Consider the following system of equivalences of integers.
x  2 mod 15
x  4 mod 21:
The number of solutions in x, where 1  x  315, to the above system of
equivalences is
Option
(A) 0
(B) 1
(C) 2
(D) 3.

• ANS (A)
  
  x ≡ 2 mod 15 means x = 15k + 2, or x - 2 is a multiple of 15
  
  x ≡ 4 mod 21 means x = 21n + 4, or x - 4 is a multiple of 21
  
  List the multiples of each number less than 315:
  
  Possible values of x-2 LE 315
  15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300 [[we should not include 315 as [1 LE x-2 LE 315 ===>1 LE x LE 317 ] which doesnt satisfy the rule... here LE represents less than r equal
  
  Possible values of x
  17,32,47,62,77,92,107,122,137,152,...
  
  Possible values of x-4
  21, 42, 63, 84, 105, 126, 147, 168, 189, 210, 231, 252, 273, 294 [Here also 315 not allowed]
  
  Possible values of x
  25,46,67,88,109,130,151,172,193,214,..
  
  There is no common value of x in both the cases .
  So, the number of solutions is 0
  
  ANS (A)
  
  
  
  
• 10 years agoHelpfull: Yes(18) No(0)
• x ≡ 2 mod 15 means x = 15k + 2, or x - 2 is a multiple of 15
  
  x ≡ 4 mod 21 means x = 21n + 4, or x - 4 is a multiple of 21
  
  List the multiples of each number less than 315:
  
  15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195, 210, 225, 240, 255, 270, 285, 300 [we should not include 315 as [1
• 10 years agoHelpfull: Yes(0) No(0)