I\'m not looking for a solution, just something to nudge me in the right direction. I\'m not sure how to proceed - I\'ve been trying to work it out for the past few hours but I have not made any progress. I know I should have to make use of gcd / Euclid\'s algorithm.
From Introduction to Mathematical Thinking (Gilbert & Vanstone), pg.50, exercise 98
If a and b are odd positive integers, and the sum of the integers, less than a and greater than b , is 1000, then find a and b.
Many thanks.
Solution
The average of the integers between a and b is (a+b)/2 and there are a-b-1 of them, so the sum of the integers between a and b is equal to (a+b)(a-b-1)/2. Therefore, (a+b)(a-b-1)/2 = 1000, or (a+b)(a-b-1) = 2000. Knowing that a and b are odd means that a+b must be even, and a-b-1 must be odd. Hopefully from there you should be able to solve it. (I don\'t see how you would use GCD or the Euclidean Algorithm to do this, however.)
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I-'m not looking for a solution- just something to nudge me in the rig.docx
1. I'm not looking for a solution, just something to nudge me in the right direction. I'm not sure
how to proceed - I've been trying to work it out for the past few hours but I have not made any
progress. I know I should have to make use of gcd / Euclid's algorithm.
From Introduction to Mathematical Thinking (Gilbert & Vanstone), pg.50, exercise 98
If a and b are odd positive integers, and the sum of the integers, less than a and greater than b , is
1000, then find a and b.
Many thanks.
Solution
The average of the integers between a and b is (a+b)/2 and there are a-b-1 of them, so the sum of
the integers between a and b is equal to (a+b)(a-b-1)/2. Therefore, (a+b)(a-b-1)/2 = 1000, or
(a+b)(a-b-1) = 2000. Knowing that a and b are odd means that a+b must be even, and a-b-1 must
be odd. Hopefully from there you should be able to solve it. (I don't see how you would use
GCD or the Euclidean Algorithm to do this, however.)
