If 50 × 51 × 52 × 53 ×…× 200 is completely divisible by 9^a, then find the maximum possible value of ‘a’. Its answer is 37. But I don't know how ??

• 50 × 51 × 52 × 53 ×…× 200 => 200! - 49!
  
  9 => 3^2
  [9 is not a prime factor... So we have to find the highest power of 3... From that we can calculate power of 9...]
  No. of 3's in 200! is
  200/3^1 + 200/3^2 + 200/3^3 + 200^3^4 = 66+22+7+2 = 97
  No. of 3's in 49! is
  49/3^1 + 49/3^2 + 49/3^3 = 16+5+1 = 22
  No. of 3's in 50 × 51 × 52 × 53 ×…× 200 is (97-22) = 75
  We have to find the highest power of 9 not 3 ... So 75/2 = 37
  
  Ans : 37
• 10 years agoHelpfull: Yes(18) No(0)
• 50 × 51 × 52 × 53 ×…× 200 = 200! / 49!
  
  9^a = 3^2a
  no. of 3's in 200! =
  [200/3^1] + [200/3^2] + [200/3^3] + [200^3^4] = 66+22+7+2 = 97
  no. of 3's in 49! =
  [49/3^1] + [49/3^2] + [49/3^3] = 16+5+1 = 22
  
  no. of 3's in 50 × 51 × 52 × 53 ×…× 200 = 200! / 49! = 3^97/3^22 = 3^75
  for this to be completely divisible by 9^a = 3^2a , value of 2a should less or equal to 75
  => 2a = 75
  => a = 37
  
• 10 years agoHelpfull: Yes(4) No(0)