Consider three prime numbers A, B and C with A ≤ B ≤ C ≤ 20, and determine the total number of triplets (A, B, C) such that the respective last digits of A+B+C and A*B*C are the same. No computer programs!

The are 39 triplets (A, B, C), such that A <= B <= C <= 20 and each of A, B and C are primes, and the terminal digit of A+B+C equals the terminal digit of A*B*C.
The triplets are:

( 2, 3, 3), ( 2, 3, 5), ( 2, 3, 7), ( 2, 3,11),
( 2, 3,13), ( 2, 3,17), ( 2, 3,19), ( 2, 5,13),
( 2, 7,13), ( 2,11,13), ( 2,13,13), ( 2,13,17), ( 2,13,19), ( 3, 3, 7), ( 3, 3,17), ( 3, 5, 7),
( 3, 5,17), ( 3, 7, 7), ( 3, 7,11), ( 3, 7,13),
( 3, 7,17), ( 3, 7,19), ( 3,11,17), ( 3,13,17), ( 3,17,17), ( 3,17,19), ( 5, 5 ,5), ( 5, 7,13), ( 5,11,19), ( 5,13,17), ( 7, 7,13), ( 7,11,13), ( 7,13,13), ( 7,13,17), ( 7,13,19), (11,13,17), (13,13,17), (13,17,17), (13,17,19)

Posted by Dej Mar on 2011-11-14 13:49:28