How many integer solutions exist for the equation 8x 5y = 221 such

Question

How many integer solutions exist for the equation 8x – 5y = 221 such that x*y < 0

A. 4
B. 5
C. 6
D. 7
E. 8

Are You Up For the Challenge: 700 Level Questions

GMATinsight · Accepted Answer

Integer solutions for 8x – 5y = 221
 x*y < 0 i.e. one of x and y must be positive and other negative

5y = 8x - 221

for integer solution, 8x must have unit digit either 1 or 6 but 8x is Even so the unit digit must be 6 and not 1

for 8x to have unit digit 6, x must be {2, 7, 12, 17...}

But since we are taking x positive so y must be negative i.e. 8x < 221 i.e. x < 27.6
i.e. x maybe {2, 7, 12, 17, 22, 27} i.e. 6 solutions
 
FOr any negative value of x, y is never positive hence no solution this way

Total Solutions obtained = 6

Answer: Option C

shameekv1989 · Answer

This equation is of the form Linear Diophantine Equations - courtesy  AnthonyRitz  -  https://www.youtube.com/watch?v=SCf-HrtYUMg

Find the first pair for the equation 8x – 5y = 221 such that xy < 0

x has to be such that 8x-221 is divisible by 5 for y to be integer => Let's try x=2 => Will give us y=-41

Now we just need to add a value to x and subtract from y or vice versa (In this case we subtract value to x, thus add 5 i.e. -(-5) and add in y - +8)

1) x=2; y=-41
2) x=2+5 =7 (Because of -5 in y); y=-41+8 = -33 (Because of 8 in x) 
3) x=7+5 =12 (Because of -5 in y); y=-33+8 = -25 (Because of 8 in x)
4) x=12+5 =17 (Because of -5 in y); y=-25+8 = -17 (Because of 8 in x)
5) x=17+5 =22 (Because of -5 in y); y=-17+8 = -9 (Because of 8 in x)
6) x=22+5 =27 (Because of -5 in y); y=-9+8 = -1 (Because of 8 in x)

If we add any further value to y => y will become positive and xy > 0 which is a constraint in the question stem

Hence 6 integer solutions

Answer - C

lacktutor · Answer

How many integer solutions exist for the equation 8x – 5y = 221 such that x*y < 0

\(y =\frac{( 8x —221)}{5}\)

—>\(\frac{ x( 8x—221)}{5}< 0\)

The value of x must be integer between 0 and 221/8 (27.6..)

—> x= 2 —> y = —41
—> x= 7 —> y= —33
—> x= 12 —> y = —25
—> x= 17 —> y = —17
—> x= 22 —> y = —9
—> x= 27 —> y = —1

There are 6 integer solutions.
The answer is C.

Posted from my mobile device

newyork2012 · Answer

(5n+2)*8<=216
5n+2<=27
n<=5
0-5 ...we have 6

Posted from my mobile device

AnthonyRitz · Answer

Bravo! A nice solution; well done.

Also, thanks for the mention. I'm glad the video helped you solve this, and it's great for others to have a chance to check it out. I actually had a problem with a negative coefficient in the slides of that lecture, but I ran out of time to cover it in the video. I'm glad that people get a chance to see a nice one here. Kudos also to its author!

ScottTargetTestPrep · Answer

If x*y < 0, then x and y must have opposite signs (i.e., one is positive and the other is negative).

If y is positive, then -5y is negative. Therefore, 8x must be greater than 221 in order for 8x - 5y = 221. However, in that case, x will also be positive, and we won’t have x*y < 0.

If x (and hence 8x) is positive, then y must be negative and -5y will be positive. Therefore, 8x < 221. Furthermore, the units digit of x must be either 2 or 7 so that y could be an integer (notice that the units digit of 8x will be 6 and that of -5y will be 5). Therefore, x can be 2, 7, 12, 17, 22 or 27 (we stop at 27 since 27 * 8 = 216 but 32 * 8 = 256). So we see that x can have 6 values (and any one of these will yield a negative value for y).

Answer: C

Kinshook · Answer

Asked: How many integer solutions exist for the equation 8x – 5y = 221 such that x*y < 0
Is equivalent to question
How many integer solutions exist for the equation 8x + 5y = 221 such that x & y > 0

y = (221 - 8x)/5

x = 0; y = 221/5 ; Not feasible
x = 1 ; y = 213/5: Not feasible
x =2 ; y = 205/5 = 41; Solution
x = 7; y = 165/5 = 33; Solution
x = 12; y = 125/5 = 25; Solution
x = 17; y = 85/5 = 17; Solution
x = 22; y = 45/5 = 9; Solution
x = 27; y = 5/5 = 1; Solution

Number of solutions = 6

IMO C

saurabhbajpai · Answer

Why did we start with 2 and not 1? And why did we go for increments of 5? It would be great if you could help with the principal

bumpbot · Answer

Automated notice from GMAT Club BumpBot:

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

This post was generated automatically.

How many integer solutions exist for the equation 8x 5y = 221 such

Prep Toolkit

Top 5 GMAT Debrief Videos

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

GMAT Problem Solving (PS) Questions

How many integer solutions exist for the equation 8x 5y = 221 such

Prep Toolkit

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

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia