Digit Sum, What's Z?

Looking at units digits, $X+Y+Z\equiv Z\pmod{10}$ so $X+Y=10$

Note that none of $X$ and $Y$ can be zero, as the other would have to be 10.

Now note that $20>X+Y+X\ge 10$ so a one carries onto the tens digit. Note that $Z\neq 9$ or else $X=0$ . We deduce that only a one carries from place value to place value, thus a one caries to the ten-thousands and $Y=1$

Finally, note that $Y+Z\equiv 9 \pmod{10}$ considering some place value adjusted with carrying, so $Z=8$

If we look at the first column of xyz and z, we see z is coming after adding x, y and z as unit digits. This means x+y must be equal to ten, so that z + 0[zero of ten] = z as unit digit. But in the left of the total sum, we see y is coming in the ten thousands position.

The number to come after adding three single digit numbers will have 1 or 2 in its ten's digit. This means y is either 1 or 2.

Now, if y is 1, then x is 9 or if y is 2, then x is 8 [both should add up to ten]. Checking both of these, the correct solution is x=9, y=1 and z=8.

Start with the Y at the beginning. Three distinct digits can only add up to a number in the 20s at most (9+8+7=24). Factor in that one of those digits is Y itself and this means Y must be 1.

If we then look at the Z at the end of the solution. For the sum of three digits to give a Z for this digit, and the next digit to be different X, this means that the three digits add up to (Z + 10). It also means that X is one more than Z.

This means that X+Y=10, therefore X is 9.

We know that X is one more than Z, therefore Z is 8.

1111 X + 1111 Y + 1111 Z = 10000 Y + 1000 X + 100 X + 10 X + Z

=> 8889 Y = 1110 Z + X

With Y = 1 and Z = 8,

X = 9

9999

1111

8888

19998

Z confirmed to be 8.

This is going to be a very alternative solution, so bear with me.

Any number composed by four repeated digits is divisible by 11 - for instance, X0X * 11 = XXXX. Therefore, we are actually adding up three products of 11: X0X * 11 + Y0Y * 11 + Z0Z * 11.

If all of the three numbers are divisible by 11, then so must be the sum result. In that case, if we divide YXXXZ by 11, we get the following process:

Step 1: Y goes to the quotient. In the numerator, we subtract YY from YX, getting (X-Y) as the partial remainder.

Step 2: (X-Y) goes to the quotient. In the numerator we subtract (X-Y)(X-Y) from (X-Y)X, getting Y as the partial remainder.

Step 3: Y goes to the quotient. In the numerator, we subtract YY from YX, getting (X-Y) as the partial remainder.

Step 4: This is the important part. Since the original number YXXXZ is divisible by zero, then our final remainder must be zero. What we still have in the numerator is (X-Y)Z, which means that (X-Y) = Z. To wrap it up, let's divide the rest of our number by (X-Y) or Z.

Ergo, YXXXZ divided by 11 is Y(X-Y)Y(X-Y), or YZYZ. For the sake of simplicity, I'll choose the latter form.

By dividing everything by 11, we get a much easier summation: X0X + Y0Y + Z0Z = YZYZ. Now, remember how Z is equivalent to X-Y? Therefore, the addition of the one's place (X + Y + Z) is equivalent to 2*X, if we substitute Z for X-Y. Therefore, we have that X multiplied by 2 must yield a number that ends in Z.

Let us now take a look at the ten's place. We have that 0 + 0 + 0 + whatever is carried from the one's place is equivalent to Y. Now, Y cannot be 0, assuming no numbers can start with 0 (YYYY must be non-zero). Now, remember how the addition in the one's place was equivalent to 2*X? Therefore, the number that's carried over can only be 1, since the product of 2 and a single-digit number can be at its highest 18.

We shall step back to our rule. We have ascertained that (X-Y) = Z; substituting Y for 1, we have that (X-1) = Z. Now we have two rules to determine these numbers: X must be one unit larger than Z and 2*X must give us a result equivalent to the number 1Z.

The only possibility for this combination is X = 9 and Z = 8, since only the duplication of the number 9 will reduce the one's place by one from the original number. Therefore, Z must be 8.

1. X+Y+Z= Z+0 or Z+10 Since the result is of five digits, it must be X+Y+Z=Z+10

2. Looking at the leftmost column, we can infer that X+Y+Z= 10Y+X

3. Looking at the tens column, we can infer that X+Y+Z+1= X

So I have three equations. Please tell me if they are correct or not because I am not able to arrive at an answer using this method.

