the sum of the digits of a number is subtraced from the number. the resulting number is always divisible by
Answers
Answer:
The resulting number is always divisible by 3 and 9.
Step-by-step explanation:
Let's take a two digit number : ab
ab can also be written as: 10a + b
The sum of the digits of ab = a + b
The number - sum of digits
⇛ ( 10a + b ) - ( a + b )
⇛ 10a + b - a - b
⇛ 10a - a
⇛ 9a
9a is a multiple of 3 and 9. So, it will be divisible by 3 and 9.
Let's take a three digit number : abc
abc can also be written as: 100a + 10b + c
The sum of the digits of abc = a + b + c
The number - sum of digits
⇛ 100a + 10b + c - (a + b + c)
⇛ 100a + 10b + c - a - b - c
⇛ 99a + 9b
99a is divisible by 9 and 3 as 99 is a multiple of 9 and 3, 9b is also divisible by 9 and 3. So, 99a + 9b is also divisible by 9 and 3.
The same holds true for any number. Note that in a single digit number, the resultant number will be zero which is also divisible by 3 and 9.