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Sum from Arithmetic and Geometric (Posted on 2016-03-24) Difficulty: 3 of 5
Each of A, B and C is a positive integer such that:
20*A, 6*B and C are in arithmetic sequence, and:
20*A, 6*B and C+1 are in geometric sequence

Find the six smallest values of A+B+C

No Solution Yet Submitted by K Sengupta    
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Solution Possible Solution | Comment 1 of 5
Call the difference d so 20A=6B-d, C=6B+d
Call the ratio r so that 20A=6B/r, C+1=6Br
Equating the 20A terms gives
6B=dr/(r-1)
Equating the C terms gives
6B=(d+1)/(r-1)
Equating these 6B terms gives
r=(d+1)/d
so r is determined by d but so are A,B,C as
A=d²/20
B=d(d+1)/6
C=d(d+2)
C+1=(d+1)²
A will only be an integer if d is a multiple of 10.
B will only be an integer if d is a multiple of 3 or 1 less than a multiple of 3.
So d must be 20, 30, 50, 60, 80, 90, etc...
and since A+B+C=(73d²+130d)/60 it's clear the smallest 6 valid values of d will give the 6 smallest values of A+B+C
 d  20A   6B    C  C+1 A+B+C
20 400 420 440 441 530
30 900 930 960 961 1160
50 2500 2550 2600 2601 3150
60 3600 3660 3720 3721 4510
80 6400 6480 6560 6561 7960
90 8100 8190 8280 8281 10050
-10 100 90 80 81 100
-30 900 870 840 841 1030
-40 1600 1560 1520 1521 1860
-60 3600 3540 3480 3481 4250
Edit:  I made the error of assuming d had to be positive.  There are other small solutions with negative d as appended to the table.  Bolding changed to the actual 6 smallest.  I also fixed a small typo.



Edited on March 24, 2016, 8:17 pm
  Posted by Jer on 2016-03-24 13:22:49

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