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| Math Central | Quandaries & Queries |
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Question from daniel, a parent: find three consecutive odd integers such that three times the second minus the third is 11 more than the first. |
Hi Daniel,
You're close! The difficulty with your three integers x + 1, x + 3, x + 5 is that you need x to be even. If x is odd, say x = 3, then you get x + 1 = 4, x + 3 = 6 and x + 5 = 8. An even number is a multiple of 2 so you can force x to be even by writing it 2 times something, for example 2n where n is some integer. I would take the three consecutive odd integers to be 2n + 1, 2n + 3 and 2n + 5.
Can you complete the problem now? If you have difficulties write back.
Penny
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