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 $Question from Lucy, a student: Hello, I have a question about consecutive numbers. So lets say a question tells you to find what three consecutive numbers are that add up to a certain number. To calculate this, would you use 2x+1, 2x+3, 2x+5 or just x, x+2, x+4. Also, the same question for even numbers. Would you use: 2x, 2x+2, 2x+4 or just x, x+2, x+4. My math book is confusing because it uses the first strategy(the 2x+…) in some situations and the second strategy(x+…) in other situations. Thus, can you provide me an explanation of why you would use the strategy that is correct? Specifically, my book answered this question: If “n” is the first of five consecutive odd numbers, what is their average, using n,n+2….=5n+20 and average is n+4(this is the answer),. For the question: the sum of 4 consecutive odd integers must be: Divisible by 8 but not necessarily 16(this is the answer), it used the strategy 2k+1, 2k+3…=8k+16. Thank you so much for taking the time to answer my question. Hi Lucy, The two examples you sent could be solved either way. For the first question, suppose it were worded "What is the average of five consecutive odd integers?" It could be answered as in your text but you could also say Let the smallest of the odd integers be$2k + 1$then the consecutive integers are$2k + 1, 2k + 3, 2k + 5, 2k + 7 \mbox{ and } 2k + 9.$the average is then$\large \frac{10k + 25}{5} \normalsize = 2k + 5.$Our answers look different but they are really the same, the answer is "4 more than the smallest of the 5 integers". For the second question you could say Let$n$be the smallest of the 4 consecutive odd integers then their sum is$n + (n + 2) + (n + 4) + (n + 6) = 4n + 12.$It's not clear that this is divisible by 8 but I haven't used the fact that$n$is odd. Since$n$is odd there is an integer$k$so that$n = 2k + 1$and hence the sum is$4n + 12 = 4(2k + 1) + 12 = 8k + 16.$Now it is clear that the sum is divisible by 8, but if$k = 1$is is not divisible by 16. Which strategy would I use? If I am to consider an odd integer I would write it as$2k + 1$so the fact that it is odd is contained in the algebra and not just in my memory. Likewise an even integer I would write$2k.\$

I hope this helps,
Harley

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