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 Question from kim, a student: Determine three consecutive odd integers such that twice the sum of the second and third is 29 less than 5 times the first.

Hi Kim,

Every odd number is one more than an even number and hence you can write an odd number as 2n + 1 for some number n. Suppose this is the middle number of the three odd numbers then one of then is 2 more than 2n + 1 and the other is 2 less than 2n + 1. Thus the largest number is 2n + 1 + 2 = 2(n + 1) + 1 and the smallest is 2n + 1 - 2 = 2(n - 1) + 1. Hence the three numbers are

2(n - 1) + 1, 2n + 1 and 2(n + 1) + 1

The question says "twice the sum of the second and third is 29 less than 5 times the first". Twice the sum of the second and third is

2[2n + 1 + 2(n + 1) + 1]

This quantity is 29 less than 5 times the first. Solve for n.

Once you have n find the three numbers 2(n - 1) + 1, 2n + 1 and 2(n + 1) + 1 and check that they satisfy "twice the sum of the second and third is 29 less than 5 times the first".

Penny

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