Question
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2. If \( c \) is any positive integer, which of the following expressions must be an odd integer F. \( \frac{c}{7} \) G. \( c+7 \) H. \( 7 c \) J. \( c^{7} \) (K. \( 7^{c} \) \( 7^{2}=49 \)

Ask by Bob Cummings. in the United States
Feb 27,2025

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Tutor-Verified Answer

Answer

J. \( 7^{c} \) must be an odd integer for any positive integer \( c \).

Solution

To determine which of the given expressions must be an odd integer when \( c \) is any positive integer, let's analyze each option step by step. ### Option F: \( \frac{c}{7} \) - This expression will only be an integer if \( c \) is a multiple of 7. If \( c \) is not a multiple of 7, \( \frac{c}{7} \) will not be an integer. Therefore, it cannot be guaranteed to be an odd integer for all positive integers \( c \). ### Option G: \( c + 7 \) - The sum of an odd integer and an even integer is odd, and the sum of two odd integers is even. Since \( c \) can be either odd or even, \( c + 7 \) can be either odd or even depending on the parity of \( c \). Thus, this expression does not have to be an odd integer. ### Option H: \( 7c \) - The product of an odd integer (7) and any integer \( c \) will always be odd if \( c \) is odd, and even if \( c \) is even. Therefore, \( 7c \) can be either odd or even depending on \( c \). This expression does not have to be an odd integer. ### Option I: \( c^{7} \) - The power of an odd integer is always odd, and the power of an even integer is always even. Since \( c \) can be either odd or even, \( c^{7} \) can also be either odd or even. Thus, this expression does not have to be an odd integer. ### Option J: \( 7^{c} \) - The base 7 is an odd integer. Any power of an odd integer is also odd. Therefore, \( 7^{c} \) will always be an odd integer for any positive integer \( c \). ### Conclusion The only expression that must be an odd integer for any positive integer \( c \) is: **J. \( 7^{c} \)**

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Mind Expander

Let's have some fun with this number game! For expression G, \( c + 7 \), if \( c \) is an odd integer, adding 7 keeps it odd. If \( c \) is even, then \( c + 7 \) becomes odd too. So, G can definitely be odd depending on \( c \)! Moving on to expression H, \( 7c \), since 7 is odd and multiplying it with any positive integer \( c \) (even or odd) gives us an odd product if \( c \) is odd and an even product if \( c \) is even. Thus, H isn’t guaranteed to be odd. Putting it all together, the only expression assured to be odd for all positive integers \( c \) is \( J) c^7 \) because any positive integer raised to an odd exponent stays true to its odd/even nature – if \( c \) is odd, so is \( c^7\), and if \( c \) is even, then \( c^7 \) would still yield an even outcome. So, if c = 1, 2, or 3, both odd and even integers abound in all of our original options!

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