Understanding Three-Digit Number Divisibility

Let's explore the property of a three-digit number when the sum of its digits is subtracted from it. We want to find out which numbers always divide the result of this subtraction.

Representing a Three-Digit Number

A three-digit number can be represented using its place values. Let the digits of the three-digit number be \(a\), \(b\), and \(c\), where \(a\) is the hundreds digit, \(b\) is the tens digit, and \(c\) is the units digit. Since it's a three-digit number, the hundreds digit \(a\) cannot be zero (\(a \neq 0\)).

The number itself can be written as:

\( \text{Three-digit number} = 100a + 10b + c \)

Calculating the Sum of Digits

The sum of the digits of this number is simply the sum of its individual digits:

\( \text{Sum of digits} = a + b + c \)

Performing the Subtraction

Now, according to the question, we need to subtract the sum of the digits from the number. Let's perform this subtraction:

\( \text{Result} = (\text{Three-digit number}) - (\text{Sum of digits}) \)

\( \text{Result} = (100a + 10b + c) - (a + b + c) \)

Simplifying the Expression

Let's simplify the expression by removing the parentheses and combining like terms:

\( \text{Result} = 100a + 10b + c - a - b - c \)

Combine the terms with \(a\), terms with \(b\), and terms with \(c\):

\( \text{Result} = (100a - a) + (10b - b) + (c - c) \)

\( \text{Result} = 99a + 9b + 0 \)

\( \text{Result} = 99a + 9b \)

Factoring and Analyzing Divisibility

The simplified result is \(99a + 9b\). We can factor out the common factor from this expression, which is 9:

\( \text{Result} = 9(11a + b) \)

Since the result can be expressed as 9 multiplied by some integer (\(11a + b\) is an integer because \(a\) and \(b\) are integers), this means the result is always a multiple of 9. Any number that is a multiple of 9 is also a multiple of 3 (since \(9 = 3 \times 3\)).

Therefore, the result of subtracting the sum of the digits from a three-digit number is always divisible by both 3 and 9.

Checking Divisibility by 6

For a number to be divisible by 6, it must be divisible by both 2 and 3. We know the result \(9(11a + b)\) is always divisible by 3.

Is it always divisible by 2? Consider the term \(11a + b\). Whether this term is even or odd depends on the values of \(a\) and \(b\).

• If \(11a + b\) is even, then \(9(11a + b)\) will be even and thus divisible by 2.
• If \(11a + b\) is odd, then \(9(11a + b)\) will be odd (since 9 is odd and odd \(\times\) odd = odd) and thus not divisible by 2.

Let's take an example: Suppose the number is 123. Sum of digits = \(1 + 2 + 3 = 6\). Number - Sum of digits = \(123 - 6 = 117\). Is 117 divisible by 6? \(117 \div 6 \approx 19.5\). No, it's not. In this case, \(a=1, b=2, c=3\). The result is \(99(1) + 9(2) = 99 + 18 = 117\). Using the factored form: \(9(11a + b) = 9(11(1) + 2) = 9(11 + 2) = 9(13) = 117\). Here, \(11a + b = 13\), which is odd. The result 117 is odd and not divisible by 2, therefore not divisible by 6.

Since the result is not always divisible by 6, it is not always divisible by "all of 3, 6, and 9".

Conclusion on Divisibility

Based on our algebraic simplification and analysis, the result \(9(11a + b)\) is always a multiple of 9. A number that is a multiple of 9 is also always a multiple of 3. Thus, the result is always divisible by both 3 and 9.

Let's check the given options:

• Option 1: 3 only (Incorrect, it's also divisible by 9)
• Option 2: 9 only (Incorrect, if it's divisible by 9, it must also be divisible by 3)
• Option 3: both 3 and 9 (Correct, as derived from \(9(11a + b)\))
• Option 4: all of 3, 6 and 9 (Incorrect, not always divisible by 6)

Examples

The examples support the finding that the result is always divisible by both 3 and 9, but not always by 6.

Revision Table: Key Concepts

Additional Information: Divisibility Rules

Understanding divisibility rules helps in solving problems like this one quickly.

• Divisibility by 3: A number is divisible by 3 if the sum of its digits is divisible by 3.
• Divisibility by 9: A number is divisible by 9 if the sum of its digits is divisible by 9.
• Divisibility by 6: A number is divisible by 6 if it is divisible by both 2 and 3. This means the number must be even (divisible by 2) and the sum of its digits must be divisible by 3.

In our derivation, the result \(9(11a + b)\) is explicitly a multiple of 9. This guarantees divisibility by 9. Since any multiple of 9 is also a multiple of 3, it guarantees divisibility by 3 as well. The divisibility by 6 depends on whether \(11a + b\) makes the entire expression even, which is not always the case.