In the first column from right, if we add X,Y and Z., we should have Z as the one's place digit. It means (X+Y+Z)mod(10) = Z and hence X+Y = 10(Since adding two single digit number can produce a maximum number 18(9+9)). Now, on looking at the second column we can observe that Sum of all the numbers is obviously equal to sum all the numbers in first column, but the first column will generate a carry. This means X = Z+1 and hence (Z+1) + Y = 10
=> Y+Z = 9. Further as @Reeshabh said Y is coming at 1000th position has to be 1 or 2 because the ten's place digit after adding three single digit numbers would be 1 or 2. This means Y can be 1 or 2, so Z can be 8 or 7 and X can be 9 or 8 respectively. Now you can check:- Z=8,X=9,Y=1

Thanks :)

Since you can only deal with single digit numbers, and you can not have a duplicate number, you have to figure out what puts you over 10,000 and the only possibility would be 9^4 * 2 or 9^4 + 8^4, and the first one has duplicate numbers besides the last integer because it is going to change as we add Z and since each number has to be distinct we can take out the other 3 options (8^4+9^4=10657, 2^4=8, 3^4=1, 8^4=6, 9^4=1, and when you add each one to the previous number only one will give us a number that hasn't already been listed)

$X+Y+Z\equiv Z \pmod{10}$ but they are all distinct positive integers so $X+Y=10$ . Then in the second column we have $X+Y+Z+c_{1}\equiv X\pmod{10} \implies Z+c_{1}=X \implies c_{1}=1 \implies Z=X-1$

Clearly $11\vert YXXXZ$ so $X-Y-Z\equiv 0 \pmod{11} \implies X-9\equiv 0 \pmod{11}$

So we are left with $X=9$ , $Y=1$ , $Z=8$ .

If we write in line this sum :

$(X+Y+Z)\times 1111 = X\times10^4+X\times1110+Z \Leftrightarrow 1110\times Z+X = Y \times 8889$

Which give us immediately $Z=8\times Y$ and $X=9$

And we naturally conclude that $Y=1, Z=8, X=9$

$X + Y + Z = Z$ , therefore $X + Y = 10$ since ten is the only multiple of ten which can be made from single digits. Also, $X + Y$ can't be zero as they are distinct.

Then, $XXXX + YYYY = 11110$

and $11110 + ZZZZ = YXXXZ$ .

Therefore $Y$ can only be 1 or 2 and $X$ can be only 8 or 9.

From the last sum we can also assume $1111 + ZZZ = YXXX$ .

From this it follows that $Z + 1 = X$ thus making $Z$ smaller than 9. However, when $Z$ is smaller than 9, there is no remainder in the ten-thousands column meaning $Y$ must be 1. This excludes the solution $X$ = 8 and therefore

$Z + 1 = 9$

$Z = 8$

We have XXXX + YYYY + ZZZZ = YXXXZ (1) so the sum of digits X , Y is X + Y = 10 (2) . Then XXXX + YYYY = 11110
so from (1) we get 11110 + ZZZZ = YXXXZ (3) . From (3) : Z + 1 = X with Z < 9 and Y = 1 so (2) became Z + 1 + 1 = 10 so Z = 8 .

1. 1111 X + 1111 Y + 1111 Z = 10000 Y + 1110 X + Z
2. X – 8889 Y + 1110 Z = 0
3. 1110 Z = 8889 Y – X
4. Y can only be 1 to keep the multiple of 8889 within one digit of a multiple of 1110
5. then Z must be 8 to make the difference a single digit and that digit is 9

brute force

https://goo.gl/PVqQqQ

Looking at the unit digits, we see that X + Y + Z = Z and hence X + Y equals 10. Since 10 + Z cannot be greater than 19, just a one is carried to the tens. Therefore X + Y + Z + 1 = X and hence X - 1 = Z. Since just a one is carried from place to place, Y equals 1. So: X + 1 = 10 and hence X = 9 X - 1 = Z thus 9 - 1 = Z and so Z equals 8

The first thing to note, as others have, is that X + Y = 10, because X + Y + Z = Z (mod 10).

The maximum value of X + Y + Z is therefore 19. Trying Z = 9 gives:

11110 + 9999 = 21109

which is clearly incorrect.

Trying Z = 8, however, gives:

11110 + 8888 = 19998

which holds for Y = 1 and X = 9.

$\large\boxed{\color{#3D99F6}{Z=8}}$

$\large\color{maroon}{YXXXZ}=\color{#3D99F6}{19998}$

1-The summation of 3 distinct digits won't exceed 24 $x+y+z\le24$ 3-Since the last number of the result is "y" that means -according to step"1" $y\le2 ..$ 3-As a result of step "2" .. step"1" will be altered as follows $x+y+z\le19$ .. and due to this "y" can only be =1 so; $x+y+z\le18$ 4-Non of them can be equal "0".. because $x+y+z=z$ ; that means that $x+y=10$ Now we have $\text{y=1 \& x+y=10 so x=9, but why z=8?}$ $\textbf{If Z<8 then the 2nd level of summation of x,y,z won't equal x}$ $x+y+z+1\ne x$ $\text{As an example take z=7}$ $x+y+z+1=9+1+7+1=18 \text{and we know x=9}$ So the final result would have been "18887" and represented as "yvvvz" and the problem will be insolvable

as "v" will be any number in [3,7]

In 3 steps:

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

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

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

Proof: 9999 + 1111 + 8888 = 19998