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

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Discussion Overview

The discussion revolves around calculating the number of 5-digit numbers that can be formed using specific digits and are divisible by 8. The two scenarios considered are: one where repetition of digits is not allowed (using the digits 1, 2, 3, 4, 5) and another where repetition is allowed (using the digits 0, 1, 2, 3, 4, 5).

Discussion Character

  • Mathematical reasoning, Debate/contested

Main Points Raised

  • Some participants propose that for 5-digit numbers formed from the digits {1, 2, 3, 4, 5} and divisible by 8 (without repetition), the valid endings are limited, leading to a calculation of 10 such numbers.
  • Others argue that for 5-digit numbers formed from the digits {0, 1, 2, 3, 4, 5} and divisible by 8 (with repetition allowed), additional valid endings increase the total to 185, based on the combinations of digits.
  • A later reply questions the complexity introduced by the inclusion of the digit 0 in the second scenario, suggesting that the number of valid endings is greater than initially considered, with a calculation yielding 24 possible endings based on even and odd digit placements.
  • One participant acknowledges a mistake regarding the consideration of the digit 0 in their previous calculations.

Areas of Agreement / Disagreement

Participants generally agree on the approach for the first scenario but express differing views on the second scenario, particularly regarding the impact of including the digit 0 and the resulting calculations. The discussion remains unresolved with multiple competing views on the second part.

Contextual Notes

There are unresolved assumptions regarding the specific combinations of digits and the conditions for divisibility by 8, particularly in the second scenario where the presence of 0 complicates the calculations.

juantheron
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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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