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

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The discussion focuses on calculating the number of five-digit numbers divisible by 8 using specific digits. For five-digit numbers formed from the digits {1,2,3,4,5} without repetition, there are 10 valid combinations based on the last three digits. In contrast, for five-digit numbers formed from the digits {0,1,2,3,4,5} with repetition allowed, the total increases to 185 combinations due to the inclusion of zero and additional valid endings. The complexity arises from ensuring the leading digit is not zero while considering various possible endings. Overall, the calculations highlight the impact of digit repetition and inclusion on the total count of valid numbers.
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(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)
 
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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.
 
jacks said:
(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).
 

Hello, everyone!

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