Difference between revisions of "2018 AMC 10A Problems/Problem 18"

Revision as of 19:36, 8 February 2018

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

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 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 $\boxed{D}$ .

$QED\blacksquare$

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 $i=1, 2, 3, ... 7$ is $3^8=6561$ . $\frac{6561-1}{2}=3280$ gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives $6561-3280=\boxed{3281}$ . ~RegularHexagon

Solution 3 (Quick Solution)

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.

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

2018 AMC 10A (Problems • Answer Key • Resources)
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All AMC 10 Problems and Solutions

2018 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 ==Solution 3 (Quick Solution)==
-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 the number of total possibilities (ignoring the conditions set by the problem) is <math>3^8=6561</math>. So, E is clearly unrealistic.
-Note that if <math>a_7</math> is 1, then it's impossible for <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> to be negative. Therefore, if a_7 is 1, there are 3^7=2187 possibilities.
+Note that if <math>a_7</math> is 1, then it's impossible for <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> to be negative. Therefore, if <math>a_7</math> is 1, there are <math>3^7=2187</math> possibilities.
 As A, B, and C are all less than 2187, the answer must be <math>\boxed{(D) 3281}</math>

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