Prove Sum of Digits Divisible by 9 iff Num is Div by 9

In summary, the conversation discusses proving that the sum of a number's digits is a multiple of 9 if and only if it is divisible by 9. The participant proposes a proof using modular arithmetic, breaking down the number into its digits and showing that if the sum of the digits is divisible by 9, then the number itself is also divisible by 9. The other participant agrees that the proof is sound and that more rigour would be unnecessary.
  • #1
RK1992
89
0

Homework Statement


prove that the sum of a number's digits is a multiple of 9 if and only if it is divisible by 9


Homework Equations


none that i can think of


The Attempt at a Solution



say any real integer can be written as:
anan-1...a2a1a0
where a0 is the units, a1 is the tens digit and so on.

if we have the number x=anan-1...a2a1a0

we can say that x (mod 9) = 0 [tex]\Leftrightarrow[/tex] x divisible by 9

rewriting the number x:
anan-1...a2a1a0 = anan-1...a2a10 + a0 = anan-1...a200 + a0 + 10a1

x= 10nan + 10n-1an-1 + ... +100a2 + 10a1 + a0

we can then break up the so 10a1 = 9a1 + a1
and likewise the 100a2 = 99a2 + a2

however, working mod 9, 9a1=99a2 = 0
so we can rewrite x (mod 9) as:
x (mod 9) = an + an-1 + ... + a2 + a1 + a0

as we said that:
x (mod 9) = 0 [tex]\Leftrightarrow[/tex] x divisible by 9

we can say that:
an + an-1 + ... + a2 + a1 + a0 = 0 (mod 9)
but 0 (mod 9) = 9k
so if:
x = anan-1...a2a1a0
is divisible by 9
an + an-1 + ... + a2 + a1 + a0 = 9k


is this sound reasoning?

apologies if I've messed up with subscripts anywhere, i did check but its hurting my eyes :p

thanks
 
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  • #2


This sounds like a good proof!
 
  • #3


okay thanks :) i know its not rigourous but i think its watertight enough for the level I am working at
 
  • #4


I think the proof IS rigourous enough. Much more rigour would be too much, in my opinion :smile:
 

1. How does one prove that the sum of digits of a number is divisible by 9?

The proof relies on the fact that any number can be written in expanded form as a sum of its digits multiplied by powers of 10. For example, 456 can be written as 4*100 + 5*10 + 6*1. Since 100, 10, and 1 are all divisible by 9, the sum of digits will also be divisible by 9.

2. Is the converse true? If a number is divisible by 9, does that mean the sum of its digits is also divisible by 9?

Yes, the converse is also true. Any number that is divisible by 9 can be written as a sum of its digits multiplied by powers of 10. Therefore, the sum of digits will also be divisible by 9.

3. How does this relate to divisibility by 3?

This proof also applies to divisibility by 3. Since 3 is a factor of 9, any number that is divisible by 9 will also be divisible by 3. Therefore, the sum of digits will also be divisible by 3.

4. Can this proof be extended to other numbers besides 9?

Yes, this proof can be extended to any number that is a factor of 10, since the expanded form of a number will always involve powers of 10. For example, to prove that a number is divisible by 4, one can show that the last two digits of the number are divisible by 4.

5. Does this proof have any applications in real-life situations?

Yes, this proof has applications in various fields such as mathematics, computer science, and cryptography. For example, in computer science, this property is used in algorithms for error detection. In cryptography, it is used in the creation of check digits to ensure the accuracy of numerical data.

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