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How many 13 digit no are possible by using the digits 12345 which are divisible by 4 if repetition of digits is allowed?
Wiki User
∙ 12y ago
5 ^12
Wiki User
∙ 12y ago
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How many 13 digit numbers are possible by using the digits 12345 which are divisible by 4 if repetition of digits is not allowed?
If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.
How many four digit passwords are possible if repetition of digits is not allowed?
5040
How many 3-digit codes using the digits 0-9 are possible if repetition are allowed?
1000
How many 3-digit numbers using the digits 0-7 are possible if repetition are allowed?
290
How many 5 digits can be formed using the digits 0 2 3 4 5 when repetition is allowed that the number formed is divisible by 2 and 5 or both?
There are 2000 such numbers.
Where d represents a digit from 0 9 how many different ISBN numbers are possible if repetition of digits is allowed?
There are 10 to the 10th power possibilities of ISBN numbers if d represents a digit from 0 to 9 and repetition of digits are allowed. That means there are 10,000,000,000 ISBN numbers possible.
How many 4-digit numbers are possible if the hundreds digit is 8 and if repetition of digits is allowed?
-4
How many 4 digit numbers can be made to give the sum of 6?
If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.
How many 3 digit numbers can be formed from the digits 1234?
64 if repetition is allowed.24 if repetition is not allowed.
How many 5 digit number can be formed using the digits 02345when repetition is allowed such that the number formed is divisible by 2 or 5 or both?
There are 2000 such numbers.
How many 2-digit numbers can be formed using the digits 1 3 7 if repetition of digits is allowed?
Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77
How many 4-digit numbers can be formed by the digits 1 2 3 4 5 with repetition is allowed and that are divisible by 4?
125 different 4 digits numbers. To be divisible by 4, the last two digits must be divisible by 4, which means they must be: 12, 24, 32, 44 or 52 - 5 possible choices. For each of these there are 5 choices for the first digit and 5 for the second, meaning: total = 5 x 5 x 5 = 125
