Scrambled Additions

In the ten's place, we have $Y+Z=Z$ , this means that $Y=0$ or $Y=9$ . (9 being the case of having a carry forward from the unit's place).

However, because the result is three digit number, $Y=9$ .

From this, we can say that $X=Z-1$ (seeing the units place)

and also $X=4$ (observing the hundred's place) and finally $Z=5$ .

Which gives $\overline{XYZ}=\boxed{495}$ .

My solution is not going to be very well formated but here is my solution. XYZ + XZY = YZX XYZ = 100X + 10Y + Z XZY = 100X + 10Z + Y YZX = 100Y + 10Z + X Simplifying produces 200X + 11Y + 11Z = 100Y + 10Z + X Bring all the terms to one side 199X + Z - 89Y I don't know if there is a way to directly figure it out from here but after placing numbers in the equation 199x4=796 X=4 Z=5 89x9=801 Y=9 Placing them back in the problem gives you: 459+495=954 the correct answer!

I am never good at cracking cryptograms so I did it this way:

We have,

$\overline { XYZ } =100X+10Y+Z\quad \rightarrow (1)\\ \overline { XZY } =100X+10Z+Y\quad \rightarrow (2)\\ \overline { YZX } =100Y+10Z+X\quad \rightarrow (3)\\ \overline { XYZ } +\overline { XZY } =\overline { YZX } \quad \quad \rightarrow (4)$

Substituting equations $(1)$ , $(2)$ and $(3)$ into equation $(4)$ ,

$\quad \quad 100X+10Y+Z+100X+10Z+Y=100Y+10Z+X\\ \rightarrow 199X-89Y+Z\quad \quad \quad \quad \quad \quad \quad \quad \quad =0\\ \rightarrow 199X\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =89Y-Z \quad \quad \rightarrow (5)$

$X$ , $Y$ and $Z$ must be single digit integers.

As they are all distinct positive single digit integers, $89Y-Z$ would be slightly bigger than $199X$ .

By logic and a bit of arithmetic, the value of $Y$ would be more than $\frac { 199+0 }{ 89 } =\frac { 200 }{ 89 } \approx 2.23$ times but less than $\frac { 199+10 }{ 89 } =\frac { 209 }{ 89 } \approx 2.35$ times the value of $X$ .

In other words, $\frac { 199 }{ 89 } \le \frac { Y }{ X } <\frac { 209 }{ 89 }$ .

The only pair of single digit integers where $\frac { 199 }{ 89 } \le \frac { Y }{ X } <\frac { 209 }{ 89 }$ is when $(X,Y)$ is equivalent to $4,9$ . Thus we can conclude $X=4$ and $Y=9$ .

Substituting $X=4$ and $Y=9$ into equation $(5)$ ,

$\quad \quad 199X\quad \quad \quad \quad =89Y-Z\\ \rightarrow 199(4)\quad \quad \quad \quad =89(9)-Z\\ \rightarrow Z=501-496\quad =5$

Hence,

$\overline { XYZ } =\boxed { 495 }$

Consider that X not equal to Z Z+Y = X+10 --------------------- (1) Z+Y+1 = Z+10 --------------------- (2) X+X+1 = Y

2X+1 = Y --------------------- (3) Z+2X+1 = X+10

X+Z = 9 ---------------------- (4)

Substituting (1)-(2), X-Z = -1 ---------------------- (5) 2X = 8 X = 4 Substituting X=4 in (3) Y = 9 Substituing X=4 in (4) Z = 9-4 = 5

495+459=954

• (A) z+y=x+CO(A) x cannot be 0; x, y, and z cannot be > 9; x cannot = y cannot = z
• (B) y+z+CO(A)=z+CO(B) y and z cannot be > 9, y cannot = z
• (C) x+x+CO(B)=y x cannot be 0; x and y cannot be > 9, x cannot = y
• Any carryovers (CO) are equal to 100, 10, or 0 depending on positions. If CO(B) exists it would be 10 x CO(A) if it exists or they coexist.
• I left out the math for the conversions.
• From (C) x=(y-(CO(B))/2
• From (B) y+z+CO(A)=z+CO(B)
• y=CO(B) - CO(A)
• CO(A) can be 1 or 0 and CO(B) can be 10 or 0 for (B)
• therefore y can equal 10,9,0,and -1
• 10 can be removed because y cannot be > 9
• 0 can be removed because x=(0-0)/2 and x cannot equal 0
• -1 can be removed because x=(-1-0)/2=-1/2; x has to be a whole number
• So, due to positions, CO(A) = 10 for (A), CO(B) = 10 and CO(A) = 1 for (B), and CO(B) = 1 for (C) and
• y=9;
• x=(9-1)/2=4;
• z=4-9+10=5;
• So, 495.

• Z+Y=X+10 eq1
• Y+Z+1=Z+10 eq2 then Y= (9) eq3
• 2X+1=Y eq4
• from 3,4 2X+1=9 then x=4 eq5
• from 3,5 Z+9=4+10
• Z=5
• XYZ = 495

From the (1) column we know that Z+Y = X, from the (2) column we know that Z+Y = Z. This is not possibile for hypothesis, unless we admit that Z+Y>10 so we have carry-over and X={Z-1, Y-1} o X=min{Z,Y}-1. Since the same sum appears in the (2) column, we have the carry-over also for this column, and, because the result under column (3) have one digit, it means that X≤4. So we have that 10<Z+Y≤14. We have some two possible set {Z, Y, X} = {5, 9, 4}, {9, 5, 4}. All the other set must be excluded because don't respect the rules. In particular we must remove also the second set, because if we have two number, x and y, the sum x+y≤17, so we can't have 9. This means that the correct choice is {Z, Y, X} = {5, 9, 4}

Well, we know that X must be < 5, since the answer is a 3-digit number.

We also know that Y has to equal either 0 or 9, since Y + Z = Z.

However, since X + X = Y, and X =/= 0, Y cannot be 0 (and therefore must be 9).

Now that we know Y is 9, X + X must equal 9. This is only possible with a carry from the tens place (4 + 4 (+1) = 9). Therefore X must equal 4.

For Z, we know that Z + 9 = 4. Z cannot be -5, and therefore must be 5 (5 + 9 = 4 with a carry of 1). This also works with the tens place.

Therefore XYZ = 495.

X must be smaller than 5. At each step assume x=i with i=1,..,4. For each x fixed, y could be 2x or 2x+1. For each value of y fixed, z is obtained looking at the units' column. Iterating this procedure at the step i=4 we have that y could be 8 or 9. For y=9 results z=5, finally 495