The Repetitious Sum

$\overline {XYZ}+\overline {XYZ}+\overline {XYZ}=\overline {ZZZ}$

The Last Digit, That is Formed By adding 3 $Zs$ together, Is Z.

The only possible number Where this is possible is 5. Any number ending with 5 when added 3 times , will have the unit's digit as 5

So, $Z=5$

Now, $\overline {XY5}+\overline {XY5}+\overline {XY5}=555$

$3 \times \overline {XY5}=555$

$\overline {XY5}=185$

$X=1,Y=8,Z=5$

$1 \times 8 \times 5= \boxed {40}$

Z+Z+Z has its last digit equal to Z. The only solution to that is Z=5. Carry a 1 over to the Y column because 3(5)=15

Y+Y+Y+1 has its last digit equal to 5. The only solution to that is Y=8. Carry a 2 to the X column because 3(8)+1=25

X+X+X+2 has to be a one digit number and has to equal 5. The only solution to that is X=1

5 is the only single digit number which when multiplied by 3 results in a number whose units digit is 5 hence z=5 now therefore zzz = 555 and we know that xyz X 3 =zzz therefore xyz X 3=555 which implies that xyz = 555/3 = 185. Hence x=1 y=8 and z=5 therefore x X y X z=40

So, firstly we have

$Z \times Z \times Z = a \times 10 + Z$ , where a is non-negative digit

If a = 0, then Z = 0. Wrong.

If a = 1, then Z = 5. Good.

If a = 2, then Z = 10. Wrong.

So Z = $\boxed{5}$

Next, we have:

$Y \times Y \times Y + 1 = b \times 10 + Z$ , where b is non-negative digit and 1 comes from 5 + 5 +5 = $\boxed{1}$ 5

If b = 0, Y = 4/3. Wrong.

If b = 1, Y = 14/3. Wrong.

If b = 2, Y = 8. Good.

If b =3, Y = 34/3>10. Wrong.

So Y = $\boxed{8}$

Finaly, we have:

$X \times X \times X + 2 = c \times 10 + Z$ , where c is non-negative digit and 2 comes from 8 + 8 +8 +1 = $\boxed{2}$ 5

If c = 0, X = 1. Good

If c = 1, X = 13/3. Wrong.

if c = 2, X = 23/3. Wrong

If c = 3, X = 33/3 > 10. Wrong.

So X = $\boxed{1}$

Then we have $X \times Y \times Z$ = $1 \times 8 \times 5$ = $\boxed{40}$

The game here is a game of digits ,right? if we try to do this sum : 3Z = Z+10a : where a is a normal number {1,2,3,4....} whatever. we will find that the only possible solution for it is that Z = 5 and Z = 0. and we start solving on those 2 ways: Z = 0 then 3Y+0=Z+10b. where be is in the same domain of a. where Z = 0 , 3Y = Z+ 10b, and the b is determined logically. then we find that Y = Z/3 + 10b. the digit Y = the digit Z and that's unaccepted, so we move on in the other way where Z = 5: then Z*3 = Z+10a , a = 1.

lets do what we did in elementary school: 1 + 3Y = Z+10b, then 3Y = 4+10b. lets do what we did in the top, and multiply 3 by each number between 0 and 9 and see which resultant number has the digit "4" as its last digit. and the only possible number we used that gets this result is number "8" and the result is 24. then 24 = 4 + 10b > b=2.

the same elementary thingy, 2 + 3X= Z + nothing because we have no more digits then 2 + 3X = 5 > 3X = 3 > X=1.

^_^

What, when added to its self, is its self? That is five. Divide by three to get X and Y

XYZ} + XYZ + XYZ = ZZZ Therefore we can say 3 $boxed{XYZ}$ = $boxed{ZZZ}$ Given that X, Y and Z are single digit positive integers, we can say:

300X + 30Y + 3Z = 100Z + 10Z + Z

300X + 30Y = 108Z

100x + 10Y = 36Z

As all digits are integers, we know the variables:

36Z > 110 and 36Z </= 324 thus 110 > 36Z >/= 324

36Z must be a multiple of 10, only value for that within the range 110<36Z</=324 is 180 where Z = 5 , therefore as X and Y are single digit positive integers (not 0, negative or more than 9) X=1 and Y=8

Therefore XYZ =1 x 8 x 5 = 40

If X could equal Y and Z, then the product could also be 0 because X, Y, Z = 0 satisfies the summation as well.

