What is the HCF of 329 - 9 and 338 - 9 ?
311 - 9
Finding the HCF of Numbers with Exponents
We are asked to find the Highest Common Factor (HCF) of two numbers: \(3^{29} - 9\) and \(3^{38} - 9\). The HCF is the largest positive integer that divides both numbers without leaving a remainder.
Let's look at the two numbers:
- First number: \(3^{29} - 9\)
- Second number: \(3^{38} - 9\)
We can rewrite the number 9 as \(3^2\). So, the numbers become:
- First number: \(3^{29} - 3^2\)
- Second number: \(3^{38} - 3^2\)
Notice that both expressions have \(3^2\) as a common factor. We can factor out \(3^2\) from both terms:
- First number: \(3^2(3^{29-2} - 1) = 3^2(3^{27} - 1)\)
- Second number: \(3^2(3^{38-2} - 1) = 3^2(3^{36} - 1)\)
Now, the HCF of the original two numbers will be \(3^2\) multiplied by the HCF of \((3^{27} - 1)\) and \((3^{36} - 1)\).
There is a useful property for finding the HCF of expressions of the form \(a^m - 1\) and \(a^n - 1\).
Property: The HCF of \((a^m - 1)\) and \((a^n - 1)\) is \((a^{\text{HCF}(m, n)} - 1)\).
In our case, for the terms \((3^{27} - 1)\) and \((3^{36} - 1)\):
- Base \(a = 3\)
- Exponent \(m = 27\)
- Exponent \(n = 36\)
We need to find the HCF of the exponents, i.e., HCF(27, 36).
Calculating HCF(27, 36)
Let's find the prime factorization of 27 and 36.
- Prime factors of 27: \(27 = 3 \times 3 \times 3 = 3^3\)
- Prime factors of 36: \(36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2\)
The common prime factors with the lowest powers are \(3^2\).
So, HCF(27, 36) = \(3^2 = 9\).
Applying the HCF Property
Using the property \( \text{HCF}(a^m - 1, a^n - 1) = a^{\text{HCF}(m, n)} - 1 \), we get:
\[ \text{HCF}(3^{27} - 1, 3^{36} - 1) = 3^{\text{HCF}(27, 36)} - 1 = 3^9 - 1 \]Now, we combine this with the \(3^2\) that we factored out earlier.
The HCF of the original numbers \(3^{29} - 9\) and \(3^{38} - 9\) is:
\[ 3^2 \times \text{HCF}(3^{27} - 1, 3^{36} - 1) = 3^2 \times (3^9 - 1) \]Let's simplify this expression:
\[ 3^2 \times (3^9 - 1) = (3^2 \times 3^9) - (3^2 \times 1) \]Using the rule of exponents \(a^x \times a^y = a^{x+y}\):
\[ 3^2 \times 3^9 = 3^{2+9} = 3^{11} \]So, the expression becomes:
\[ 3^{11} - 3^2 \] \[ 3^{11} - 9 \]Thus, the HCF of \(3^{29} - 9\) and \(3^{38} - 9\) is \(3^{11} - 9\).
Checking the Options
Let's compare our result with the given options:
- Option 1: \(3^9 - 1\)
- Option 2: \(3^{11} - 1\)
- Option 3: \(3^{11} - 3\)
- Option 4: \(3^{11} - 9\)
Our calculated HCF, \(3^{11} - 9\), matches Option 4.
| Number 1 | Number 2 | Factored Form | HCF Calculation |
|---|---|---|---|
| \(3^{29} - 9\) | \(3^{38} - 9\) | \(3^2(3^{27} - 1)\), \(3^2(3^{36} - 1)\) | \(3^2 \times \text{HCF}(3^{27} - 1, 3^{36} - 1)\) |
| Using \( \text{HCF}(a^m - 1, a^n - 1) = a^{\text{HCF}(m, n)} - 1 \) | |||
| m = 27 | n = 36 | HCF(27, 36) = 9 | \( \text{HCF}(3^{27} - 1, 3^{36} - 1) = 3^9 - 1 \) |
| Final HCF = \(3^2 \times (3^9 - 1) = 3^{11} - 9\) | |||
The Highest Common Factor of \(3^{29} - 9\) and \(3^{38} - 9\) is \(3^{11} - 9\).
Revision Table: HCF and Exponents
| Concept | Description | Relevance to Problem |
|---|---|---|
| HCF (Highest Common Factor) | The largest number that divides two or more numbers exactly. | The goal is to find the HCF of the given expressions. |
| Exponents | A number raised to a power (e.g., \(a^n = a \times a \times ... \times a\) (n times)). | The given numbers involve exponents (powers of 3). |
| Factoring | Expressing a number or algebraic expression as a product of its factors. | We factored out \(3^2\) from both expressions. |
| HCF Property for \(a^m-1\) and \(a^n-1\) | \( \text{HCF}(a^m - 1, a^n - 1) = a^{\text{HCF}(m, n)} - 1 \) | This property is key to solving the main part of the problem. |
| Exponent Rule: \(a^x \times a^y = a^{x+y}\) | When multiplying powers with the same base, add the exponents. | Used in the final simplification: \(3^2 \times 3^9 = 3^{11}\). |
Additional Information: Understanding HCF with Powers
Finding the HCF of numbers that involve powers can sometimes be simplified using algebraic properties or properties related to the structure of the numbers. In this problem, recognizing that 9 is \(3^2\) was the first crucial step.
The property \( \text{HCF}(a^m - 1, a^n - 1) = a^{\text{HCF}(m, n)} - 1 \) is very useful for finding the HCF of differences of powers when the base is the same and the numbers are one less than a power of the base. This property arises from the division algorithm and properties of modular arithmetic, but for problem-solving, understanding and applying the formula directly is sufficient.
For example, to find HCF of \(2^6 - 1\) and \(2^9 - 1\):
- Here, \(a=2\), \(m=6\), \(n=9\).
- First, find HCF(6, 9). Factors of 6 are 1, 2, 3, 6. Factors of 9 are 1, 3, 9. The highest common factor is 3.
- Using the property, HCF is \(2^{\text{HCF}(6, 9)} - 1 = 2^3 - 1 = 8 - 1 = 7\).
In our problem, the numbers were not exactly in the form \(a^m - 1\), but \(a^m - a^k\). By factoring out the common \(a^k\) term (\(3^2\) in this case), we transformed the problem into finding the HCF of terms in the \(a^m - 1\) form, multiplied by the factored term. This is a common technique when dealing with HCF of expressions involving powers with common bases and offsets.
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