Show that pos int divisible by 3 iff sum of digits divisible by 3

**oldguynewstudent** · November 18th 2009, 07:48 PM

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.

**Drexel28** · November 18th 2009, 07:51 PM

Originally Posted by oldguynewstudent

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.

What part of the proof eludes you?

**Sampras** · November 18th 2009, 08:17 PM

Suppose $n = a_0 + a_{1}10+ a_{2}10^{2} + \cdots + a_{n} 10^n$ .

Consider $s = a_0 + \cdots a_n$ and look at n-s.

**oldguynewstudent** · November 18th 2009, 08:42 PM

Originally Posted by Drexel28

What part of the proof eludes you?

Since 10 $\equiv$ 1 (mod 3) we have a $\equiv$ $a_n$ $_-$ $_1$ + ... + $a_1$ + $a_0$ (mod 3). Therefore a $\equiv$ 0 (mod 3) iff the sum of the digits is congruent to 0 (mod 3).

I don't follow the part we have a $\equiv$ 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.

**gmatt** · November 19th 2009, 04:20 PM

Let $x = \sum _{i=0} ^n a_i 10^i$ .

Suppose $3 | x$ then $0 \equiv x \equiv \sum _{i=0} ^n a_i {10} ^n \equiv \sum _{i=0} ^n a_i {1} ^n \bmod{3}$

**pspss** · February 20th 2013, 09:32 AM

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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