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Find all pairs of positive integers (a, n) such that n ≥ 2 and
 

                     a + (a+1) + (a+2) + ... + (a+n-1) = 100.

 Jan 17, 2019
 #1
avatar+533 
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here is hint: when n is even, you can pair up numbers, when n is odd, find the middle term, because it will be the average.

 

HOPE THIS HELPED!

 Jan 17, 2019
edited by asdf335  Jan 17, 2019
 #2
avatar+106539 
+1

18 + 19 + 20 + 21 + 22 =

 

18 + 22 + 19 + 21 + 20 =

 

40 + 40 + 20 =

 

100

 

So   ( a, n)  =  (18, 5)

 

 

cool cool cool

 Jan 17, 2019
 #3
avatar+107053 
+2

How did you work that out Chris?

Melody  Jan 17, 2019
 #4
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Also, the following: 9 + 10 + 11 + 12 + 13 + 14 + 15 +  16 = 100

 Jan 17, 2019
 #5
avatar+107053 
+2

Are you both doing it by trial and error?  The question does say ALL

 Jan 18, 2019
edited by Melody  Jan 18, 2019
 #6
avatar+28359 
+3

We can write:

 

a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2

 

Hence we have:  na + (n-1)n/2 = 100, which we can rewite as:  n2 + (2a-1)n - 200 = 0 

 

Solve this for n in terms of a: \(n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}\)  (the other solution results in negative values for n).

 

Let a run from 1 to 100 to find the only integer solutions are:  (a,n) = (9,8) and (18,5) as found by Chris and Guest.

 

(There is also a = 100 and n = 1, but the question requires n>=2).

 

(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)

Alan  Jan 18, 2019
edited by Alan  Jan 18, 2019
 #7
avatar+107053 
+2

Thanks Alan :)

I did that, I just didn't think to finish it with a run. 

I was looking for some other way to do it. 

But I guess I could do it in EXCEL easily enough. :/

Melody  Jan 18, 2019

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