Let X be a 5 digit number such that the three consecutive digits are the same number

• Ans is 2700 as there will be 3 cases each consists 900 bcoz we have to find 5 digit n
• 10 years agoHelpfull: Yes(26) No(4)
• possible cases : aaabc baaac bcaaa
  for the first case we cannot take zero
  therefore total number : 9(for a)* 10(for b) *10(for c)=9*10*10=900
  second case: 10*10*10=1000
  third case: 10*10*10=1000
  total number of numbers: 900+1000+1000=2900
• 10 years agoHelpfull: Yes(18) No(20)
• "six" combinations for the 5 digit number: aaabc aaacb baaac caaab bcaaa cbaaa
  for any number the first digit cannot be "0"
  therefore for every combination only 900 combinations are possible.
  therefore ["6" combinations for the 5 digit number * 900 combinations]=54000
• 10 years agoHelpfull: Yes(5) No(6)
• let us assume a pack of 3 numbers ....either 0,1,2,3,4,5,6,7,8,9 that means first we have to choose location of pack 5c3=6 ways but if the pack comes at first place we would not start it with zero so ..for that case we chan choose 9 values from set n rest other 2 digits in 10*10 ways so....9*10*10= 900
  
  now rest 5 cases we can choose any value for that pack so 10 ways n for rest 2 digits it would be 9*10....so....9*10*10
  
  adding both 900+900=1800 numers are possible...
  
• 10 years agoHelpfull: Yes(1) No(9)
• plz explain clearly
• 10 years agoHelpfull: Yes(1) No(0)
• The question is
  Let X be a four digit number with exactly three consecutive digits being same and is a multiple of 9. How many such X’s are possible?
  
  
  “Sum of all digits must be divisible by 9” - for a number to be multiple of 9.
  
  The way to proceed is - considering three consecutive same digits and for the selection of fourth digit, checking divisibility by ‘9’ for the complete number.
  
  Let us see case by case:
  
  Three consecutive 0’s: 9000 – 1 case
  (Note here that as the sum of Three 0’s is ‘0’, the fourth digit must be ‘9’)
  Three consecutive 1’s: 1116, 6111 – 2 cases
  (Note here that as the sum of Three 1’s is ‘3’, the fourth digit must be ‘6’)
  Three consecutive 2’s: 2223, 3222 – 2 cases
  Three consecutive 3’s: 3330, 3339, 9333 – 3 cases
  (Note here that as the sum of Three 3’s is ‘9’, the fourth digit can be ‘0’ or ‘9’)
  Three consecutive 4’s: 4446, 6444 – 2 cases
  Three consecutive 5’s: 5553, 3555 – 2 cases
  Three consecutive 6’s: 6660, 6669, 9666 – 3 cases
  Three consecutive 7’s: 7776, 6777 – 2 cases
  Three consecutive 8’s: 8883, 3888 – 2 cases
  Three consecutive 9’s: 9990 – 1 case
  (Note that 9999 won’t be considered as it is asked for only three consecutive numbers case and not for four)
  
  If we sum up all, we get a total of 20 possibilities
• 10 years agoHelpfull: Yes(1) No(0)
• can u xpln in shortcut method
• 10 years agoHelpfull: Yes(0) No(1)