As easy as XYZ

In the ones digit, we know $x + y = 10$ since the three digits add to $z.$ Carry the $1$ into the tens digit, and we have $x + y + z + 1 = \overline{xy} = 10x + y.$ (see end of proof for clarification) Therefore $9x = z + 1.$

Since $9x = z + 1$ and we're dealing only with natural numbers, $z + 1$ is a multiple of $9.$ The only natural number $z \leq 9$ That satisfies that condition is $z =8.$ This is enough to solve the problem: $z = 8 \\ x = \frac{z + 1}{9} = \frac{9}{9} = 1 \\ y = 10 - x = 10 - 1 = 9 \\ \overline{xyz} = \boxed{198}$

(clarification: $x + y + z + 1$ is the sum of the tens digits; equivalently, under the bar (once again ignoring ones digit) the sum of the tens digits is $\overline{xy}$ , the decimal representation of $10x + y.$ There are other ways to find the sum of the tens digits; this approach is decimal-based, but at this step, there are multiple logical approaches)

Writing out the numbers as expressions (e.g., $\overline{xy} = 10x + y$ ) gives $11x + 11y + 11z = 100x + 10y + z.$

Rearranging and canceling gives us:

$y + 10z = 89x$

Since $y$ and $z$ cannot be greater than 9, $89x$ cannot be greater than 99, so $x=1$ and $y + 10z = 89.$

If $z>8,$ $y$ would be negative. But if $z<8,$ $y>9.$ Since they are all digits, we must then have $z=8$ and $y=9.$

Thus, the 3 digit number is 198.

This solution relies more on logic and deduction rather than algebra -- hopefully it gives some insight into an alternative thought process.

Step 1: What could X possibly be? X is carried forward after adding together the previous two columns. Taking the lowest values of 00 + 11 + 22 would give 33, so the minimum value of X is 0. Taking the highest values of 99 + 88 + 77 gives 264, so the maximum value of X is 2.

However, if the maximum value of X is 2, that means that taking the highest value of the equation gives 22 + 99 + 88 = 209 but these values do not match up with the needed answer. Going down a step and taking 22 + 99 + 77 = 198 shows that X cannot possibly equal 2 but will equal 1 at most.

X cannot be 0 either. If X is zero, this means that Y + Z < 10. Why? Because if X+Y was more than 10, this would carry forward and give an answer in the hundreds which contradicts X = 0. Why is the issue of Y + Z < 10 a problem? Because X+Y cannot equal both Y and Z unless there is a carry forward from the units column into the tens column.

So, with absolute certainty, X = 1.

Step 2: Make a Few Deductions

So far, we should be able to conclude three things: 1) X = 1 2) If X = 1, then the sum of X, Y and Z must be more than 10 and less than 20 in order for a carry forward of 1 to happen and ensure X=1. 3) From the above point, we can deduce that Y = Z + 1. Why? Because in the units column, we add up X, Y and Z and get Z, but in the tens column we add up the same numbers and get Y, which is a result from a carry forward of 1 from addition in the units column.

Step 3: Try Adding Up the Units Column

If 1 + Y + Z = Z, then 1 + Y = 0 (or rather, 10, given the carry forward). Therefore, Y must equal 9.

Step 4: Try Adding up the Tens Column

We know that X=1, Y=9 and there's a carry forward of 1. Therefore, 1 + 9 + Z + 1 (carry forward) = 19 (not 9, given that 1 has to carry forward to the hundreds column)

Solving, Z= 8

And from there you, have your answer of XYZ = 198

$11(x+y+z) = \overline{xyz} \Rightarrow \overline{xyz}$ is a multiple of $11$ . So, either $x+z = y$ or $(x-1)+z = 10+y$ .

First, assume the former:

$\begin{aligned}&\overline{xyz} = 11\cdot\overline{xz}\\ \Rightarrow & 10x+z=x+y+z\\ \Rightarrow & 10x+z=2x+2z\\ \Rightarrow & 8x=z\end{aligned}$ .

Thus, $x = 1, z = 8, y = 9.$ and $\overline{xyz} =198$

Now, let's consider the latter:

$\begin{aligned}&\overline{xyz} = 11\cdot\overline{(x-1)z}\\ \Rightarrow & 10(x-1)+z=x+y+z \\ \Rightarrow & 10x-10+z=x+(x-1+z-10)+z.\\ \Rightarrow & 8x+1 = z \end{aligned}$

$x=0$ or $x=1$ both imply $y$ is negative, which it can't be. The contradiction establishes that $198$ is the only solution.

Z=8, in 3 steps:

➊ In ones column, to have Z in both equation and result, X + Y should be equal to 10

➋ In tens column, as X + Y = 10, and 8 is the max value for Z, so the maximum value for the first digit in result is 1 so X = 1 therefore Y = 9

➌ In the tens, column, I have a carry of 1 from the ones, and 10 (as x+y) and I need 9 in result so Z = 8.

