It is given that x, y and z are three consecutive multiples of 3.
So lets take x=3, y=6 and z=9.
Option A gives us certain conditions, x+1 = 4, y-2= 4, z+3 = 12. On adding them, we get 4+4+12= 20.
On dividing 20 by 9, we get remainder as 2.
Thus, the answer is C.

You can start by simplifying the summation.

1 - 2 + 3 = 2

Since the consecutive multiples of 3 can be 3,6,9 or 6,9,12 etc..

3 + 6 + 9 = 18 + 2 = 20/9 = 18/9 + 2/9. Here the remainder is 2

Nothing will change because we're basically increasing the values by 9 every time we increase. (since we have 3 multiples) 29/9 = 27/9 + 2/9

38/9 = 36/9 + 2/9 etc.

Since the remainder is 2 from A and it is given as 2 from B, the answer choice is C.

If x, y ,z are consecutive multiples of 3 then y = x + 3 and z = x + 3 + 3 = x + 6

Sum of x+1, y-2 and z+3 = x+y+z+2

Substituting values of y and z interms of x above we get sum = 3x + 11

Remainder when \(\frac{3x + 11 }{ 9}\) = Remainder of \(\frac{3x }{ 9 }\) + \(\frac{11 }{ 9}\).

As x is already a multiple of 3 hence 3x will be exactly divisible by 9. Remainder =0

Remainder of \(\frac{11}{9}\) = 2 , Remainder = 0 + 2

Hence answer is C

x; y; z are 3 consecutive multipliers of 3 ==> y = x+3; z = y+3 = x+6

<==> x + y + z = x + (x+3) + (x+6) = 3x + 9
==> x+1 + y-2 + z+3 = x+y+z +1-2+3 = 3x+9 +2

Dividing to 9, we have
(3x+9 +2)/9 = (3x+9)/9 +2/9 = x/3+ 1 = 2/9

As x is multiplier of 3, so x/3 is integer; therefore, the remainder will be the numerator of the 2/9, which equals 2 and equal the value in volume B

Hence, the answer is C

What happens if the #s are for example:
x = 9, y = 12, z = 15 though?
Then (x+1)(y-2)(z+3) = 10*10*18 = 1800 = 3^2 * 2^3 * 5^2, so it is divisible by 9.
1800/9 = 200, remainder = 0

In that case, then I think the answer would be D - cannot be determined.

Can someone weigh in? Am I wrong?

Given that $\(\mathrm{x}, \mathrm{y}\)$, and \(z\) are three consecutive multiples of 3 such that $\(0<x<y<z\)$.
We can express $x, y$, and $z$ in terms of an integer $n$. Since $x$ is a multiple of 3 and $x>0$, let $x=3 n$ where $n$ is a positive integer ( $\(n \geq 1\)$ ). Since $\(x, y, z\)$ are consecutive multiples of 3 :

$$
\(\begin{gathered}
y=x+3=3 n+3 \\
z=y+3=x+6=3 n+6
\end{gathered}\)
$$


Quantity A is the remainder when the sum of $\(x+1, y-2\)$, and $z+3$ is divided by 9 . Let $S$ be the sum:

$$
\(S=(x+1)+(y-2)+(z+3)\)
$$


Substitute the expressions for $x, y$ and $z$ :

$$
\(\begin{gathered}
S=(3 n+1)+((3 n+3)-2)+((3 n+6)+3) \\
S=(3 n+1)+(3 n+1)+(3 n+9)
\end{gathered}\)
$$


Combine the terms:

$$
\(\begin{gathered}
S=(3 n+3 n+3 n)+(1+1+9) \\
S=9 n+11
\end{gathered}\)
$$


We need to find the remainder when $\(S=9 n+11\)$ is divided by 9 . We can rewrite 11 as $9+2$ :

$$
\(S=9 n+9+2\)
$$


Factor out 9:

$$
\(S=9(n+1)+2\)
$$

Since $n$ is an integer, $n+1$ is also an integer. Thus, $\(9(n+1)\)$ is a multiple of 9 . When $S$ is divided by 9 , the term $9(n+1)$ has a remainder of 0 , and the remainder of $S$ is the remainder of 2 .

Therefore, Quantity $\(\mathbf{A}\)$ is 2 .
Quantity B is 2 .

Comparing Quantity A and Quantity B:

Quantity $\(\mathrm{A}=2\)$

Quantity $\(\mathrm{B}=2\)$
The two quantities are equal.

The correct choice is C .

For this, you might need to take two cases.

When x is 3 and when x is 6, and thereafter solve for both.