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I need to prove that, given integers $a$, $b$, and $c$ such that

$a\mod6=3$, $b\mod8=3$, and $c\mod10=3$ then,

$(a + b + c) \mod6$ is odd.

Here's what I have tried.

By using the definition of mod and quotient remainder theorem, I get

$a = 6x + 3$

$b = 8y + 3$

$c = 10z + 3$

I then consider $a + b + c$:

$a + b + c = 6x + 3 + 8y + 3 + 10z + 3 = 6x + 8y + 10z + 9$

I know that I need to get that in the form $6q + r $ where q is an integer and r is odd. But it doesn't look like I can manipulate that equation to get it to be in the form $6q$ ... unless I'm not seeing something or I need to somehow use the definition of odd somewhere.

Update: I was able to show that $a + b + c$ is odd, by getting this:

$a + b + c = 2(3x + 4y + 5z + 4) + 1$

But now I'm stuck on how I would show that "an odd integer" mod 6 = an odd integer.

user129354
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  • Showing that something is "odd mod 6" is the same as just showing that it is odd. – 2'5 9'2 Feb 17 '14 at 05:16
  • @alex.jordan Ok, I was able to show that a + b + c is odd. But how would I go about showing "odd mod 6" = an odd integer? – user129354 Feb 17 '14 at 05:48
  • You have some number — you want this number modulo 6. You subtract (or add) 6 until you have something between 0 and 5. But when you add or subtract even number from some other number, parity doesn't change. – sas Feb 17 '14 at 06:06

3 Answers3

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Notice that $a+b+c\equiv 2y+4z+3\equiv r\pmod6,$ where $r$ is odd. Hope this helps.

Luitgard
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An integer is "odd mod 6" if and only if it is just plain odd. The same is true for 8 and 10. So you have 3 odd numbers, which add up to an odd number, which is "odd mod 6".

2'5 9'2
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First of all $6x=0\mod 6$

Then, $8y+10z = 2y+4z \mod 6$ and for any $y,z$ — $2y+4z$ is even number. So $2y+4z \mod 6$ is even too.

And of course $9 =3 \mod 6$

sas
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