Question

the standard addition algorithm below is being used to add three two-digit numbers 4z + 27 + x 5 equals y 14. If x, y, and z each represent a different digit from 0 to 9 what is the value of (x)(y)(z)?

Answer 1

For the units positions of all numbers we have:
$z+7+5=4 \\ z+12=4$
From this we can conclude that total sum of these three numbers is 14. Number 1 we carry to next step. So we have:
$z+12=14 \\ z=2$

For the tens positions of all numbers we have:
$4+2+x+1=1 \\ 7+x=1$
The extra number 1 on left side comes from the carry from last step. Similar to ones position we know that total sum is 11.
$7+x=11 \\ x=4$

Now we insert x and z to find out y:
$42+27+45=y14 \\ 114=y14 \\ y=1$

Now we need to find out the product of these three numbers:
$x*y*z=4*1*2=8$