Three lines intersect at point P, creating 6 angles. 3 adjacent angles

Question

Three lines intersect at point  O , creating 6 angles. Three adjacent angles are  19X, 3Y,  and  19Z . If  X, Y,  and  Z  are positive integers, how many different values of  X  can have?
A. 2
B. 3
C. 5
D. 8
E. 13

Source: All questions I posted are adapted or enhanced from OG or OG Quant .
 Kudos for a good or creative solution.  Also, give +1 kudos if you think this is a good question.

nick1816 · Answer

Sum of any 3 adjacent angle is 180 degrees

19X+3Y+19Z=180
19(X+Z)+3Y=180
19(X+Z)= 3(60-Y)

Hence, X+Z is a multiple of 3.
As X, Y and Z are positive integers, X+Z can be 3, 6 or 9

When X+Z=9
X can take any integral value from 1 to 8 (both inclusive)

D

Peddi · Answer

Given, 19(x+z)+3y=180, maximum value x+z can take is 9, we know that 3y is a multiple of 3 ,for 3y to be a multiple of 3 ,x+z should be multiple of three remember x+z max value is nine ,so x+z = 3,6,9 , substituting integers which add up to 9 , we get x=1,2,3,4,5,6,7,8 total possible values of x=8 , it took me more than four minutes can anyone suggest shorter method

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Posted from my mobile device

freedom128 · Answer

Peddi , well done. I think your solution is very efficient (multiple of 3 and range of possible x). But, people tend to overevaluate so as not to miss out any possibilities.

Fatineel , try this question. Good or creative solution (other than posted one) will be given kudo. Also, please kudo if this is good question.

nick1816 · Answer

X+Z=9 is a straight line in 2-D plane, and x,z≥1. Hence x will take all integral values in range (9-1)≥x≥1. You don't have to find all the combinations.

Prasannathawait · Answer

I solved it this way,
19x+3y+19z=180
We have to find max value of X so Y and Z should be minimum. If y and z are 1 then eq will be = 19x+22=180
19x=158
We can get 19*8=152. 158-152=6 (multiple of 3)
Hence ans should be 8.
13 is anyways not possible as 19*13 will be greater than 180.

freedom128 · Answer

Hahaha, it is true that maximum value of integer x is 8. But, it doesn't directly translate to 8 nos. of solutions that x can have (1,2,...,8) - though in this question, yes. It could be 4 only (e.g. 2,4,6,8).

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unraveled · Answer

X, Y and Z all are positive integers. X > 0, Y > 0 and Z > 0

Since the point has three lines intersecting each other the angles made sum to 360 deg. Thus,

19X + 19X + 19Z + 19Z + 3Y + 3Y = 360

38(X + Z) + 6Y = 360

  Y = \frac{(180 -19(X + Z))}{3} 
  Y = 60 – \frac{19(X + Z)}{3}

i.e. the maximum value of  \frac{19(X + Z)}{3} = 57  and minimum value of  \frac{19(X + Z)}{3} = 19  so that Y remains positive.
  Max(X + Z) = 9  and  Min(X + Z) = 3

But we are concerned with finding maximum values of X so considering  X + Z = 9 ,

1 ≤ X ≤ 8  and  1 ≤ Z ≤ 8  complementing X

Hence X can have 8 values.

Answer (D).

unraveled · Answer

chondro48  looks like  Prasannathawait  missed writing down 'multiply it by 2'.
Otherwise it's also good solution condition being Y ≠ 1 which would make X a non integer.

Doesn't saying "13 is anyways not possible as 19*13 will be greater than 180" enough as rightly pointed out by him.

Kinshook · Answer

19X + 3Y + 19Z = 180
Since last digit of 180 is 0, last digit of 19X + 3Y + 19Z must be 0
Y=1; Not feasible
Y=2; Not feasible
Y=3; X + Z = 9; Number of solutions = 8 = {(1,8),(2,7),(3,6),(4,5),(5,4),(6,3),(7,2),(8,1)}; X can have 8 different values = {1,2,3,4,5,6,7,8}
Y=22; X + Z = 6; Number of solutions = 5 = {(1,5),(2,4),(3,3),(4,2),(5,1)}; X can have 5 different values = {1,2,3,4,5}
Y=41; X + Z = 3; Number of solutions = 2 = {(1,2),(2,1)}; X can have 2 different values = {1,2}

Overall X have have different values = {1,2,3,4,5,6,7,8} = 8 different values

IMO D

Three lines intersect at point P, creating 6 angles. 3 adjacent angles

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