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805+ (Hard)|   Algebra|      
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Bunuel
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 09 Dec 2021, 08:45
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95% (hard)

Question Stats:

53% (02:41) correct 47% (02:47) wrong based on 213 sessions

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How many pairs of positive integers (x, y) satisfy the equation 3x + 2y = 120 such that y > x.

A. 7
B. 8
C. 10
D. 11
E. 12



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D
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Anurocks1233
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 14 Dec 2021, 11:09
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X= 40 - 2/3 y
Y =3 to 60 so total 20 integers are possible.
10 cases where y>x
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 14 Dec 2021, 17:06
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Using the exchange rate method, x=y= 24 satisfies.

Exchange x (@-2) and y (@+3) so y>x satisfies. When x is reduced by 2, we get 22, 20, 18.... 2 (giving 11 pairs).

IMO: D

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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 05 Feb 2023, 05:32
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Bunuel
How many pairs of positive integers (x, y) satisfy the equation 3x + 2y = 120 such that y > x.

A. 7
B. 8
C. 10
D. 11
E. 12
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Solution:

  • Given equation \(3x+2y=120\)
    \(⇒2y=120-3x\)
    \(⇒y=\frac{120-3x}{2}\)
    \(⇒y=60-\frac{3x}{2}\)
  • Since x and y are positive integers, x has to be even number
  • If \(x=2\) (minimum value), then \(y=60-\frac{3\times 2}{2}=60-3=57\)

  • So, \(x=2\) and \(y=57\) is one of the value
  • For other possible values, we can add values of x with the coefficient of y i.e., 2 and value of y with the coefficient of x i.e., 3
    • This is the concept of special equation which we have discussed at length in this blog
    Attachment:
    special1.png
    special1.png [ 8.52 KiB | Viewed 4018 times ]
  • Since y has to be greater than x, there are total 11 possible pairs of possible values

Hence the right answer is Option D
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 08 Jun 2025, 23:23
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We are given the linear equation: 3x+2y=120,

Step 1: Solve for x
x= ( 120−2y ) / 3 => x = 60 - 2/3 y

Since we are only interested in positive integer pairs, both x>0 and y>0

To keep x >0,
( 120−2y ) / 3 > 0 => 0< y < 60

Step 2: Apply the constraint y>x
y > (120−2y) / 3 => y > 24

Now combine both conditions for y: 60> y > 24

Now, from step 1 expression
For x to be an integer, y must be multiple of 3

So valid y values between 24 and 60 that are divisible by 3 are:
y=27,30,...,57

This is an arithmetic sequence. Count the terms = ( 57 - 27 ) / 3 + 1 = 11

There are 11 values of y that satisfy all conditions, so there are 11 valid (x, y) pairs.
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 13 Jun 2025, 03:00
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Isn't x = 40 - 2/3y ?

Ashwi
We are given the linear equation: 3x+2y=120,

Step 1: Solve for x
x= ( 120−2y ) / 3 => x = 60 - 2/3 y

Since we are only interested in positive integer pairs, both x>0 and y>0

To keep x >0,
( 120−2y ) / 3 > 0 => 0< y < 60

Step 2: Apply the constraint y>x
y > (120−2y) / 3 => y > 24

Now combine both conditions for y: 60> y > 24

Now, from step 1 expression
For x to be an integer, y must be multiple of 3

So valid y values between 24 and 60 that are divisible by 3 are:
y=27,30,...,57

This is an arithmetic sequence. Count the terms = ( 57 - 27 ) / 3 + 1 = 11

There are 11 values of y that satisfy all conditions, so there are 11 valid (x, y) pairs.
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 02 Nov 2025, 17:18
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StacyArko
Using the exchange rate method, x=y= 24 satisfies.

Exchange x (@-2) and y (@+3) so y>x satisfies. When x is reduced by 2, we get 22, 20, 18.... 2 (giving 11 pairs).

IMO: D

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Hey stacy, this looks like a nifty trick. Could you elaborate on this exchnage rate method.
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 06 Nov 2025, 08:32
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3X + 2Y = 120 ........ Y>X ....... X and Y integer.......
Now, Y has 2 multiplied with it.... So 120 - 3X must be even so that when we divide 120 - 3X by 2 (another even).....the result is even and integer...... So 120 - 3X must be even......now, 120 is even..... So 3X must be even so that when we subtract 3X from 120.....the result is even ....... So, 3X must be even...... Then X must be even because 3 is odd....and (odd)×(odd) = (odd)......(odd)×(even) = (even)........so X must be even.......so least positive even integer is 2..... So if we plug X = 2......then Y = 57.....now we gotta test X = 2,4,6......but the options in the question say highest 12 values for X and Y......so if X = 2,4,6...... Then 12th value for X must be 24.....but if X = 24...... Then Y = 24.....but Y ≠ X so X cant have 12th value....so X has 11 possible values.....then Y also has 11 possible corresponding values......so answer is 11......
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 06 Nov 2025, 18:13
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2y = 120 such that y > x.

y = (120-3x)/2 = 60 - 1.5x

x = 2; y = 57
x = 4; y = 54
x = 6; y = 51
x = 8; y = 48
x = 10; y = 45
x = 12; y = 42
x = 14; y = 39
x = 16; y = 36
x = 18; y = 33
x = 20; y = 30
x = 22; y = 27
x = 24; y = 24 ; Not valid since y > x

11 pairs

IMO D
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How many pairs of positive integers (x, y) satisfy the equation 3x + 2 06 Nov 2025, 18:15
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Another way to solve

x = 2k
y = 60 - 3k

y > x ; 60 - 3k > 2k; 5k < 60 ; k < 12

k = 1 to 11 ; 11 pairs

Kinshook
How many pairs of positive integers (x, y) satisfy the equation 3x + 2y = 120 such that y > x.

y = (120-3x)/2 = 60 - 1.5x

x = 2; y = 57
x = 4; y = 54
x = 6; y = 51
x = 8; y = 48
x = 10; y = 45
x = 12; y = 42
x = 14; y = 39
x = 16; y = 36
x = 18; y = 33
x = 20; y = 30
x = 22; y = 27
x = 24; y = 24 ; Not valid since y > x

11 pairs

IMO D
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