textbf{(C) } 1094 \qquad  \textbf{(D) } 3281 \qquad  \textbf{(E) } 59,048$

Solution 1

This looks like balanced ternary, in which all the integers with absolute values less than $\frac{3^n}{2}$ are represented in $n$ digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of $|x|=3280.5$, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are $3280+1=\boxed{3281}$ integers or $\boxed{\textbf{D}}$.

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 $a_i=0$. The total number of ways to pick $a_i$ from $i=0, 1, 2, 3, ... 7$ is $3^8=6561$. $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives $6561-3280=\boxed{\textbf{(D) } 3281}$. (RegularHexagon, KLBBC minor changes)

Solution 3

Note that the number of total possibilities (ignoring the conditions set by the problem) is $3^8=6561$. So, E is clearly unrealistic.

Note that if $a_7$ is 1, then it's impossible for \[a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,\] to be negative. Therefore, if $a_7$ is 1, there are $3^7=2187$ possibilities. (We also must convince ourselves that these $2187$ different sets of coefficients must necessarily yield $2187$ different integer results.)

As A, B, and C are all less than 2187, the answer must be $\boxed{\textbf{(D) } 3281}

==Solution 4== Note that we can do some simple casework: If$ (Error compiling LaTeX. Unknown error_msg)a_7=1$, then we can choose anything for the other 7 variables, so this give us$3^7$. If$a_7=0$and$a_6=1$, then we can choose anything for the other 6 variables, giving us$3^6$. If$a_7=0$,$a_6=0$, and$a_5=1$, then we have$3^5$. Continuing in this vein, we have$3^7+3^6+\cdots+3^1+3^0$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$\boxed{(D) 3281}$. 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$ (Error compiling LaTeX. Unknown error_msg)a\in\{0,1,2\}$if each value of$a$was increased by$1$, essentially making it into base$3$. Then the range would be from$0\cdot3^7+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^6+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^5+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^4+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^3+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^2+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^1+$$ (Error compiling LaTeX. Unknown error_msg)0\cdot3^0=$$ (Error compiling LaTeX. Unknown error_msg)0$to$2\cdot3^7+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^6+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^5+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^4+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^3+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^2+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^1+$$ (Error compiling LaTeX. Unknown error_msg)2\cdot3^0=$$ (Error compiling LaTeX. Unknown error_msg)3^8-1=$$ (Error compiling LaTeX. Unknown error_msg)6561-1=$$ (Error compiling LaTeX. Unknown error_msg)6560$, yielding$6561$different values. Since the distribution for all$a_i\in \{-1,0,1\}$the question originally gave is symmetrical, we retain the$3280$positive integers and one$0$but discard the$3280$negative integers. Thus, we are left with the answer,$\boxed{\textbf{(D)} 3281}\qquad$. --anna0kear

==Solution 6== First, set$ (Error compiling LaTeX. Unknown error_msg)a_i=0$for all$i\geq1$. The range would be the integers for which$[-1,1]$. If$a_i=0$for all$i\geq2$, our set expands to include all integers such that$-4\leq\mathbb{Z}\leq4$. Similarly, when$i\geq3$we get$-13\leq\mathbb{Z}\leq13$, and when$i\geq4$the range is$-40\leq\mathbb{Z}\leq40$. The pattern continues until we reach$i=7$, where$-3280\leq\mathbb{Z}\leq3280$. Because we are only looking for positive integers, we filter out all$\mathbb{Z}<0$, leaving us with all integers between$0\leq\mathbb{Z}\leq3280$, inclusive. The answer becomes$\boxed{\textbf{(D) } 3281}$. --anna0kear

==Solution 7== To get the number of integers, we can get the highest positive integer that can be represented using <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$ (Error compiling LaTeX. Unknown error_msg)a_i\in \{-1,0,1\}$for$0\le i \le 7$.

Note that the least nonnegative integer that can be represented is$ (Error compiling LaTeX. Unknown error_msg)0$, when all$a_i=0$. The highest number will be the number when all$a_i=1$. That will be <cmath>3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}</cmath> <cmath>=3280</cmath>

Therefore, there are$ (Error compiling LaTeX. Unknown error_msg)3280$positive integers and$(3280+1)$nonnegative integers (while including$0$) that can be represented. Our answer is$\boxed{\textbf{(D) } 3281}$~OlutosinNGA

==Solution 8== Notice that there are$ (Error compiling LaTeX. Unknown error_msg)3^8$options for$a_7, a_6, \cdots a_0$since each$a_i$can take the value of$-1$or$0$or$1$. Now we want to find how many of them are positive and then we can add one in the end to account for$0$(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$ (Error compiling LaTeX. Unknown error_msg)\lfloor{\tfrac{3^8}{2}}\rfloor = 3280.$So now we will add$1$because of the$0$to account for the non-negative solutions.

So our final answer is$ (Error compiling LaTeX. Unknown error_msg)3280 + 1 = 3281$which is$\boxed{\textbf{(D) } 3281}$.

Video Solution

https://youtu.be/0xmbHDcUI2w

~IceMatrix

See Also

2018 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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. AMC logo.png