7 of 100: XX YY ZZ

I used the fact that $\color{#69047E}{XX}\color{#333333}, \color{#D61F06}{YY}\color{#333333},$ and $\color{#3D99F6}{ZZ}\color{#333333}$ are multiples of 11 to solve this one real fast!

$\begin{array}{rcl} 11\color{#69047E}{X}\color{#333333}+11 \color{#D61F06}{Y}\color{#333333}+ 11\color{#3D99F6}{Z}\color{#333333} &=&100\color{#69047E}{X}\color{#333333}+10 \color{#D61F06}{Y}\color{#333333}+\color{#3D99F6}{Z} \\ & \Downarrow & \\ \color{#D61F06}{Y}\color{#333333}+ 10\color{#3D99F6}{Z}\color{#333333}&=&89\color{#69047E}{X} \end{array}$

Therefore, $\color{#69047E}{X}$ must be equal to 1. Why? Because we know that $\color{#69047E}{X}\color{#333333}, \color{#D61F06}{Y}\color{#333333},$ and $\color{#3D99F6}{Z}\color{#333333}$ are non-zero digits, meaning that they are whole numbers greater than 0 and less than 10. Therefore, $\color{#69047E}{X}$ cannot be >1 because the largest the left side of the equation above can be is 99 (or 98 if the digits have to all be different). Now we know that:

$\color{#D61F06}{Y}\color{#333333}+ 10\color{#3D99F6}{Z}\color{#333333}=89$

Using similar reasoning, because $\color{#D61F06}{Y}$ cannot be greater than 9, $10\color{#3D99F6}{Z}\color{#333333}$ must be greater than or equal to 80. Therefore, we know that $\color{#D61F06}{Y}$ must equal 9 and $\color{#3D99F6}{Z}\color{#333333} = \fbox{8}$

Adding columnwise: $1\text{st column: } \color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\color{#333333}+\bcancel{\color{#3D99F6}{Z}}\color{#333333}=\bcancel{\color{#3D99F6}{Z}}\color{#333333}+10 ~(**)$ $\begin{aligned}2\text{nd column: } & \color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\color{#333333}+\color{#3D99F6}{Z}\color{#333333}+\underbrace{1}_{\small{\text{carry}}}=10\color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\\&\implies\color{#3D99F6}Z\color{#333333}=9\color{#69047E}{X}\color{#333333}-1\end{aligned}$ Since $\color{#3D99F6}Z$ is a digit implies $\color{#69047E}{X}\color{#333333}{=1}$ and $\color{#3D99F6}Z\color{#333333}=8$ ( $\because$ for $\color{#69047E}X\color{#333333}\ge 2\implies \color{#3D99F6}{Z}\color{#333333}>10$ )and for these values $\color{#D61F06}{Y}\color{#333333}=9$ . Hence $\boxed{\color{#3D99F6}{Z=8}}$ .

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$**$ NOTE:- Here carry could be $0,1,2$ accordingly $\color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\color{#333333}=0,10,20$ respectively but since $1\le \color{#69047E}{X}\color{#333333},\color{#D61F06}{Y}\color{#333333}\le 9$ we are only left to take $\color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\color{#333333}=10$ .

First column says that X+Y=10. This tells us that the second says Y=Z+1. And as X+Z=9, the X in the third column must be 1. Hence, X=1, Y=9, Z=8.

We know that X+Y=10(X&Y are single digit number and X+Y+Z=10+Z), at max if we put Z=9 also we will get X+Y+Z=19 & X+Y+Z+1=20 which is not allowed as in the problem statement. (X,Y,Z all are different nonzero digit)

So looking at the units column:

$X+Y+Z=Zmod10$

So it follows, given the single digit nature of X, Y and Z that:

$X+Y = 10$

now looking at the tens column only we have

$X+Y+Z+1=Y$

The 1 appeared due to the carry over from the units column:

as we know X+Y=10 and therefore will give a zero (in this column) we have:

$Z+1=Y$

now looking at the hundreds column we have only the 1 carry over, so:

$1=X$

therefore given we know:

$X+Y=10$

we can deduce:

$Y=9$

and from:

$Z+1=Y$

we know now know:

$Z=8$

lets test out answer:

$11+99+88=198$

CORRECT!

so we can confidently say

$Z=8$

We'll consider each numbers separately.

