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In order to be divisible by both 2 and 5, the number must be divisible by 10.

That pretty well means that the last digit always has to be zero.

We're also going to assume that since each number must be a 5-digit, leading zeros

are ruled out.

First digit . . . . . 2, 3, 4, 5 . . . . . 4 possibilities

Second digit. . . 0, 2, 3, 4, 5 . . . 5 possibilities

Third digit. . . . . 0, 2, 3, 4, 5 . . . 5

Fourth digit . . . 0, 2, 3, 4, 5 . . . 5

Fifth digit . . . . . 0

Total number of possibilities = 4 x 5 x 5 x 5 = 500

Q: How many 5 digit numbers can be formed using the digits 0 2 3 4 5 when repetition is allowed that the number is divisiible by 2 or 5 or both?

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64 if repetition is allowed.24 if repetition is not allowed.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

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.

10^7 if the repetition of digits is allowed. 9*8*7*6*5*4*3 , if the repetition of digits is not allowed.

125

Related questions

64 if repetition is allowed.24 if repetition is not allowed.

If repetition of digits is allowed, then 56 can.If repetition of digits is not allowed, then only 18 can.

If repetition of digits isn't allowed, then no13-digit sequencescan be formed from only 5 digits.

24 three digit numbers if repetition of digits is not allowed. 4P3 = 24.If repetition of digits is allowed then we have:For 3 repetitions, 4 three digit numbers.For 2 repetitions, 36 three digit numbers.So we have a total of 64 three digit numbers if repetition of digits is allowed.

-123456787

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.

10^7 if the repetition of digits is allowed. 9*8*7*6*5*4*3 , if the repetition of digits is not allowed.

125

290

-4

125 There are five choices for each of the three digits (since repetition is allowed). So there are 5*5*5=125 combinations.

Possible solutions - using your rules are:- 11,13,17,31,33,37,71,73 &77