Alok is attending a workshop How to do more with less and today's theme is working with fever digits. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fever digits. The problem posed at
the end of the workshop is, How many 10 digit numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4? Can you help Alok to find the answer?
(a) 1953125
(b) 781250
(c) 2441407
(d) 2441406

• 5^n-1
  5^9=1953125
  ans is a.
• 13 years agoHelpfull: Yes(17) No(5)
• no divisible by 4 means its last two digits are divisible by 4 ...
  
  now last two digits which are divisible by 4 are 5..
  
  means for 10 digits no.. we have first 8 digits with any of the no..
  so total nos with repetition are 5^8 (ctrl + shift + = not working)
  now last two digits can have 5 choices ... so 5^8 with 12 , 5^8 with 32 same way for the next three nos ..
  
  so total is 5 times 5^8...means 5^9 is the answer
• 13 years agoHelpfull: Yes(7) No(1)
• According to me answer will be- 1953125 (A)
  
  Lets mark 10 places for 10 digits..
  - - - - -
  Now at 1st place the no. of ways of selecting 1 digit from availalbe 5 digits is 5C1
  As it is given Repeatation is allowed.
  
  For 2nd place it is again 5C1 as repeatation is allowed.
  similarly for 3rd 4th....8th place...
  5c1 5c1 5c1 5c1 5c1 5c1 5c1 5c1 - -
  Now for last 2 places in order to be divisible by 4....last 2 places must always end with 12, 24, 32, 44, 52 (you can see it by writing the counting table of 4= 4, 8, 12, 16...)
  So No. of ways of choosing 1 combination from 5 marked combinations is 5C1
  
  So finally, No of digits divisible by 4 are= 5C1*5C1*5C1*5C1*5C1*5C1*5C1*5C1*5C1
  = 5*5*5*5*5*5*5*5*5
  = 1953125 (A)
• 12 years agoHelpfull: Yes(6) No(0)