3y + 2|x| = 33. How many positive integral values of (x, y) are

Question

3y + 2|x| = 33. How many positive integral values of (x, y) are possible?

A. 5
B. 6
C. 10
D. 11
E. 12

AvidDreamer09 · Accepted Answer

How many positive integral values of (x, y) are possible?

Solution:
x,y must be positive and has to be an integer
 3y + 2|x| = 33

2|x| = 33-3y 
in the equation Right hand side must be an even number to balance LHS, for which y must be an odd number 
=>since,  odd-odd=even  => 33 is odd... 3y must be odd

y cannot be >10 because |x| has to be >0 ( positive and integer)

y can be

Answer A

To explain further  2|x| = 33-3y  =>  |x| = \frac{{33-3y}}{{2}}  (To make |x| an even integer,  \frac{{33-3y}}{{2}}  should be equal to an even integer

2|x| = 33-3y 
2|x| = 33-3=30=2*15
2|x| = 33-9=24=2*12
2|x| = 33-15=18=2*9
2|x| = 33-21=12=2*6
2|x| = 33-27=6=2*3

BrushMyQuant · Answer

3y + 2|x| = 33 and we need to find positive integral values of (x,y)

Since, x is positive so |x| = x

=> 3y + 2|x| = 33 will become
3y + 2x = 33
=> 2x = 33-3y
=> x =
[ltr]33−3y233−3y2[/ltr]

Now, for x to be an integer 33-3y should be divisible by 2
33 is odd so 3y also has to be odd to make 33-3y as even.
=> y will be odd number and 0 < 3y < 33
=> 0 < y < 11
So there are 5 possible values of y (1,3,5,7,9) (so there will be 5 positive value of x which are possible too)

So, answer will be A
Hope it helps!

CareerGeek

3y + 2|x| = 33. How many positive integral values of (x, y) are possible?

A. 5
B. 6
C. 10
D. 11
E. 12

MathRevolution · Answer

3y + 2|x| = 33

For any value of y, if the difference of 33 - 3y is divisible by 2, then we will get the value of 'x'.

For y = 1 and x = 15
For y = 3 and x = 12
For y = 5 and x = 9
For y = 7 and x = 6 
For y = 9 and x = 3

Five values.

Answer A

IanStewart · Answer

If x is positive, the absolute value isn't doing anything, so we want to know for how many values of x it is true that 3y + 2x = 33. If that's true, then 2x = 33 - 3y, and since 33 - 3y is clearly divisible by 3, and since 33 - 3y is the same number as 2x, it must be true that 2x is divisible by 3, and x must be divisible by 3. Certainly x can't be larger than 15, because then 2x alone is bigger than 33, but x can be any multiple of 3 between 3 and 15 inclusive, for five possible values.

Kinshook · Answer

Given: 3y + 2|x| = 33. 
Asked: How many positive integral values of (x, y) are possible?

Since x & y are positive integers, x>=0
3y + 2|x| = 33
3y + 2x = 33
y = (33-2x)/3 = (11-2x/3)
x should be a multiple of 3

(x,y) = {(3,9),(6,7),(9,5),(12,3),(15,1)}

IMO A

jyotirmayeenaik24 · Answer

Solving the equation we derive,
y= 11-2/3 |x|
x, y being positive integer we can have (3, 6, 9,12 & 15 as |x|)

GMATGuruNY · Answer

Consider the following equation: 
2x + 3y = 30

If x and y are nonnegative integers, the following solutions are possible: 
x=15, y=0 
x=12, y=2 
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following: 
The value of x changes in increments of 3 (the coefficient for y). 
The value of y changes in increments of 2 (the coefficient for x). 
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

Since x must be positive, the absolute value is irrelevant.
Question stem, rephrased:
If 3y+2x = 33, how many positive values of (x, y) are possible?

One easy-to-see solution is as follows:
y=11, x=0

Since the value of y can change only in increments of 2 (the cooefficient for x), while the value of x can change only in increments of 3 (the coefficieent for y), we get the following options:
y=9, x=3
y=7, x=6
y=5, x=9
y=3, x=12
y=1, x=15

A

Suyash1331 · Answer

3y + 2|x| = 33 and we need to find positive integral values of (x,y)

Since, x is positive so |x| = x

=> 3y + 2|x| = 33 will become
3y + 2x = 33
=> 2x = 33-3y
=> x =
[ltr]33−3y233−3y2[/ltr]

Now, for x to be an integer 33-3y should be divisible by 2
33 is odd so 3y also has to be odd to make 33-3y as even.
=> y will be odd number and 0 < 3y < 33
=> 0 < y < 11
So there are 5 possible values of y (1,3,5,7,9) (so there will be 5 positive value of x which are possible too)

So, answer will be A
Hope it helps!

CareerGeek

3y + 2|x| = 33. How many positive integral values of (x, y) are possible?

A. 5
B. 6
C. 10
D. 11
E. 12

LongTermWarrior · Answer

Can someone explain why the solution to the equation (x,y) = (0,11) is not valid? Why would the absolute value of x discard this?

Bunuel · Answer

Check the highlighted part:

3y + 2|x| = 33. How many  positive  integral values of (x, y) are possible?

So, x and y can take only  positive  integer values. Since 0 is not positive x cannot be 0.

3y + 2|x| = 33. How many positive integral values of (x, y) are

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3y + 2|x| = 33. How many positive integral values of (x, y) are

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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My Rewards