Bunuel
For positive integers m and n, is \(7+7^2+7^3+….+7^{mn}\) divisible by 14?
(1) \(n=even\)
(2) \(m = \frac{a}{b}\), where one of a and b is an odd and the other is an even
\(7+7^2+7^3+….+7^{mn}\)
\(7(1+7^1+7^2+….+7^{mn-1})\)
\(7(1+7^1+7^2+….+7^{mn-1})\) is a multiple of 7, hence \(7+7^2+7^3+….+7^{mn}\) is divisible by 14 if \((1+7^1+7^2+….+7^{mn-1})\) is even.
Hence, the question can be reframed as
Is \(1+7^1+7^2+….+7^{mn-1}\) even ?Let's write a few terms to understand the pattern
\(1+7^1+7^2+7^3\)= odd + odd + odd + odd = even
\(1+7^1+7^2+7^3+7^4+7^5\)even+odd+odd = even
\(1+7^1+7^2+7^3+7^4+7^5+7^6+7^7\)even+odd+odd = even
Inference:
- Whenever the value of the exponent of 7 is odd, the value of the sum is even.
- Whenever the value of the exponent of 7 is even, the value of the sum is odd.
Statement 1(1) \(n=even\)As n is even, mn-1 is odd.
Hence the sum \(1+7^1+7^2+….+7^{mn-1}\) will result in an even value.
Therefore, we can conclude that \(7+7^2+7^3+….+7^{mn}\) is divisible by 14.
The statement alone is sufficient. Eliminate B, C, and E.
Statement 2(2) \(m = \frac{a}{b}\), where one of a and b is an odd and the other is an evenVery cleverly crafted statement
We are given that \(m = \frac{a}{b}\) and one of a and b is an odd and the other is an even
Therefore \(m = \frac{\text{odd}}{\text{even}}\) or \(m = \frac{\text{even}}{\text{odd}}\).
The question premise states that m is an integer. Hence, m cannot be of the form
\(m = \frac{\text{odd}}{\text{even}}\)
We can conclude that \(m = \frac{\text{even}}{\text{odd}}\). The result of this will always be even.
Hence, m = even.
If m = even, mn-1 is odd. Now the solution is similar to that of Statement 1.
From our previous work, we know that when mn-1 = odd, the sum \(1+7^1+7^2+….+7^{mn-1}\) will result in an even value and the sum \(7+7^2+7^3+….+7^{mn}\) is divisible by 14.
Statement 2 is also sufficient to answer the target question.
Option D