$XX = 10X+X$

$YY = 10Y+Y$

$ZZ = 10Z + Z$

$XYZ = 100X+10Y+Z$

$10X+X+10Y+Y+10Z+Z=100X+10Y+Z$

Simplifying this we get,

$11X+Y+10Z=100X$

$\therefore Y+10Z=89X$

Clearly, $X$ can only take the value of 1 which implies that $Y=9$ and $Z$ = $\boxed{8}$ .

if my girl and papa john both drowning and i can only save 1.... catch me at my girls funeral with betr ingridates and bter pezza, prapy jhon

Method 1 : Brute force algebra! $11 X + 11 Y + 11 Z = 100 X + 10 Y + Z \\ 89 X = Y + 10 Z.$ Since $Y, Z \leq 9$ , $89 X \leq 99$ which implies $X = 1$ .

$89 = Y + 10 Z$ has the obvious solution $Y = 9, Z = \boxed{8}$ .

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Method 2 : Reasoning through the process. Since $X + Y + Z = Z$ , $X + Y$ must end in a zero, therefore $X + Y = 10$ .

The carry from the units to the tens is at most one, therefore the sum of the tens is at most $X + Y + Z + c \leq 10 + 9 + 1 = 20$ . But $Y \not= 0$ , which means that the sum of the tens is no more than 19, and $X = 1$ . It follows that $Y = 10 - 1 = 9$ .

Finally, comparing the units and tens columns we have $Y = Z + 1$ because of carry. Therefore $Z = 9 - 1 = 8$ .

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Method 3 : Just because nobody posted anything along these lines yet...

Since the three summands are multiples of 11, $XYZ$ is a multiple of eleven. This means that $X - Y + Z$ is a multiple of eleven. Also, $X \leq 2$ , $Z \leq 9$ , and $Y \geq 1$ implies that $X - Y + Z < 11$ , and obviously $X - Y + Z > -11$ . Thus $X - Y + Z = 0$ .

We rewrite this as $Y = X + Z,$ and the sum becomes $2\overline{YY} = \overline{XYZ}.$ The presence of a hundreds place implies $X = 1$ , $Y \geq 5$ ;

moreover, $Z =$ even, so $Y = X + Z$ odd. $Y = 5$ is ruled out because $Z \not= 0$ .

We are left trying $Y = 7:\ \ 2\cdot 77 = 154;\ \ \ \text{but}\ 7 \not= 5;$ and $X = 9:\ \ 2\cdot 99 = 198;\ \ \ \text{which works}.$

X+Y =10 , for Z at the one's place and also, X+Z=9 FOR Y in tens place , here we get that ,Y=1+Z, and possibilities of X are now only 1&2.

I took a numerical approach by solving in R.

I begin by making a list of all possible combinations of x, y, and z with the knowledge they can take any value in the set {1:9}. Then for every possible value of x, y, and z, I calculate the LHS: $xx + yy + zz$ and RHS: $xyz$ . Using MSE as the error function, I then find where the error is equal to zero. And the solution is 198, so the answer is $\boxed{8}$ .

Note: MSE = $((xx + yy + zz) - xyz)^{2}$

R code:

l <- list(x = 1:9, y = 1:9, z = 1:9)
w <- do.call(expand.grid, l)  # all possible combinations

q <- matrix(0, nrow = 729, ncol = 3)
for (i in 1:729) {
  u <- w[i, ]
  x = as.numeric(u[1])
  y = as.numeric(u[2])
  z = as.numeric(u[3])

  q[i, 1] = as.numeric(paste0(x, x)) + as.numeric(paste0(y, y)) + as.numeric(paste0(z, z))  # LHS
  q[i, 2] = as.numeric(paste0(x, y, z))  # RHS
  q[i, 3] = (q[i, 1] - q[i, 2])^2  # loss function
}

q[which(q[,3] == 0), ]  # where there is no error

First column says that X+Y=10. then easily Y=Z+1. And X+Z=9, the X in the third column must be 1. so, X=1, Y=9, Z=8.

so.11+99+88=198. so Z is the answer

Base on the figure, we can say that

x + y + z = 10 + z

x + y = 10 (1)

Since 1 is added to the tens place,

x + y + z + 1 = 10 + y

10 + z + 1 = 1 + y

z + 1 = y (2)

since the remainder 1 from x + y + z + 1 is equal to x,

x + y = 10x (3)

Then we have now three equations:

x + y = 10 (1)

z + 1 = y (2)

x + y = 10x (3)

And it is easy for us to get the solution. After we solve, we get:

x=1, y=9, z=8 or (1,9,8)

So, the value of z is equal to 8. 8 is the correct answer.

