- #1
RK1992
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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