Difference between revisions of "1999 CEMC Gauss (Grade 7) Problems/Problem 5"

Latest revision as of 13:35, 15 September 2025

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-== Problem==
+#Redirect [[1999 CEMC Gauss (Grade 8) Problems/Problem 3]]
-Which one of the following gives an odd integer?
-<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>
-==Solution 1==
-Evaluating all of the answer choices, we get:
-<math>6^2 = 36</math>
-<math>23 - 17 = 6</math>
-<math>9 \times 24 = 216</math>
-<math>96 \div 8 = 12</math>
-<math>9 \times 41 = 369</math>
-The only odd number from the list is <math>\boxed {\textbf{(E)} 9 \times 41}</math>.
-==Solution 2==
-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.
-<math>9 \times 41</math> involves the multiplication of two odd numbers, meaning that it must be the odd number.
-Thus, the answer is <math>\boxed {\textbf{(E)} 9 \times 41}</math>.