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.

Sum of x + 1, y-2, and z + 3 is x+y+z+2.
Since x,y and z are consecutive multiples of 3, y=x+3 and z=x+6
x+y+z+2/9= x+x+3+x+6+2/9 =3x +11/9= x/3+11/9
Since x is a multiple of 3, the remainder of the first term will be 0, and the remainder of the second term will be 2.
Therefore option 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

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