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*

 

[blue_raver22]

(1) To find the total number of 5-digit numbers formed using the digits 1, 2, 3, 4, 5 and divisible by 8 when repetition is not allowed, we can use the following steps:

Step 1: Understand the Divisibility Rule for 8
A number is divisible by 8 if the last three digits of the number are divisible by 8.

Step 2: Determine the Possible Combinations
Since repetition is not allowed, we can use each digit only once. Therefore, the possible combinations of 5-digit numbers using the digits 1, 2, 3, 4, 5 are:
5 x 4 x 3 x 2 x 1 = 120 combinations

Step 3: Determine the Last Three Digits
Since the last three digits must be divisible by 8, we can only use the following combinations:
168, 328, 488, 528, 648

Step 4: Determine the First Two Digits
Since the first two digits can be any of the remaining digits, we have 3 x 2 = 6 combinations.

Step 5: Determine the Total Number of 5-digit Numbers
The total number of 5-digit numbers formed using the digits 1, 2, 3, 4, 5 and divisible by 8 when repetition is not allowed is:
6 x 5 = 30

Therefore, there are 30 5-digit numbers with no repetition that are divisible by 8 when formed using the digits 1, 2, 3, 4, 5.(2) To find the total number of 5-digit numbers formed using the digits 0, 1, 2, 3, 4, 5 and divisible by 8 when repetition is allowed, we can use the following steps:

Step 1: Understand the Divisibility Rule for 8
A number is divisible by 8 if the last three digits of the number are divisible by 8.

Step 2: Determine the Possible Combinations
Since repetition is allowed, we can use any of the digits multiple times. Therefore, the possible combinations of 5-digit numbers using the digits 0, 1, 2, 3, 4, 5 are:
6 x 6 x 6 x 6 x 6 = 7776 combinations