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Divisible by 28 (Posted on 2014-12-21) Difficulty: 3 of 5
Each of a, b and c is a positive integer with a ≤ b ≤ c ≤ 70.

Find total number of triplets (a, b, c) such that a2 + b2 + c2 is divisible by 28.

No Solution Yet Submitted by K Sengupta    
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re: possible solution | Comment 2 of 4 |
(In reply to possible solution by Ady TZIDON)

For the numbers whose mod 28 representation is (4,8,16), the calculation of the number of choices is correct. 103 = 1000. There are 10 different numbers mod 28 ≅ 4 × 10 different numbers mod 28 ≅ 8 × 10 different numbers mod 28 ≅ 16.

For the numbers whose mod 28 representation is (0,0,0), the calculation is in error. Though there are 5 numbers corresponding to mod 28 ≅0, the calculation is not 53 = 125 for the number of combinations. As 3 of these numbers must be chosen from the set of 5, and as repetition is allowed, the number of combinations is given by (5 + 3 - 1)!/(3!×(5 - 3)!) = 35.

Thus, the total number of possible solutions is 1000 + 35 = 1035.

Nevertheless, I found your analysis of determining the number of different numbers mod 28 brilliant.

*Corrected, as the comparison operators also include equal in addition to the less than. (Thanks, Charlie).

Edited on December 21, 2014, 10:14 pm
  Posted by Dej Mar on 2014-12-21 18:08:34

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