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$E$ is a three digit number so

$$\begin{align*}E &= 100a+10b+c\\ &= 99a+a+9b+b+c\\ &=(99a+9b)+(a+b+c)\\ &=3(33a+3b)+(a+b+c) \end{align*}$$

If $E$ is divisible by $3$ then the sum of its digits must also be divisible by $3$.

How does the last equation prove this?

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"How does the last equation prove this?"

I number $n$ divides $m*n + k$ if and only if $n$ divides $k$. If $n$ divides $k$ than it will divide $m*n + k$ because that's a multiple of $n$ more. As if $n$ doesn't divide $k$ then $n$ won't divide $m*n + k$ because .... that's a multiple of $n$ more.

So, $3$ divides $33(33a + b)$ and so $3$ divides $33(33a+b) + (a+b+c)$ if and only if $3$ divides $a+b+c$.

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Hint: For integers $x,r,s$, if $x = r + s$ and $x,r$ are divisible by 3, then $s$ has to be divisible by 3.

Edit: I should also note that if you replace $3$ by any non-zero integer, the statement still holds.

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We know that $$E=3(33a+3b)+(a+b+c)=3k+(a+b+c)$$

Since 3k is divisible by $3$, the only way that $E$ is divisible by $3$ is that $(a+b+c)$ be divisible by $3.$

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E=3(33a+3b)+(a+b+c)mod3

or E=0(0)+(a+b+c)mod3

or E=(a+b+c)mod3

So if E= 0mod3 then

(a+b+c)= 0mod3

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We have found $\dfrac E3 ~{= \dfrac{3(33a+b)+(a+b+c)}{3}\\ = (33a+b)+\dfrac{(a+b+c)}{3}}$

So: $E/3$ is an integer iff $(a+b+c)/3$ is an integer.

That is: $3\mid E~$ iff $~3\mid(a+b+c)$.

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