MHB Can k be equal to 4 in the equation k + 2 = 3^(n)?

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In the equation k + 2 = 3^(n), where n is a positive integer, k cannot equal 4 because it results in k + 2 equaling 6, which is not a power of 3. The discussion highlights that since 3 raised to any integer power is odd, k must also be odd to maintain the equation's integrity. Therefore, options like 1, 7, 25, and 79 are valid values for k, while 4 is not. The reasoning presented confirms that k = 4 is indeed an invalid solution. The conclusion emphasizes the importance of the parity of k in relation to the powers of 3.
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If n is a positive integer and k + 2 = 3^(n), which of the following could NOT be a value of k?

A. 1
B. 4
C. 7
D. 25
E. 79

I say the answer is B or 4.

When k = 4, k + 2 becomes 6.
However, there is no power we can raised 3 to that will yield 6.

Is my reasoning correct?
 
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RTCNTC said:
If n is a positive integer and k + 2 = 3^(n), which of the following could NOT be a value of k?

A. 1
B. 4
C. 7
D. 25
E. 79

I say the answer is B or 4.

When k = 4, k + 2 becomes 6.
However, there is no power we can raised 3 to that will yield 6.

Is my reasoning correct?
Looks good to me!

-Dan
 
Or since 3^(n) is odd, then k can't be 4.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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