MHB No. of 5 Digit Nos Divisible by 8 (With/Without Repetition)

[juantheron]

(1) Total no. of $5$ digit no. which is formed by Using the Digits $1,2,3,4,5$ and Divisible by $8$

(when repetition is not allowed)

(2) Total no. of $5$ digit no. which is formed by Using the Digits $0,1,2,3,4,5$ and Divisible by $8$

(when repetition is allowed)

 

[soroban]

Hello, jacks!

(1) Total no. of 5-digit no. formed of the digits {1,2,3,4,5} and divisible by 8
(when repetition is not allowed)

A number is divisible by 8 if its rightmost 3-digit number is divisible by 8.

There are 5 such endings: .--152, --312, --352, --432, --512

The other two spaces can be filled in 2! ways.

Therefore, there are: .5(2!) = 10 such numbers.

(2) Total no. of 5-digit no. formed of digits {0,1,2,3,4,5} and divisible by 8
(when repetition is allowed)

If repetition is allowed, there are 7 more endings:
. . --112, --224, --232, --344, --424, --552, --544

The other two spaces can be filled in 5² ways.

Therefore, there are: 10 + 7(25) = 185 such numbers.

 

[Opalg]

(1) Total no. of $5$ digit no. which is formed by Using the Digits $1,2,3,4,5$ and Divisible by $8$

(when repetition is not allowed)

(2) Total no. of $5$ digit no. which is formed by Using the Digits $0,1,2,3,4,5$ and Divisible by $8$

(when repetition is allowed)

I agree with soroban on (1). For (2), the situation is more complicated because the digit 0 is also allowed. The possible three-digit endings are $$E24,\,E32,\,E40,\,O04,\,O12,\,O20,\,O44,\,O52,$$ where $E$ stands for an even number (0,2 or 4) and $O$ stands for an odd number (1,3 or 5). That gives $3\times8=24$ possible endings. For each ending there are $5\times6=30$ choices for the first two digits (5 for the leading digit, which must not be 0, and 6 for the second digit).

 

[soroban]

Hello, everyone!

Silly me . . . I overlooked the zero in the second part. . *slap head*