Question

write the smallest digit and the largest digit in the blank space of each of the following numbers divisible by 3 a)_6724 b)4765_2​

Answer 1

Step-by-step explanation:

a) _6724

let the blank be x

sum of digits = multiple of 3

x+6+7+2+4=3or9or12

19 +x = 21 ( As it should not be less than 19)

x = 21- 19 = 2

Hence, x = 2

b) 4765_2

let the blank be x

sum of digits = multiple of 3

4+7+6+5+x+2 = 3or9or12...

x+22 = 24 ( As it should not be less than 22)

x = 24-22=2

Hence, x =2

Answer 2

Answer:

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Write the smallest digit and the largest digit in the blank space of each of the following numbers divisible by 3.

a) 26724 b) 476502

$Remember$

Divisibility Rule for 3 = Sum of the digits.

$How \: did \: we \: did \: that?$

a) As _6724 was given then 6 + 7 + 2 + 4 = 19 but 19 isn't divisible by 3. So, to make it divisible if we add 2 to it ( 19 + 2 = 21 ) then it's divisible by 3. ( As 21 is divisible by 3 ) So, the number is 26724.

b) As 4765_2 was given then 4 + 7 + 6 + 5 + 2 = 24 which is divisible by 3. So, let's add 0 to it. So, the number is 476502.