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RMS Rigor (Posted on 2014-11-04) Difficulty: 3 of 5
A < B < C < D are four positive integers such that:

(i) B is the arithmetic mean of A and C and:
(ii) C is the rms of B and D, and:
(iii) D-A = 50

Determine all possible quadruplets satisfying the given conditions and prove that there are no others.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution Analytical solution (without quadratic equation) | Comment 3 of 4 |
from (ii),    2*C^2 = B^2 + D^2

Let    B = A + n, where n > 0
Then C = A + 2n
        D = A + 50

Substituting gives

2*(A+2n)^2 = (A+n)^2 + (A+50)^2 

Expanding and solving for A gives

  A = (2500 - 7n^2) / (6n - 100)

The denominator is even, so the numerator must be even, so n must be even.

A is positive, so the numerator and the denominator must be of the same sign.

The numerator is negative if n > 18.9, and the denominator is negative if n < 16.6, and this is impossible, so they are not both negative.

The numerator is positive if n < 18.9, and the denominator is positive if n > 16.6, and we have already said that n must be even, so the only possible n is 18.

This leads to an integral A, as (2500 - 7*18^2)/(6*18-100) =
232/8 = 29.

So the only solution is the one already posted, 
A = 29, n = 18 -->  (29, 47, 65, 79)
  

  Posted by Steve Herman on 2014-11-05 18:18:05
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