In the addition shown, each of $X, Y$ and $Z$ represents a digit. What is the value of $X+Y+Z$ ?
(A) 10
(B) 15
(C) 22
(D) 20
(E) 8

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-039.jpg?height=203&width=336&top_left_y=2381&top_left_x=1382)
First, we consider the column of units (ones) digits.

From this column, we see that the units digit of $Z+Z+Z$ (or $3 Z$ ) must be 5 .

By trying the possible digits from 0 to 9 , we find that $Z$ must equal 5 .

When $Z=5$, we get $3 Z=15$, and so there is a "carry" of 1 to the column of tens digits.

From this column, we see that the units digit of $1+Y+Y+Y($ or $3 Y+1)$ must be 7 .

By trying the possible digits from 0 to 9 , we find that $Y$ must equal 2 .

There is no carry created when $Y=2$.

Looking at the remaining digits, we see that $2 X=16$ and so $X=8$.

Checking, if $X=8$ and $Y=2$ and $Z=5$, we obtain $825+825+25$ which equals 1675 , as required.

Therefore, $X+Y+Z=8+2+5=15$.

ANSWER: (B)