If x & y are integers, how many solution pairs (x,y) satisfy the equat

Can someone explain me how to tackle this one please ?

Can you help with the explanation?? Thanks in advance

72 = {1,2,3,4,6,8,9,12,18,24,72}

A x B = 72

Case 1

A=1 B=72
Then (x,y) will be (13,104),(13,-112),(-11,104) and (-11,-112)

and so on for A=72 and B=1

Total such 12 A B pair cases will be there

And 4 integral pair will come as a result of each A B pair

Hence 48 is the answer in my opinion.

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A very good attempt but x & y may be negative integers.
Therefore, consider that case.

Given: x & y are integers
Asked: How many ordered pairs (x,y) satisfy the equation (|x-2|-12)(|y+4|-36)=72?

\(72 = 2^3*3^2\)
72 has 4*3 = 12 factors
Factors of 72 = {1,2,3,4,6,8,9,12,18,24,36,72}

If x & y are positive integers, we will have 12 cases

72=1*72
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=1 => |x-2|=12+1=13 =>x=2+-13 => x=15 or x=-11
|y+4|-36=72 =>|y+4|=36+72=108 =>y=-4+-108 => y=112 or y=-104
This case will give 4 ordered pairs = (15,112),(-11,112),(15,-104) & (-11,-104)

72=2*36
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=2 => |x-2|=12+2=14 =>x=2+-14 => x=16 or x=-12
|y+4|-36=36 =>|y+4|=36+36=72 =>y=-4+-72 => y=-76 or y=68
This case will give 4 ordered pairs = (16,68),(-12,-76),(16,68) & (-12,-76)

72=3*24
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=3 =>|x-2|=12+3=15 => x=2+-15 => x=17 or x=-13
|y+4|-36=24 =>|y+4|=36+24=60 =>y=-4+-60 => y=-64 or y=56
This case will give 4 ordered pairs = (17,56),(-13,-64),(17,56) & (-13,-64)

72=4*18
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=4 => |x-2|=12+4=16 => x=2+-16 => x=18 or x=-14
|y+4|-36=18 =>|y+4|=36+18= 54 => y=-4+-54 => y=-58 or y=50
This case will give 4 ordered pairs = (18,50),(-14,-58),(18,50) & (-14,-58)

72=6*12
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=6 => |x-2|=12+6=18 => x=2+-18 => x=20 or x=-16
|y+4|-36=12 =>|y+4|=36+12= 48 => y=-4+-48 => y=-52 or y=44
This case will give 4 ordered pairs = (20,44),(-16,-52),(20,44) & (-16,-52)

72=8*9
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=8 => |x-2|=12+8=20 => x=2+-20 => x=22 or x=-18
|y+4|-36=9 =>|y+4|=36+9= 45 => y=-4+-45 => y=-49 or y=41
This case will give 4 ordered pairs = (22,41),(-18,-49),(22,41) & (-18,-49)

72=9*8
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=9 => |x-2|=12+9=21 => x=2+-21 => x=23 or x=-19
|y+4|-36=8 =>|y+4|=36+8= 44 => y=-4+-44 => y=-48 or y=40
This case will give 4 ordered pairs = (23,40),(-19,-48),(23,40) & (-19,-48)

72=12*6
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=12 => |x-2|=12+12=24 => x=2+-24 => x=26 or x=-22
|y+4|-36=6 =>|y+4|=36+6= 42 => y=-4+-42 => y=-46 or y=38
This case will give 4 ordered pairs = (26,38),(-22,-46),(26,38) & (-22,-46)

72=18*4
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=18 => |x-2|=12+18=30 => x=2+-30 => x=32 or x=-28
|y+4|-36=4 =>|y+4|=36+4= 40 => y=-4+-40 => y=-44 or y=36
This case will give 4 ordered pairs = (32,36),(-28,-44),(32,36) & (-28,-44)

72=24*3
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=24 => |x-2|=12+24=36 => x=2+-36 => x=38 or x=-34
|y+4|-36=3 =>|y+4|=36+3= 39 => y=-4+-39 => y=-43 or y=35
This case will give 4 ordered pairs = (38,35),(-34,-43),(38,35) & (-34,-43)

72=36*2
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=36 => |x-2|=12+36=48 => x=2+-48 => x=50 or x=-46
|y+4|-36=2 =>|y+4|=36+2= 38 => y=-4+-38 => y=-42 or y=34
This case will give 4 ordered pairs = (50,34),(-46,-46),(50,34) & (-46,-46)

