P is an integer P883 If P-7 is a multiple of 11 then the largest number that will always divide P4 P15 is Op 1 11 Op 2 121Op 3 242 Op 4 None of these

• OPTION C: 242
  
  P > 883
  P − 7 > 876
  
  The smallest number after 876 that is a multiple of 11 is 880 = 11*80
  So P = 11k+7, for some integer k ≥ 80
  
  (P+4)(P+15)
  = (11k+7+4) (11k+7+15)
  = (11k+11) (11k+22)
  = 11(k+1) * 11 (k+2)
  = 121 (k+1)(k+2)
  
  Now when k is even, then k+2 is even, so (k+1)(k+2) is even
  and when k is odd, then k+1 is even, so (k+1)(k+2) is even
  (k+1) (k+2) = 2n, for some integer n
  
  (P+4)(P+15)
  = 121 (k+1)(k+2)
  = 121 * 2n
  = 242n
  
• 9 years agoHelpfull: Yes(7) No(2)