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.

(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