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

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How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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26 Feb 2020, 23:32
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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

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Re: How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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26 Feb 2020, 23:51
3
2
Bunuel wrote:
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

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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
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How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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Updated on: 27 Feb 2020, 17:31
4
1
Bunuel wrote:
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

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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

Originally posted by shameekv1989 on 27 Feb 2020, 16:24.
Last edited by shameekv1989 on 27 Feb 2020, 17:31, edited 1 time in total.
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Re: How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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27 Feb 2020, 00:01
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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.

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Re: How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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27 Feb 2020, 08:51
(5n+2)*8<=216
5n+2<=27
n<=5
0-5 ...we have 6

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How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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27 Feb 2020, 17:18
3
shameekv1989 wrote:
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 y

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

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!
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How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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01 Mar 2020, 12:49
2
Bunuel wrote:
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

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
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Re: How many integer solutions exist for the equation 8x – 5y = 221 such  [#permalink]

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28 Mar 2020, 04:53
Bunuel wrote:
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

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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
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Re: How many integer solutions exist for the equation 8x – 5y = 221 such   [#permalink] 28 Mar 2020, 04:53

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

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