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Question 374811: how many 13 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if repetition of digits is allowed?
Found 3 solutions by vasumathi, kutekarthik, richard1234:
Answer by vasumathi(46)   (Show Source):
You can put this solution on YOUR website! The given digits are 1,2,3,4,5
For the number to be divisible by 4 we should have the last two digits to be a multiple of 4
That is we should have the last two digits to be 2 and 4
so let us write the last two digits of as 2 4
now we are left with 11 places
these places are to be filled with any of the digits 1,2,3,4,5
so each place can be filled in 5 ways since repetition is allowed
so we have 5 * 5* 5* .... 11 times
so the total number = 5^11=48828125
similarly the last two digits may be 12 and 44
so with 12 and 44 also we get 5^11 = 48828125
so total = 3* 48828125=146484375
This is the final answer
Answer by kutekarthik(3) (Show Source):
You can put this solution on YOUR website! If the 13 digit number to be divisble by 4 then it must contain the last two digits that must be divisble by 4
Given digits are 1,2,3,4,5
Last two digits possibilities are 12,24,32,44,54
The remaining 11 positions can have values 1-5 since repetition is allowed then we have 5^11
Total 5*5^11=244140625
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! The last two digits must be a multiple of 4. This means that the last two digits must be either 12, 24, 32, 44, or 52 (as these are the only two digit multiples of 4 possible with these digits)
There are five ways to choose one of these 2-digit combinations. The other 11 digits don't matter, and are mutually exclusive to each other so there are 5^11 ways to choose these 11 digits.
Therefore the total number of ways is 5*(5^11) = 5^12 = 244,140,625 ways.
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