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26 January, 21:06

Determine the sum of all single-digit replacements for z such that the number 24, z38 is divisible by 6.

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  1. 26 January, 21:34
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    The sum of all single-digit replacements for z is 12

    Step-by-step explanation:

    * Lets explain how to solve the problem

    - The number is divisible by 6 if it divisible by 2 and 3

    - Any even number divisible by 2

    - The number is divisible by 3 is the sum of its digits divisible by 3

    * Now lets solve the problem

    - The number 24, z38 is divisible by 6

    - We need to find all the possible values of z which keep the number

    divisible by 6

    ∵ Lets add the sum of the digits without z

    ∵ 2 + 4 + 3 + 8 = 17

    ∵ 18 is the nearest number to 17

    ∵ 18 is divisible by 3

    ∴ Add 17 by 1 to get 18

    ∴ z = 1

    - Lets check the number

    ∵ The number is 24,138

    ∵ 24,138 : 6 = 4023

    ∴ The number is divisible by 6

    ∵ 21 is the next number after 18 and divisible by 3

    ∴ We must add 17 by 4 to get 21

    ∴ z = 4

    - Lets check the number

    ∵ The number is 24,438

    ∵ 24,438 : 6 = 4073

    ∴ The number is divisible by 6

    ∵ 24 is the next number after 18 and divisible by 3

    ∴ We must add 17 by 7 to get 24

    ∴ z = 7

    - Lets check the number

    ∵ The number is 24,738

    ∵ 24,738 : 6 = 4123

    ∴ The number is divisible by 6

    - There is no other value for z because if we take the next number

    of 24 divisible by 3 it will be 27, then we must add 17 by 10 but

    10 not a single digit

    ∴ The possible values of z are 1, 4, 7

    ∴ The sum of them = 1 + 4 + 7 = 12

    ∴ The sum of all single-digit replacements for z is 12
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