X, Y and Z

First, we consider the column of units (ones) digits.

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

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

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

column, we see that the units digit of $1+Y +Y +Y ( 3Y +1)$ 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 $2X = 16$ 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, $X +Y +Z = 8+2+5 = 15.$

First let us take the first column and as Z + Z + Z =3 Z and the units digit of the number is 5 and hence 3 Z should be equal to a multiple of 5 and as Z is a one digit number ,from 0 to 9 we find that Z is 5 . Now in the tens digit 3 Y+ 1 should end with the digit 7 and as Y is a one digit number from 0 to 9 ,we find that Y = 2 and now 2 X = 16 and X is a one digit number from 0 to 9 and we find that X = 8. Hence X + Y + Z = 8 + 2 +5 = 15.

3Z = 15 Z = 5 / 7-1 = 6 3Y = 6 Y = 2 / 2X = 16 X = 8 / X + Y + Z = 5 + 2 + 8 = 15

If X=8, Y=2 Z=5

8+2+5= 15

825+825+25=1675

$3Z =$ a number ending in $5$ . $Z$ must be $5$ . $3 \times 5 = 15$ . The $1$ carries over to the tens.

$30Y + 10 = 70 \Longrightarrow Y = 2$

$NOTE:$ $30Y + 10 = 70$ because $10Y + 10Y + 10Y + 10$ (carried over from $3 \times 5 = 15$ ) equals $7 \times 10$ . $Y$ and $7$ are multiplied by $10$ because they are in the tens place. In addition, $Y$ must be an integer $< 10$ because each place in a number's decimal representation can only be a one-digit integer.

$30Y + 10 = 70 \Longrightarrow Y = 2 < 10$

$30Y + 10 = 170 \Longrightarrow Y = \frac{16}{3} < 10$ but does not belong to $\mathbb{Z}$

$30Y + 10 = 270 \Longrightarrow Y = \frac{26}{3} < 10$ but does not belong to $\mathbb{Z}$

$30Y + 10 = 370 \Longrightarrow Y = 12 > 10$

Thus the expression $30Y + 10 = 70$ can only equal $70$ .

$200X = 1600 \Longrightarrow X = 8$

$X + Y + Z = 8 + 2 + 5 = \boxed{15}$

$According \space to \space above \space cryptogram \space,$

$2 \space ( \space 100X \space+ \space10Y \space+\space Z) \space + \space10Y \space+ \space Z \space= \space1675$

$\rightarrow\space 200X \space+ \space 30Y \space +\space3Z\space= \space 200 \times 8 \space+ \space30 \times 2 \space+ \space 3 \times 5$

$Comparing \space L.H.S. \space and \space R.H.S.$

$\Rightarrow \space X \space= \space8 \space, \space Y \space= \space2 \space, \space Z \space= \space5$

$\therefore \space X \space+ \space Y \space+ \space Z \space= \space 8 \space+ \space2 \space+ \space 5 \space=\boxed{15}$