Respuesta :
Answer:
Proof is shown in the Explanation
Step-by-step explanation:
Suppose that you have a three-digit number natural number n that is written xyz. Then
[tex]n=10^2x+10y+z\\=(99+1)x+(9+1)y+z\\=(99x+9y)+(x+y+z)\\=3(33x+3y)+(x+y+z)[/tex]
So when we divide n by 3, we obtain:
[tex]\frac{n}{3} =\frac{3(33x+3y)+(x+y+z)}{3} \\=33x+3y+\frac{x+y+z}{3}[/tex]
The remainder is clearly going to come from the division [tex]\frac{x+y+z}{3}[/tex], since 33x+3y is an integer.
Since 3 divides n, the remainder is going to be a multiple of 3 and so we have shown that 3|(xyz).
Example:
Consider the number 336
[tex]336=(100X3)+(10X3)+6\\=(99+1)3+(9+1)3+6=[99*3+9*3]+3+3+6\\=3(99+9)+12\\\frac{336}{3} =99+9 +\frac{12}{3} \\=99+9+4\\=112[/tex]
