Question

Ow 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?

Answer 1

Answer:

244140625 ways

Step-by-step explanation:

Given numbers to be used;

1, 2, 3, 4, 5

To be noted that repetition of digits is allowed.

A number is divisible by 4 if the last two digits of the number is divisible by 4

If the number is to be divisible by 4, the last digit should either be 2 or 4

Case 1: If the last digit is 2

The last but one digit should either be 1 or 3 or 5 = 3 ways

The remaining 11 spaces can be filled by any of the 5 digits = 5 x 5 x 5 x 5 x …….11 times = $5^{11}$

So, the number of ways if the last digit is 2 = $5^{11}$ x 3

Case 2: If the last digit is 4

The last but one digit should either be 2 or 4 = 2 ways

The remaining 11 spaces can be filled by any of the 5 digits = 5 x 5 x 5 x 5 x …….11 times = $5^{11}$

So, the number of ways if the last digit is 2 = $5^{11}$ x 2

Therefore, the number of possible 13 digit numbers using 1, 2, 3, 4, 5 which are divisible by 4

= $5^{11}$ x 3 + $5^{11}$ x 2

= $5^{11}$ (3 + 2)

= $5^{12}$

= 244140625 ways