41 43 + 43 43 is divisible by
84
Understanding the Divisibility of \(41^{43} + 43^{43}\)
The question asks us to determine which of the given options divides the expression \(41^{43} + 43^{43}\).
To solve this, we can use a standard property related to the sum of powers. The property states that if \(n\) is a positive odd integer, then \(a^n + b^n\) is always divisible by \(a+b\).
Let's look at our expression: \(41^{43} + 43^{43}\). Here, we have:
- \(a = 41\)
- \(b = 43\)
- \(n = 43\)
The exponent \(n = 43\) is an odd integer. Therefore, according to the property mentioned above, the expression \(41^{43} + 43^{43}\) must be divisible by \(a+b\).
Let's calculate \(a+b\):
\(a+b = 41 + 43 = 84\)
So, \(41^{43} + 43^{43}\) is divisible by 84.
Now we check the given options to see which one is 84 or a factor of 84:
- 80
- 84
- 86
- 88
Our calculated value, 84, is one of the options.
Thus, \(41^{43} + 43^{43}\) is divisible by 84.
Step-by-Step Solution
- Identify the form of the expression: The expression is \(41^{43} + 43^{43}\), which is in the form \(a^n + b^n\) with \(a=41\), \(b=43\), and \(n=43\).
- Check the exponent \(n\): The exponent \(n=43\) is an odd number.
- Recall the divisibility property: For any positive odd integer \(n\), \(a^n + b^n\) is divisible by \(a+b\).
- Apply the property: Since \(n=43\) is odd, \(41^{43} + 43^{43}\) is divisible by \(41 + 43\).
- Calculate the sum \(a+b\): \(41 + 43 = 84\).
- Conclusion: \(41^{43} + 43^{43}\) is divisible by 84.
- Match with options: 84 is one of the provided options.
Divisibility Property Used
The key property used is: For any positive odd integer \(n\), \(a^n + b^n\) is divisible by \(a+b\). This is a consequence of the factor theorem, which states that if \(P(x)\) is a polynomial, then \(x-c\) is a factor of \(P(x)\) if and only if \(P(c)=0\). Consider the polynomial \(P(x) = x^n + b^n\). If \(n\) is odd, then \(P(-b) = (-b)^n + b^n = -b^n + b^n = 0\). Since \(P(-b) = 0\), \(x - (-b) = x+b\) is a factor of \(x^n + b^n\). Replacing \(x\) with \(a\) shows that \(a+b\) is a factor of \(a^n + b^n\).
| Expression Form | Condition on \(n\) | Divisible By |
|---|---|---|
| \(a^n + b^n\) | \(n\) is odd | \(a+b\) |
| \(a^n - b^n\) | \(n\) is any positive integer | \(a-b\) |
| \(a^n - b^n\) | \(n\) is even | \(a+b\) and \(a-b\) (hence by \(a^2-b^2\)) |
Revision Table: Exponent Divisibility Rules
| Rule | Expression | Condition | Divisor |
|---|---|---|---|
| Sum of Powers | \(a^n + b^n\) | \(n\) is a positive odd integer | \(a+b\) |
| Difference of Powers | \(a^n - b^n\) | \(n\) is a positive integer | \(a-b\) |
| Difference of Powers | \(a^n - b^n\) | \(n\) is a positive even integer | \(a+b\) and \(a-b\) |
Additional Information: Applying Divisibility Rules
Understanding divisibility rules for exponents is crucial for quickly solving problems like this one. These rules are derived from algebraic factorization properties.
- For \(n\) odd, \(a^n + b^n = (a+b)(a^{n-1} - a^{n-2}b + a^{n-3}b^2 - \dots - ab^{n-2} + b^{n-1})\). This clearly shows \(a+b\) as a factor.
- For any positive integer \(n\), \(a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + ab^{n-2} + b^{n-1})\). This shows \(a-b\) as a factor.
- For \(n\) even, let \(n=2k\). Then \(a^n - b^n = a^{2k} - b^{2k} = (a^k)^2 - (b^k)^2 = (a^k - b^k)(a^k + b^k)\). This expression is divisible by \(a-b\) (from the \(a^k-b^k\) term if \(k\) is any integer) and by \(a+b\) (specifically from \(a^k+b^k\) term if \(k\) is odd, or \(a^2+b^2\) related terms if k is even). A simpler way is \(a^n - b^n = (a^{n/2})^2 - (b^{n/2})^2 = (a^{n/2} - b^{n/2})(a^{n/2} + b^{n/2})\). Since \(n\) is even, \(n/2\) is an integer. The term \(a^{n/2} - b^{n/2}\) is divisible by \(a-b\). The term \(a^n - b^n\) can also be factored as \((a+b)(a^{n-1} - a^{n-2}b + \dots + ab^{n-2} - b^{n-1})\) when \(n\) is even, but with alternating signs, so it is divisible by \(a+b\) as well.
In summary, knowing these basic factorization and divisibility rules simplifies problems involving sums and differences of powers significantly.
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