How many five digit numbers can be formed using the digits 02345 when repetition is allowed such that the number formed is divisible by 2 or 5 or both?

Answer 1

There are 2000 possible five digit numbers that can be formed from the digits 02345 that are divisible by 2 or 5 or both.

To be divisible by 2, the last digit must be even, namely 0, 2 or 4 (in the digits allowed).

To be divisible by 5, the last digit must be 0 or 5.

Thus to be divisible by 2 or 5 or both, the last digit must be 0, 2, 4 or 5 (a choice of 4).

Presuming that a 5 digit number must be at least 10000, then:

For the first digit there is a choice of 4 digits (2345);

for each of these there is a choice of 5 digits (02345) for the second, making a total so far of 4 x 5 numbers;

for each of these choices for the first and second digits there is a choice of 5 digits (02345) for the third digit making the total so far (4 x 5) x 5 numbers;

for each of these choices for the first three digits there is a choice of 5 digits (02345) for the fourth digit making the total so far (4 x 5 x 5) x 5 numbers;

for each of these choices for the first four digits there is a choice of 4 digits (0245 - as discussed above) for the last digit, giving a total of (4 x 5 x 5 x 5) x 4 numbers.

So the total number of five digit numbers so formed is:

number = 4 x 5 x 5 x 5 x 4

= 2000.