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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)

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Solution General solution ( D-A = K ) Comment 4 of 4 |
More generally, instead of D - A = 50, let D - A = K 

Then

(1) 2*(A+2n)^2 = (A+n)^2 + (A+K)^2 

Expanding and solving for A gives

(2) A = (K^2 - 7n^2) / (6n - 2K)

The denominator is even, so the numerator must be even, so n must be the same parity (odd or even) as K

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

The numerator is negative if n > K/sqrt(7) ~ K*.378, and the denominator is negative if n < K/3 ~ K*.333, and this is impossible, so they are not both negative.

The numerator is positive if n < K*.378, and the denominator is positive if n > K*.333, and so the only possible n is between K/3 and k/sqrt(7).  Numerically, n must be between K*.333 and K*.378

For instance, instead of K = 50, let K = 176
Then n must be even and between 58.7 and 66.5.
The only possible n are 60, 62, 64, and 66

Substituting in (2) shows that there are actually 3 solutions if K = 176
n = 60 --> (722, 782, 842, 898)
n = 64 --> ( 72, 136, 200, 248)
n = 66 --> ( 11,  77, 143, 187)

Edited on November 6, 2014, 5:09 pm
  Posted by Steve Herman on 2014-11-05 21:23:55

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