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Difficulty:
85%
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Question Stats:
46% (02:14) correct
54%
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wrong
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The lengths of the sides of a triangle are x, y, and z, where x, y, and z are integers. If xz = 12, how many such triangles are possible?
A. 6
B. 7
C. 8
D. 9
E. 10
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A. 6
B. 7
C. 8
D. 9
E. 10
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Bunuel
xy = 12
12 =
1*12
2*6
3*4
CONCEPT: SUm of two sides > Third Side
Case 1: when x = 1, y = 12, 11 < z < 13 i.e. z = 12 so 1 triangle
Case 2: when x = 2, y = 6, 4 < z < 8 i.e. z = {4, 5, 6} so 3 triangles
Case 3: when x = 3, y = 4, 1 < z < 7 i.e. z = {2, 3, 4, 5, 6} so 5 triangles
Total Possible Triangles = 1+3+5 = 9
Answer: Option D
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Bunuel
x,y and z are integers
x*z = 12
--> 1*12 = 12
--> 2*6 = 12
--> 3*4= 12
In order x,y and z to be sides of a triangle, The sum of two sides should be greater than third side
(Case1): y < 1+12 --> y< 13
12 < 1+y --> y >11
---> y could be 12
(1,12,12)
(Case2): y < 2+6 --> y <8
6 < y+2 --> y >4
y could be 5,6,7
(2,5,6)
(2,6,6)
(2,7,6)
(Case3): y < 3+4 --> y< 7
4< y+3 --> y > 1
---> y could be 2,3,4,5,6
(3,2,4)
(3,3,4)
(3,4,4)
(3,5,4)
(3,6,4)
In Total, there are \(9\) possible triangles
Answer (D).
GMAT 1: 720 Q49 V40
Posts: 939
A triangle with integer side lengths (x, y, z) such that x.z=12
Possible sets of (x,z) = (1,12), (2,6), and (3,4)
If (x,z)=(1,12) --> 11<y<13. So, y can be 12.
(x,y,z) = (1,12,12)
... 1 nos.
If (x,z)=(2,6) --> 4<y<8. So, y can be 5,6,7.
(x,y,z) = (2,6,5), (2,6,6), (2,6,7)
...3 nos.
If (x,z)=(3,4) --> 1<y<7. So, y can be 2,3,...,6.
(x,y,z) = (3,4,2), (3,4,3), (3,4,4), (3,4,5), (3,4,6)
...5 nos.
So, there are 1+3+5 = 9 possible triangles.
FINAL ANSWER IS (D)
Posted from my mobile device
Possible sets of (x,z) = (1,12), (2,6), and (3,4)
If (x,z)=(1,12) --> 11<y<13. So, y can be 12.
(x,y,z) = (1,12,12)
... 1 nos.
If (x,z)=(2,6) --> 4<y<8. So, y can be 5,6,7.
(x,y,z) = (2,6,5), (2,6,6), (2,6,7)
...3 nos.
If (x,z)=(3,4) --> 1<y<7. So, y can be 2,3,...,6.
(x,y,z) = (3,4,2), (3,4,3), (3,4,4), (3,4,5), (3,4,6)
...5 nos.
So, there are 1+3+5 = 9 possible triangles.
FINAL ANSWER IS (D)
Posted from my mobile device
wickedvikram
Joined: 08 Jun 2008
Last visit: 21 Jul 2026
Posts: 198
Location: India
Concentration: Finance, General Management
Schools: INSEAD Jan '26
GPA: 6.5
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Schools: INSEAD Jan '26
Posts: 198
Bunuel
Hi Bunuel
Although not mentioned in the options, shouldn't 18 be the answer.
if x*z=12, possible values
x. z. y range # of y
1 12 11<y<13 1
2 6 4<y<8 3
3 4 1<y<7 5
4 3 1<y<7 5
6 2 4<y<8 3
12 1 11<y<13 1
I understand that top half and bottom half represents triangles with same dimensions, but values of x, y & z are different.
How do we resolve such confusions during the exam?
wickedvikram
No, because the triangles you get with the last three cases are the same as the triangles you get with the first three. Meaning that 1-12-12 triangle is the same as 12-1-12 triangle.
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