To check, 11+99+88=198, right?

XX+YY+ZZ = 11+99+88 = 198

Z=8

X could be at most 2. Writing the problem out substituting what I can, I begin with X=1, and Z as each of the given answer choices, with Y unknown initially. Quickly it is seen that in order for the Z in the sum to equal the Z addend, Y has to be 9, and, Z cannot equal 9. Only three attempts in, and the solution is found.

e.g.

   11+YY+66   --> 11+99+66 = 176

   11+YY+77   --> 11+99+77 = 187

   11+YY+88   --> 11+99+88 = 198 = Answer

I made cases and cancel out the wrong ones one by one. 1. X can be 1 or 2, but 2 is possible only and only when all 3 digits are non eqaully more than 5 i.e., X should also be greater than 5, hence X can be "1" and not 2.

1. Choose Y in such an order that X + Y= 10, because only in this case adding Z will give Z at units place and 1 will go carry to tens place, which will satisfy my 3rd case. So this statement deduces that Y = 9.

2. Y = Z + 1, this statment can be seen by that when we are adding X, Y and Z in ones place, they are giving Z (obviously with a carry), so that carry is gonna sum to tens digit X, Y and Z and produce Y (with carry 1).Hence Z = 8

Its easy if you figure out the basic rule that y is z + 1 and that x must be equal to 1 or 2 as its maximum and minimum this means that y + x is equal to z this means it's equal to 8 or 9 however we can eliminate 8 as 2 + 8 + 9 is equal to 19 which makes it impossible for it to Equal 2 which means x is equal to 1 y is equal to 9 which leaves x as equalling 8 as y - 1 equals z as previously established we can test this through substituting 11 +88 +99 =189 There are probulary easier solutions but i like to over complicate problems(don't die reading it punctuation is terrible)

Being Z the first digit in the result necessitates that X+Y = 0 or 10 and as we know "each letter represents a different nonzero digit",SO (X+Y must equal 10 ) then on the other side X+Y+Z+1 must equal XY which means (11+Z = XY) trying different values for Z keeping in mind that X+Y must equal 10, we find Z must be 8 ...So XY = 19 which is X+Y = 1 +9 = 10 .

Just use hit and trial with some logic.JAI MATA DI

It's easy to see that $\color{#69047E}{X}\color{#333333} +\color{#D61F06}{Y}\color{#333333}=10$ and $\color{#3D99F6}{Z}\color{#333333}+1=\color{#D61F06}{Y}\color{#333333} \Rightarrow 1 \leq \color{#3D99F6}{Z}\color{#333333} \leq 8$ . $\Rightarrow 11 \leq \color{#69047E}{X}\color{#333333}+\color{#D61F06}{Y}\color{#333333}+\color{#3D99F6}{Z}\color{#333333} \leq 18$ $\Rightarrow \color{#69047E}{X}\color{#333333}=1 \Rightarrow \color{#D61F06}{Y}\color{#333333}=9 \Rightarrow \color{#3D99F6}{Z}\color{#333333}=8$

• If you look at the digits column, X + Y + Z = A||Z therefore, X + Y = 10 . This means that the highest that X + Y can be is 19 and so looking at the tens column, Y must be 1 more than Z. Again, as the highest that X + Y can be is 19, X has to be 1. Therefore Y = 9 , substituting it in, 11 + 99 + Z||Z = 19||Z .

Therefore, Z has to equal 8.

My first observation was that z is both in the sum and the addends. This led me to believe the sum of x+y was 10. Next the logic was y was a result of regrouping a plus 1. Therefore Z+1= y. Using these observations I concluded x equal to 1. It followed that 9+1 =10. If y is one more than z, the z is 8. Concluding x y z is 198

Suppose that X, Y and Z were the maximum possible values, i.e. 9, 8 and 7. This give 99 + 88 + 77 = 264. Since X corresponds to the hundreds digit of the sum, X has to be less than or equal to 2. Now suppose that X = 2 and find the maximum possible sum i.e. 22 + 99 + 88 = 209. Since Y is a nonzero digit, X can't be 2 and therefore equals 1.

The above reduces the first column (of the sum) to 1 + Y + Z = 10 + Z. Solving for Y gives Y = 9.

