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| == Problem==
| | #Redirect [[1999 CEMC Gauss (Grade 8) Problems/Problem 3]] |
| Which one of the following gives an odd integer?
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| <math>\text{(A)}\ 6^2 \qquad \text{(B)}\ 23-17 \qquad \text{(C)}\ 9\times 24 \qquad \text{(D)}\ 96\div 8 \qquad \text{(E)}\ 9\times 41 </math>
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| ==Solution 1==
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| Evaluating all of the answer choices, we get:
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| <math>6^2 = 36</math>
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| <math>23 - 17 = 6</math>
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| <math>9 \times 24 = 216</math>
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| <math>96 \div 8 = 12</math>
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| <math>9 \times 41 = 369</math>
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| The only odd number from the list is <math>\boxed {\textbf{(E)} 9 \times 41}</math>.
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| ==Solution 2==
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| Without evaluating the answers, we can see that <math>6^2</math> is the square of an even number, <math>9 \times 24</math> involves multiplication with an even number, and <math>23 - 17</math> involves two odd numbers, so those are even. This means that we can eliminate those answers.
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| <math>9 \times 41</math> involves the multiplication of two odd numbers, meaning that it must be the odd number.
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| Thus, the answer is <math>\boxed {\textbf{(E)} 9 \times 41}</math>.
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