Alok is attending a workshop 'How to do more with less and today's theme is Working with fewer digits. The speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind) had only worked with fewer digits. The problem posed at the end of the workshop is 'How many 5 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 find the answer?

• 5 possible combinations for last 2 digits are 2,24,32,44,52.
  First 3 digits can have any of 5 values ( 1,2,3,4,5)
  
  so reqd 5 digit numbers are 5*5*5*5 = 625
• 12 years agoHelpfull: Yes(6) No(1)
• SORRY I DID MISTAKE IN LAST WAYS 5*5*5*5=625
• 12 years agoHelpfull: Yes(3) No(2)
• DIVISIBILITY RULE OF 4 IS LAST TWO DIGITS ARE DIVISIBLE BY 4
  SO IN 5 DIGIT NO LAST TWO DIGIT SHOULD ALWAYS BE 12,24,32,44,52;
  (ANY ONE OF THEM MAY BE)
  SO FOR LAST TWO DIGITS 5 WAYS ARE POSSIBLE ;
  SINCE REPETION ARE ALLOWED THEN FOR 3RS PLACE FORM RIGHT THERE ARE ALSO 5 WAYS SO IN 3RD PLACE EITHER OF 1,2,3,4,5 IS POSSIBLE;
  SIMILLARLY AT 4 RTH PLACE EITHER OF 1,2,3,4,5 IS POSSIBLE SO AGAIN 5 WAYS ARE POSSIBLE
  SIMILLARLY AT 5TH PLACE ALSO 5 WAYS ARE POSSIBLE
  SO TOTAL NO ARE: 5*5*5*5*5=3125
• 12 years agoHelpfull: Yes(1) No(10)
• ANS : 625
  
  
  ...
• 12 years agoHelpfull: Yes(0) No(1)
• answer is 625-30=595
  is this truee
• 6 years agoHelpfull: Yes(0) No(0)