1 ? 3 Find the missing digit if it has 11 and 13 as factors?
4
Finding the Missing Digit with Factors 11 and 13
The problem asks us to find a missing digit in the number 1?3 such that the resulting three-digit number is divisible by both 11 and 13. Let the missing digit be represented by the variable \(x\).
Representing the Number
The number 1?3 can be written in terms of place values. The digit 1 is in the hundreds place, the missing digit \(x\) is in the tens place, and 3 is in the units place.
So, the number can be expressed as:
\(1 \times 100 + x \times 10 + 3 \times 1 = 100 + 10x + 3 = 103 + 10x\)
Since \(x\) is a digit, it must be an integer value from 0 to 9.
Divisibility by 11 and 13
We are told that the number \(103 + 10x\) is divisible by both 11 and 13. When a number is divisible by two different prime numbers, it must also be divisible by their product.
The numbers 11 and 13 are both prime numbers. Their product is:
\(11 \times 13 = 143\)
Therefore, the number \(103 + 10x\) must be a multiple of 143.
Finding the Relevant Multiple of 143
The number \(103 + 10x\) is a three-digit number that starts with 1 and ends with 3. The possible range for this number, considering \(x\) is a digit from 0 to 9, is:
- If \(x=0\), the number is \(103 + 10(0) = 103\).
- If \(x=9\), the number is \(103 + 10(9) = 103 + 90 = 193\).
So, the number \(103 + 10x\) is a number between 103 and 193 (inclusive of 103 and 193).
Now, let's list the multiples of 143:
- \(143 \times 1 = 143\)
- \(143 \times 2 = 286\)
- \(143 \times 3 = 429\)
We are looking for a multiple of 143 that falls within the range 103 to 193. The only multiple in this range is 143.
Solving for the Missing Digit
We now know that the number \(103 + 10x\) must be equal to 143. We can set up an equation and solve for \(x\):
\(103 + 10x = 143\)
Subtract 103 from both sides of the equation:
\(10x = 143 - 103\)
\(10x = 40\)
Divide by 10:
\(x = \frac{40}{10}\)
\(x = 4\)
The value of \(x\) is 4, which is a single digit between 0 and 9. This means the missing digit is 4.
The number is 143. Let's quickly verify if 143 is divisible by 11 and 13:
- \(143 \div 11 = 13\)
- \(143 \div 13 = 11\)
It is divisible by both. Therefore, the missing digit is 4.
| Step | Description | Calculation/Result |
|---|---|---|
| 1 | Represent the number 1?3 | \(103 + 10x\) |
| 2 | Find the product of factors 11 and 13 | \(11 \times 13 = 143\) |
| 3 | State the number must be a multiple of the product | \(103 + 10x = 143k\) |
| 4 | Determine the range of the number | 103 to 193 |
| 5 | Find the multiple of 143 in the range | 143 |
| 6 | Solve for \(x\) | \(103 + 10x = 143 \Rightarrow 10x = 40 \Rightarrow x = 4\) |
| 7 | Missing digit | 4 |
Revision Table: Divisibility Rules and Prime Factors
Understanding divisibility rules and properties of prime numbers is crucial for solving problems like this.
| Concept | Explanation | Relevance |
|---|---|---|
| Divisibility | A number is divisible by another if dividing leaves no remainder. | The number 1?3 must be perfectly divisible by 11 and 13. |
| Prime Number | A natural number greater than 1 that has no positive divisors other than 1 and itself (e.g., 2, 3, 5, 7, 11, 13). | 11 and 13 are prime factors in this problem. |
| Coprime Numbers | Two integers a and b are coprime (or relatively prime) if the only positive integer that divides both of them is 1. Prime numbers are always coprime to other prime numbers. | 11 and 13 are coprime. If a number is divisible by two coprime numbers, it's divisible by their product. |
| Product of Factors | Multiplying two numbers together. If a number is divisible by coprime numbers, it is divisible by their product. | The number must be divisible by \(11 \times 13 = 143\). |
Additional Information: Solving Problems with Multiple Factors
When a problem states that a number has multiple factors (is divisible by multiple numbers), consider the nature of these factors:
- If the factors are coprime: The number must be divisible by their product. For example, if a number is divisible by 3 and 5 (coprime), it must be divisible by \(3 \times 5 = 15\).
- If the factors are not coprime: The number must be divisible by their Least Common Multiple (LCM). For example, if a number is divisible by 4 and 6 (not coprime, LCM is 12), it must be divisible by 12, not necessarily \(4 \times 6 = 24\). The number 12 is divisible by 4 and 6, but not 24.
In this problem, 11 and 13 are prime numbers, so they are coprime. This is why we could use their product (143) directly.
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select the correct answer using the code given below:
