The minimum number of numbers required to form a number from 9 to 9000 which are multiples of 5 is:

a) 31 b) 41 c) 42 d) 32

• The multiples of 5 must end with 0 or 5.
  Therefore, units place of a number from 9 to 9000 can be filled by only two digits 0 and 5. i.e., 2 digits.
  10's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
  100's place of a number from 9 to 9000 can be formed by 0 to 9 digits. i.e., 10 digits.
  1000's place of a number from 9 to 9000 can be formed by by 1 to 9 digits 1. i.e., 9 digits.
  Therefore, total number of numbers required is 2 + 10 + 10 + 9 = 31.
• 8 years agoHelpfull: Yes(22) No(2)
• 31 is aswer
• 8 years agoHelpfull: Yes(0) No(1)