72=72*1
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=72 => |x-2|=12+72=84 => x=2+-84 => x=86 or x=-82
|y+4|-36=1 =>|y+4|=36+1= 37 => y=-4+-37 => y=-41 or y=33
This case will give 4 ordered pairs = (86,33),(-82,-41),(86,33) & (-82,-41)

These 12 cases when x & y are positive integers will give 12*4 = 48 ordered pairs

Now let us consider negative integer cases:-

72=(-1)*(-72)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-1 => x=2+-11 => x=13 or x=-9
|y+4|-36=-72 => |y+4| = 36-72 = -36 => NOT FEASIBLE
This case will give 0 ordered pairs

72=(-2)*(-36)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-2 => x=2+-10 => x=12 or x=-8
|y+4|-36=-36 => |y+4| = 0 => y = -4
This case will give 2 ordered pairs (12,-4) & (-8,-4)
This case will give 2 ordered pairs

72=(-3)*(-24)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-3 => x=2+-9 => x=11 or x=-7
|y+4|-36=-24 => |y+4| = 36-24 = 12 => y = -4+-12 => y=8 or y=-16
This case will give 4 ordered pairs (11,8),(11,-16),(-7,8) & (-7,-16)
This case will give 4 ordered pairs

72=(-4)*(-18)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-4 => x=2+-8 => x=10 or x=-6
|y+4|-36=-18=> |y+4| = 36-18 = 18 => y = -4+-18 => y=14 or y=-22
This case will give 4 ordered pairs

72=(-6)*(-12)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-6 => x=2+-6 => x=8 or x=-4
|y+4|-36=-12=> |y+4| = 36-12 = 24 => y = -4+-24 => y=20 or y=-28
This case will give 4 ordered pairs

72=(-8)*(-9)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-8 => x=2+-4 => x=6 or x=-2
|y+4|-36=-9=> |y+4| = 36-9 = 27 => y = -4+-27 => y=23 or y=-31
This case will give 4 ordered pairs

72=(-9)*(-8)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-9 => |x-2|= 12-9=3 => x=2+-3 => x=5 or x=-1
|y+4|-36=-8=> |y+4| = 36-8 = 28 => y = -4+-28 => y=24 or y=-32
This case will give 4 ordered pairs

72=(-12)*(-6)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-12 => |x-2|= 12-12=0 => x= 2
|y+4|-36=-6=> |y+4| = 36-6 = 30 => y = -4+-30 => y=26 or y=-34
This case will give 2 ordered pairs

72=(-18)*(-4)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-18 => |x-2|= 12-18=-6 => NOT FEASIBLE
|y+4|-36=-4=> |y+4| = 36-4 = 32 => y = -4+-32 => y=28 or y=-36
This case will give 0 ordered pairs

72=(-24)*(-3)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-24 => |x-2|= 12-24=-12 => NOT FEASIBLE
|y+4|-36=-3=> |y+4| = 36-3 = 33 => y = -4+-33 => y=29 or y=-37
This case will give 0 ordered pairs

72=(-36)*(-2)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-36 => |x-2|= 12-36=-24 => NOT FEASIBLE
|y+4|-36=-2=> |y+4| = 36-2 = 34 => y = -4+-34 => y=30 or y=-38
This case will give 0 ordered pairs

72=(-72)*(-1)
72= (|x-2|-12)(|y+4|-36)
|x-2|-12=-72 => |x-2|= 12-72=-60 => NOT FEASIBLE
|y+4|-36=-1=> |y+4| = 36-1 = 35 => y = -4+-35 => y=31 or y=-39
This case will give 0 ordered pairs

Total ordered pairs for negative integers = 2+ 5*4 +2 = 24

Total integer ordered pairs = 48 + 24 = 72

IMO A

Is there a shorter way to attempt this question?

Bunuel how can this problem be solved faster?

You need to check values of modulus is positive for each case and need not solve final values of x and y.

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First determine the number of factors of 72: 2^3*3^2 => (3+1)*(2+1)=12 factors
Then, use the chart below to determine the zones for positivity of negativity of combined modulus on the number line:
______region 1____ region 2_____region 3
______________-4________2_____________
x-2 ---------------------------0++++++++++
y+4 -------------0++++++++++++++++++

Since there are 3 regions, (x,y) couple can take 3 different set of values for a couple of positive factors of 72. So we have 3*12 possibilities of couples.
Finally , the factors of 72 can be both positive or both negative, so multiply the number of couples found for positive factors by 2 =>3*12*2=72 couple possibilities.

Can you help me to see if this reasoning is correct?