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Numbers Divisible by 3
Any number is divisible by 3 if the sum of the digits is divisible by 3. For example 123 is 1+2+3 = 9 so it is
divisible by 3.
Proof:
Two Digit Proof:
Let n be the number, a and b the digits
10a b
The sum of the digits is c
b c
Re-arrange the sum of the digits
c a
Substitute into the first equation
10a c a
Simplify
9a c
Given a b %3 0 , or the sum of the digits is divisible by 3, c is divisible by 3
Both 9a and c are divisible by 3, allowing a 3 to be factored out, guaranteeing n%3 0
n
a
b
n
n
Three Digit Proof:
n
a
c
n
n
100a
b c
d b
100a
99a
Given a
Let n be the number, a, b, and c the digits
10b c
d The sum of the digits is d
a Re-arrange the sum of the digits
Substitute into the first equation
10b d b a
Simplify
9b d
b c %3 0 , or the sum of the digits is divisible by 3, d is divisible by 3
99a, 9b, and d are divisible by 3, allowing a 3 to be factored out, guaranteeing n%3
X Digit Proof:
digit0 is the least significant digit, digit x is the most significant digit
x
digiti 10i
n
Let n be the number and digiti the digits
i 0
x
s
The sum of the digits is s
digiti
i 0
x
digit0
s
digiti
Re-arrange the sum of the digits
i 1
x
n
digiti 10i
digit0100
digiti 10i
digit0
digiti 10i
s
Re-write the first equation
i 1
x
n
i 1
x
n
Simplify
x
i 1
digiti
Substitute the re-arranged sum of the digits
i 1
x
digiti 10i
n
i 1
digiti
s Combine the summations
0
x
digiti 10i 1
n
s
Factor
i 1
i 10i 1 %3 0
Need to prove
10i 1
10i
1 1
10i
1 1
10i
1
1
1 1
10i
1 1
1
9
9
9
Add and subtract 1 from the exponent
1
Multiply and divide by 9
9
10i 1 1 1
Re-write the bottom 9 as (10 – 1)
10 1
9
i 1
10i 1 1 1
9
9 10 z
10 1
z 0
x
Re-write equation as a summation
i 1
n
digiti 9
i 1
Replace original equation with summation
10 z
s
Pull the 9 out of the summation
i 1
digiti
i 1
s
z 0
x
n 9
10 z
z 0
9 and s are divisible by 3, allowing a 3 to be factored out, guaranteeing n%3
0