Show that a positive integer is divisible by 3 iff the sum of its digits is divisible by 3.
This is from Rosen but I can't follow the proof in the solutions manual (again). Section 3.6 problem 29.
Please explain the detail so I'll understand the whole process.
Since 10 1 (mod 3) we have a + ... + + (mod 3). Therefore a 0 (mod 3) iff the sum of the digits is congruent to 0 (mod 3).
I don't follow the part we have a sum (mod 3).
Is this because 10 to any power divided by 3 has a 1 remainder? Then a divided by 3 leaves the sum of the digits?
I just now saw this relationship. The solutions manual assumes this is intuitively obvious, I guess.
Let a number be abc
= 100a + 10b + c
= 99a + 9b + a + b + c
99a and 9 b are divisible by 3
so (a + b + c) must be divisible by 3.
so (a + b + c) is sum of digits must be divisible by 3.
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