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C
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Difficulty:
95%
(hard)
Question Stats:
42% (02:26) correct
58%
(02:40)
wrong
based on 105
sessions
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How many non-negative integer triplets {x, y, z} satisfy \(x+3y+8z=20\) ?
A. 12
B. 13
C. 14
D. 15
E. 16
Are You Up For the Challenge: 700 Level Questions
A. 12
B. 13
C. 14
D. 15
E. 16
Are You Up For the Challenge: 700 Level Questions
04 Jan 2023, 11:20
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x+3y+8z= 20
So x = 20-8z-3y
Now, z can take values 0,1,2
When z = 0, we have x = 20-3y
Now y can take values 0,1,2,3,4,5,6 i.e. 7 values
When z = 1, we have x = 20-8-3y
Or x = 12-3y
Now y can take values 0,1,2,3,4 i.e. 5 values
When z = 2, we have x = 20-(8*2) - 3y
Or x = 4-3y
Now y can take only 2 values 0 and 1
Therefore we have total 7+5+2 = 14 solutions. Hence C 14
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So x = 20-8z-3y
Now, z can take values 0,1,2
When z = 0, we have x = 20-3y
Now y can take values 0,1,2,3,4,5,6 i.e. 7 values
When z = 1, we have x = 20-8-3y
Or x = 12-3y
Now y can take values 0,1,2,3,4 i.e. 5 values
When z = 2, we have x = 20-(8*2) - 3y
Or x = 4-3y
Now y can take only 2 values 0 and 1
Therefore we have total 7+5+2 = 14 solutions. Hence C 14
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04 Jan 2023, 11:50
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Discover how many solutions are there based on different combinations of the variables being equal to 0 (i.e., x=0, y=z, z=0).
We know that all the variables cannot equal 0 because 0 != 20. So, look at how many solutions there are when either 0, 1, or 2 variables are set to 0.
For the case of 1 variable equals 0:
x=0 --> 3y+8z = 20
To see how many solutions satisfy this, start off with seeing the possible values for the highest coefficient (i.e., 8). There are two possible values (1 and 2) since if z =3, that side of the equation would be greater than 20. When z=1, y=4. When z=2, there are no possible solutions since 3y+8(2)=20 -> 3y=4 and 4 is not divisible by 3.
Thus, there's only 1 solution.
y=0 --> x+8z=20.
Following the steps as above, we get 2 solutions (as x can take on any value since it's coefficient is 1)
z=0 --> x+3y=20.
Starting off with the variable with the highest coefficient (i.e., 3). There are six possible values for y since y>6 would cause that side of the equation to be greater than 20. And, we know all these solutions work since x can take on any value as its coefficient (1) is divisible by every number.
Thus, we get 6 possible solution.
For the case of 2 variable equals 0:
If you set any two variables two zero, you have to see if the coefficient of the remaining variable is a factor of 20. The only situation in which this work is when y and z are equal to 0.
So, there's only 1 solution.
For the case of no variable is equal to 0:
Similar to the steps above, take the variable with the highest coefficient aka z and see what are the possible solutions. We determined the possible values were z=1 and z=2.
When z=1 --> x+3y+8(1)= 20 -> x + 3y = 12
Once again, take the variable with the highest coefficient aka y and see how many possible solutions. There are 3 values (y=1, 2, or, 3). We don't consider 4 as that will set x=0 and we've already counted that solution above when we were considering the variables that equal 0. Since x can take on any value, the total solutions are 3.
When z=2 --> x+3y + 8(2) = 20 --> x+ 3y = 4
Once again, take the variable with the highest coefficient aka y and see how many possible solutions. There is only 1 value (y=1). So, total possible solution is 1.
Combining all the possible solutions above: 1+2+6+1+3+1 = 14
We know that all the variables cannot equal 0 because 0 != 20. So, look at how many solutions there are when either 0, 1, or 2 variables are set to 0.
For the case of 1 variable equals 0:
x=0 --> 3y+8z = 20
To see how many solutions satisfy this, start off with seeing the possible values for the highest coefficient (i.e., 8). There are two possible values (1 and 2) since if z =3, that side of the equation would be greater than 20. When z=1, y=4. When z=2, there are no possible solutions since 3y+8(2)=20 -> 3y=4 and 4 is not divisible by 3.
Thus, there's only 1 solution.
y=0 --> x+8z=20.
Following the steps as above, we get 2 solutions (as x can take on any value since it's coefficient is 1)
z=0 --> x+3y=20.
Starting off with the variable with the highest coefficient (i.e., 3). There are six possible values for y since y>6 would cause that side of the equation to be greater than 20. And, we know all these solutions work since x can take on any value as its coefficient (1) is divisible by every number.
Thus, we get 6 possible solution.
For the case of 2 variable equals 0:
If you set any two variables two zero, you have to see if the coefficient of the remaining variable is a factor of 20. The only situation in which this work is when y and z are equal to 0.
So, there's only 1 solution.
For the case of no variable is equal to 0:
Similar to the steps above, take the variable with the highest coefficient aka z and see what are the possible solutions. We determined the possible values were z=1 and z=2.
When z=1 --> x+3y+8(1)= 20 -> x + 3y = 12
Once again, take the variable with the highest coefficient aka y and see how many possible solutions. There are 3 values (y=1, 2, or, 3). We don't consider 4 as that will set x=0 and we've already counted that solution above when we were considering the variables that equal 0. Since x can take on any value, the total solutions are 3.
When z=2 --> x+3y + 8(2) = 20 --> x+ 3y = 4
Once again, take the variable with the highest coefficient aka y and see how many possible solutions. There is only 1 value (y=1). So, total possible solution is 1.
Combining all the possible solutions above: 1+2+6+1+3+1 = 14
