Q. If x and y are two integers then number of possible solutions for 5|x|+2|y|=100 ?

• 5|x|+2|y|=100
  => 5|x|=100-2|y|
  (100-2|y|) is even so 5|x| should even
  for
  x=0,y=50
  x=2,y=45
  x=4,y=40
  x=6,y=35
  ....
  x=18,y=5
  x=20,y=0
  
  for x=0,y=50,(x,y)=(0,50)&(0,-50) => 2 soln
  for x=20,y=0,(x,y)=(20,0)&(-20,0) => 2 soln
  
  for remaining 9 soln, (x,y)=(2,4,6..,18, 45,40,35,..5)
  4 soln for each pair
  for x=2,y=45, (x,y)=(2,45)(-2,45)(-2,-45)(2,-45)
  similarly 4 soln for remaining 8 pair
  
  total no. of soln= 2+2+9*4= 40
  
• 10 years agoHelpfull: Yes(19) No(1)
• 5|x|+2|y|=100
  solution for integer solution x is always even
  possible value of x=(-20,-18,-16.........0,2,4,6.....20){total 21 valus
  for every value of x y have 2 solution + or - remaining at -20 and 20 (y=0)
  so total solution =21*2-2=40
  
  so ans is 40
• 10 years agoHelpfull: Yes(4) No(3)
• @Rakesh kindly solve all ques ur solution seems logical.....
• 10 years agoHelpfull: Yes(4) No(0)
• Ans:- 40
  
  I checked by programming.. they given mod it means we need to take -ve and +ve numbers.
• 10 years agoHelpfull: Yes(2) No(0)
• 100/lcm of 5&2
  100/10=10
• 10 years agoHelpfull: Yes(1) No(0)
• 2*anything is an even number.
  hence 2|y|=100-5|x| has to be an even number.
  which can be possible if and only if 5|x| is an even number. hence x can be of range 2,4,6,8,10,12,14,16,18,20 and its negative representations along with a 0(zero). hence total possible solutions is 21
• 10 years agoHelpfull: Yes(1) No(0)
• @Arun,but 21 wasn't there among options..
  
• 10 years agoHelpfull: Yes(0) No(0)
• good job rakesh
• 10 years agoHelpfull: Yes(0) No(0)