The second column becomes 1 + 9 + Z + 1 = 19 (the additional 1 on the LHS is carried over from the sum of the first column), and solving for Z gives Z = 8.

We can see immediately that X=1 or X=2 (three 2-digit numbers can't sum to 300), and that X+Y=10. Hence, we either have (X, Y) as (1, 9) or (2, 8).

(1, 9) gives 1+1+9+Z=19 from the left-hand column - remembering to include the 1 carried from the right-hand column - so Z=8.

(2, 8) gives 1+2+8+Z=28 - this cannot be the case since it would mean Z=17.

Units place column shows.
x+y = 10.
We also have.
100x+10y+z=11x+11y+11z.
Or 89x = y +10z=10-x+10z.
Or 90x = 10 + 10z.
Or 9x = 1 + z,
As 1 + z is less than or equal to 10,.
9x can only be 9, so 1 + z = 9.
=> z = 8.

Since X + Y + Z = Z plus a carrier (this carrier can only be one* ), and X +Y + Z = Y as the first digit, X = 1 **. and Y = Z + 1. so : 1 + Z + 1 + Z = Z plus carrier --> Z = 8 since 1 + 8 + 1 + 8 = 18 which has 8 as the first digit.

*the carrier is one because of this:

• it can be zero but that is impossible since the same three digits amount to different result in the two columns.
• it can be two at maximum (9+8+7 = 24) but that its also impossible cause if it had two as a carrier, that means the result will be one of (20, 21, 22, 23, 24) so y will be one of (0, 1, 2, 3, 4) but even for the max value of y = 4 and x = 3 at max, we can have z to amount to 24.

** X= 1 because of two things: - it is the carrier of the second step and it is the same as the first step since the sum of the first (X+Y+Z) cannot have 9 as the first digit because of the same previous argument in *

10x+x+10y+y+10z+z=100x+10y+z;
x+y must equal 10 so that z+0=z; 10x+10+10z=100x; 10+10z=90x; 10(z+1)=90x; z+1=9x; z=9x-1; We want (9x-1) to be a digit. This only happens when x=1. So z=8.

The first column shows that x+y=10 so that x+y+z=z+10. You carry the one and x+y+z+1=y+10 which means that y=z+1. Given these restrictions, it seems incredibly unlikely that x is anything other than 1 because the three digit number would be too large. Plugging in 1 for x and filling in the rest of the variables gives us a possible result so x=1, y=9, and z=8.

$11X + 11Y + 11Z = 100X + 10Y + Z \\ ⇒ 10Z + Y = 89X \\ ⟹ 10Z = 89X - Y$

Therefore,

$89X - Y \in \lbrace 89X-9, \ldots, 89X-1 \rbrace \cap \lbrace 10, 20, \ldots, 90 \rbrace$

But:

$\begin{cases} \text{if } X = 1, \text{ then } \lbrace 89X-9, \ldots, 89X-1 \rbrace \cap \lbrace 10, 20, \ldots, 90 \rbrace = \lbrace 80, \ldots, 88 \rbrace \cap \lbrace 10, 20, \ldots, 90 \rbrace = \lbrace 80 \rbrace \\ \text{if } X ∈ ⟦2, 10⟧, \text{ then } \lbrace 89X-9, \ldots, 89X-1 \rbrace \cap \lbrace 10, 20, \ldots, 90 \rbrace = ∅ \\ \end{cases}$

Hence

$X = 1$

thus (with $10Z = 89X - Y$ and $Y, Z \in ⟦1, 9⟧$ ):

$Y = 9 \text{ and } Z = 8$

I first figured out the constraints and knew that XYZ was between 132 and 264, which meant that X either had to be a 1 or 2. Since you're going to be carrying over from the first column, the value of y was going to be one more than the value of z. Some guessing from there with those parameters got me 1,9 and 8.

X + Y = 10 (only way that X + Y + Z = Z from first column, where X & Y are different nonzero)

X + Z = 9 (only way that X + Y + Z + 1 = Y from second column, where we know a one was carried from the first column)

X = 1 (because Z < 10 and X +Y + Z + 1 = Y from second column, so X + Y + Z + 1 is less than 20, but greater than 10)

Therefore, Z = 8 from second equation (and Y = 9 from first equation):

    11

    99

 +  88

 -----

   198

From the equation, we know that x+y+z=xy-1>10 and x=1

Since we know that 1+y+z=1y-1 then the only possible number for z is 8