Difference between revisions of "2018 AMC 10A Problems/Problem 18"
Line 11: | Line 11: | ||
==Solution 1== | ==Solution 1== | ||
− | This looks like balanced ternary, in which all the integers with absolute values less than <math>\frac{3^n}{2}</math> are represented in <math>n</math> digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of <math>|x|=3280.5</math>, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are <math>3280+1=\boxed{3281}</math> integers or | + | This looks like balanced ternary, in which all the integers with absolute values less than <math>\frac{3^n}{2}</math> are represented in <math>n</math> digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of <math>|x|=3280.5</math>, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are <math>3280+1=\boxed{3281}</math> integers or \boxed{\textbf{D}<math>. |
==Solution 2== | ==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 <math>a_i=0< | + | 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 </math>a_i=0<math>. The total number of ways to pick </math>a_i<math> from </math>i=0, 1, 2, 3, ... 7<math> is </math>3^8=6561<math>. </math>\frac{6561-1}{2}=3280<math> gives the number of possible negative integers. The question asks for the number of non-negative integers, so subtracting from the total gives </math>6561-3280=\boxed{\textbf{(D) } 3281}<math>. (RegularHexagon, KLBBC minor changes) |
==Solution 3== | ==Solution 3== | ||
− | Note that the number of total possibilities (ignoring the conditions set by the problem) is <math>3^8=6561< | + | 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< | + | 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. (We also must convince ourselves that these </math>2187<math> different sets of coefficients must necessarily yield </math>2187<math> different integer results.) |
− | As A, B, and C are all less than 2187, the answer must be <math>\boxed{(D) 3281} | + | As A, B, and C are all less than 2187, the answer must be </math>\boxed{\textbf{(D) } 3281} |
==Solution 4== | ==Solution 4== | ||
Line 37: | Line 37: | ||
==Solution 6== | ==Solution 6== | ||
− | First, set <math>a_i=0</math> for all <math>i\geq1</math>. The range would be the integers for which <math>[-1,1]</math>. If <math>a_i=0</math> for all <math>i\geq2</math>, our set expands to include all integers such that <math>-4\leq\mathbb{Z}\leq4</math>. Similarly, when <math>i\geq3</math> we get <math>-13\leq\mathbb{Z}\leq13</math>, and when <math>i\geq4</math> the range is <math>-40\leq\mathbb{Z}\leq40</math>. The pattern continues until we reach <math>i=7</math>, where <math>-3280\leq\mathbb{Z}\leq3280</math>. Because we are only looking for positive integers, we filter out all <math>\mathbb{Z}<0</math>, leaving us with all integers between <math>0\leq\mathbb{Z}\leq3280</math>, inclusive. The answer becomes <math>\boxed{(D)3281}</math>. --anna0kear | + | First, set <math>a_i=0</math> for all <math>i\geq1</math>. The range would be the integers for which <math>[-1,1]</math>. If <math>a_i=0</math> for all <math>i\geq2</math>, our set expands to include all integers such that <math>-4\leq\mathbb{Z}\leq4</math>. Similarly, when <math>i\geq3</math> we get <math>-13\leq\mathbb{Z}\leq13</math>, and when <math>i\geq4</math> the range is <math>-40\leq\mathbb{Z}\leq40</math>. The pattern continues until we reach <math>i=7</math>, where <math>-3280\leq\mathbb{Z}\leq3280</math>. Because we are only looking for positive integers, we filter out all <math>\mathbb{Z}<0</math>, leaving us with all integers between <math>0\leq\mathbb{Z}\leq3280</math>, inclusive. The answer becomes <math>\boxed{\textbf{(D) } 3281}</math>. --anna0kear |
==Solution 7== | ==Solution 7== |
Revision as of 17:32, 30 December 2020
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 \boxed{\textbf{D}$.
==Solution 2==
Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all$ (Error compiling LaTeX. Unknown error_msg)a_i=0a_i
i=0, 1, 2, 3, ... 7
3^8=6561
\frac{6561-1}{2}=3280
6561-3280=\boxed{\textbf{(D) } 3281}$. (RegularHexagon, KLBBC minor changes)
==Solution 3== Note that the number of total possibilities (ignoring the conditions set by the problem) is$ (Error compiling LaTeX. Unknown error_msg)3^8=6561$. So, E is clearly unrealistic.
Note that if$ (Error compiling LaTeX. Unknown error_msg)a_7a_7
3^7=2187
2187
2187$different integer results.)
As A, B, and C are all less than 2187, the answer must be$ (Error compiling LaTeX. Unknown error_msg)\boxed{\textbf{(D) } 3281}
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.