Proof: 11+99+88 = 198

Why making the solution so difficult to understand.By mere observation we can conclude that X=1 and the difference between Y&Z is one or is equal to the carry. Now 1+Y+Z =10+Z and only 8&9 satisfy this term .so X=1,Y=9&Z=8....

OH HI KEVIN!

We know that $x+y = 10$ becasuse the last column digits adds up to $z$ . And by divisibility rule, $\overline{XX} \equiv \overline{YY} \equiv \overline{ZZ} \equiv 0 \pmod{11} \Rightarrow \overline{XYZ} \equiv 0 \pmod{11}$ , so $x - y + z = 0 \text{ or } 11$ only. Because three two-digit numbers add up to a three digit number, $x = 1 \text{ or } 2$ . If $x = 1 \Rightarrow y = 9 \Rightarrow z = 8$ which gives $\boxed{198}$ as the answer. If $x = 2 \Rightarrow y = 8 \Rightarrow z = 6$ but couldn't attain the final value of $286$ . So the answer is $198$ only.

From the first column it is clear that $x+y = 10$ . Carrying this into the next column gives $z+1 = y$ and $x+y+z+1 = 11 +Z < 20$ So 0 < x < 2

$\rightarrow$ $x = 1 \ , \ y = 9 \ , \ z=8$

X + Y = 10 & 1+X+Z = 10 & X = 1 So Y = 9 and Z = 9-1 = 8

x + y + z + 10 x +10 y + 10 z = 100 x + 10 y + z

=> 89 x = 10 z + y

With x = 1 and z = 8,

y = 9

11 + 99 + 88 = 198 which is correct.

We know that X+Y is a number whose digit in the (10^0) column is 0 since Z needs to be added to such a number in order to produce a number whose digit in the (10^0) is Z. Since it is impossible to add two digits to produce a number whose digit in the (10^1) column is <0 or >1, either X+Y= 0 or X+Y= 10.

If we assume that X+Y= 0, then X= Y= 0, which is not possible since X and Y are distinct. Therefore, X+Y= 10 -------(1)

     1+X+Y+Z= 10X+Y

So 11+Z= 10X+Y ------(2)

We see that X represents the digit in the (10^1) column of the number formed as a result of 11+Z. So the possible digits denoted by X are 1 and 2, the latter only being the case when Z= 9. If Z= 9, then Y= 0, which has been established to not be possible. Therefore, X= 1.

Subbing X= 1 into equation (1), 1+Y= 10, implying that Y= 9. Subbing X= 1 and Y= 9 into equation (2), 11+Z= 19, implying that Z= 8. Thus, XYZ(not X times Y times Z) represents the number 198.

We get y = 9x . Hence by hit and trial method we can get x=1 and y=9 . So by assuming value of z from 1-9 (1and9 included) we get the value of z =8 . Hence 99+11+88 = 198. We get out answer

Alternative method

The 3 numbers are 10x+x=11x, 11y and 11z. Their sum is 11( x+y+z)=100x+10y+z Subtracting 10(x+y+z) from both sides of the equation, we have 11(x+y+z) - 10(x+y+z)=(100x+10y+z) - 10(x+y+z), or (x+y+z)=90x-9z= 9(10x-z) As right side is a multiple of 9, the left side or specifically the (x+y+z) should also be a multiple of 9. (x+y+z) can not be 9 because the number 11(x+y+z), in that case, will be 11×9=99, a two digit number. It can not be 27 or more, because the numbers x, y and z are different and are less than or equal to 9. Therefore, being a multiple of 9, it can only be 18 and the answer is 11(x+y+z)=11×18=198.

My solution, perhaps an unorthodox one. we'll use logic to deduce the following: {comments} - (alternative value)

X+ Y = 10 ......... {since we need X+Y+Z to have Z in the one's digit}

X = 1 (or 2)...........{max value of X+Y+Z = 19} Y = 9 (or 8)...........{same as above}

and therefore, 1{carry over} + 10{X+Y} + Z = 19 (or 28) hence Z = 8 (or 17)

well, we cant have Z = 17, and thus eliminating all the '(or)' mentioned above

we get the values: X=1 Y=9 Z=8

Think of it this way

XYZ

Since we are only adding tens and units the highest value considering distinct values for X,Y and Z is in the 200s. Try 99+88+77, thus it must be less than 3 so this means X>3

so now we know that now we have in the units collumn X+Y+Z=Z

This means Z-(X+Y) also equals Z so we can assume that its 10, since [For example 5+10=15, 15-10=5 the unit stays the same]

now we know this much.

So X+Y=10 right?

so the only solutions are, if X is 2 then Y is 8 or if X is 1 then Y is 9

Now lets do the two solutions and test them

1) If we test 11+99+88, we get 198 which also satisfies the conditions since X=1, Y=9 and Z=8

2)If we test 22+99+88=209, we will find the Y value isn't 8 thus doesn't satisfy the condition. If we take it lower than that, we get 22+99+77 which equals 198 which leaves the X value unsatified.

So the only correct answer is X=1 Y=9 Z=8 making XX+YY+ZZ=198

I hope that makes sense.