# Divisible by 3Watch

#1
Show an integer is divisible by 3 if and only if the sum of its 2 digits is divisible by 3.

How would I go about showing this?

Thanks
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13 years ago
#2
(Original post by alwaysingtogeth)
Show an integer is divisible by 3 if and only if the sum of its 2 digits is divisible by 3.

How would I go about showing this?

Thanks
it actually works for integers of any length (ie. it could be 2 digits or a 1000 digits, it doesn't matter).
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13 years ago
#3
Let us assume the two digits are 'a' and 'b'.

Assuming that 'a + b' is divisible by 3.

a + b = a(10 - 9) + b

a + b = 10a - 9a + b

a + b = -3*3a + 10a + b

10a + b = a + b - 3*3a

Since everything on the right hand side is divisible by 3, '10a + b' must also be divisible by 3. 10a + b gives the original two-digit number.

Hence, when 'a + b' is a multiple of 3, the number 'ab' (not multiplied) is also a multiple of 3.

When a + b is not a multiple of 3, 'ab' will not be a multiple of 3 because 'a + b - 3*3a' does not divide by 3.

Hope this helps,

~~Simba
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13 years ago
#4
oh yeah just remembered, you wanted the proof:

well, the proof for 2 digit numbers would probably go along the following lines:

let us say you have a 2 digit number AB.

you know that this number is 10*A + B.

we can rewrite this as 9*A + (A+B).

we know that 9*A is divisible by 3, because 9 is divisible by 3, and hence the whole lot would be divisible by 3 iff A+B is divisible by 3.

QED.

anyhows, i'm not sure if this proof is right though, but this is how i would approach it personally.
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13 years ago
#5
(Original post by pratikv)
it actually works for integers of any length (ie. it could be 2 digits or a 1000 digits, it doesn't matter).
can you prove this?
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13 years ago
#6
(Original post by mik1w)
can you prove this?
yes, but i can't be bothered. you could probably do it by induction.

i'll just show you the case for 3 digit integers:

if you have a 3 digit number ABC, then we can write it as 100*A + 10*B + C.

we can rewrite this as 99*A + 9*B + (A + B + C).

since 99*A and 9*B are obviously divisible by 3, then so is the number ABC iff A+B+C is divisible by 3.

etc etc.

i'm sure you can show it by induction with a little thought. have a go at it.
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13 years ago
#7
Let N = (a1 + 10a2 + 100a3... + 10k-1ak)
=(a1 + 9a2 + a2 + 99a3 + a3... + (10k-1-1)ak+ ak)
=(9a2+ 99a3... + (10k-1-1)ak) + (a1+a2...+ak)

(10k-1-1)ak is divisible by 3 - can be easily proved by induction (by someone else.)

Therefore (9a2+ 99a3... + (10k-1-1)ak) is divisible by 3.

Therefore, if sum of digits is divisible by 3, so is N
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13 years ago
#8
(Original post by mik1w)
can you prove this?
Any integer n can be written as
n=a+10b+100c+..., where a,b,c,.. are the digits of n read from right to left (so that a is the units, b is the tens, and so on).
Since 10=1(mod 3), 102=1(mod 3), etc...
n=a+b+c+... (mod 3)
In other words, n is divisble by 3 if and only if the sum of its digits is divisible by 3.
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