Z+Z+Z = Z , only 5 , bcaz cant be 0 , when Z = 5, 3(XYZ) =ZZZ , 555=ZZZ , ZZZ/3 = XYZ , 555/3 = 185 , 1(8(5)) =40

From the ones column, we deduce that

$3Z = 10k + Z$ where $k$ is a non-negative integer.

$2Z = 10k$

$Z = 5k$ .

Since $Z$ is single digit, $5k$ is at most $9$ so $k$ is at most $1.8$ . Since $k$ is positive, $k$ is strictly greater than $0$ so the only possible integer for $k$ is $1$ so that $Z = 5$ .

Hence $XYZ = \frac{555}{3} = 185$ and $X \times Y \times Z = 1 \times 8 \times 5 = \boxed{40}$ .

z+z+z=nz, 3y+n=rz. 3x+r=z,, n and r are carry So: z=0 or, z=5,, Then z=5, n=1,, 3y+1=r5,, Y=8,, (8×3)+1=25,, r=2,, 3x+2=5,, X=3

From the 1's digit, Z + Z + Z = Z(mod 10), so subtracting Z from each side gives 2Z = 0(mod 10), so either 2Z = 0, or 2Z = 10. (2Z = 20 or 30, etc, yields a value of Z larger than a single digit). So either Z=0 or Z=5, but since Z is a leading digit in the result, it cannot be 0. Hence Z = 5.

Now, if $N = \overline{XYZ}$ , then it follows from the sum that 3N = 555, so N = 185. Hence, X=1, Y=8, and Z=5. Hence, XYZ = (1)(8)(5) = 40.

From the 3s multiplication table:

The only number Z such that 3Z ends in Z is Z = 5.

Then, carrying the 2, 3Y + 1 ends in 5, or 3Y ends in 4. The only such single-digit number is Y = 8.

Again, carrying the 2, 3X + 2 = 5, so X = 1.

5 * 8 * 1 = 40.

We must find a digit Z so that 3xZ has a unit digit of Z. Thus, Z=5 is the only solution. Carry a 1 over to the Y column because 3x5=15. That means 3x $\overline {XY}$ +1=55. Thus, 3x $\overline {XY}$ =54, so $\overline {XY}$ =18. X=1 and Y=8 1x8x5=40, 40 is the answer.

The concept of 5 where in any number is multiplied with 5 is equal to 0 or 5 at the end of the digit. So by multiplying 3(because we have 3 digit) by 5 we can have 15 at the end of the digit Z. And substitute Z to 5. Our equation will be (XY5) x 3 = 555. X will always be 1 because if we consider it to be more than 1 if you add it up it will not be 5 it will be more than 5. We have 3 and we have 5. 3 + 5 = 8 for Y. 1 x 8 x 5 = 40

$\overline{XYZ}*3=\overline{ZZZ}$

And we know that the only possible number in which the condition $Z*3 = Z$ in the tens place is for $Z=5$ .

Therefore, we can use the equation $\frac{555}{3}$ to get the value of $\overline{XYZ}$ which is 185.

Multiplying each digit results in a product of $\boxed{40}$ .

The last number has to 5, so the answer must be 555. 555/3= 185 so 1 8 5= 40

Z has to equal 5 (only non-zero one-digit number that equals itself when multiplied by 3)

XYZ + XYZ + XYZ = 555

3(XYZ) = 555

XYZ = 185

1 x 8 x 5 = 40

z + z+z= z(in unit) the only no. < 10 that has this properties is 5 so z+z+z=15 so there is a carry up 1 here so y+y+y+1=5(in unit) so y+y+y=4(in unit) the only no. < 10 that has this properties is 8 so y+y+y=24 so we have a carry up 2 so x+x+x+2=5 and have no tens so x+x+x=3 so 3x=3 so x=3/3=1

now x y z=1 5 8=40

Z is a number when multiplied by 3 is Z it self. What could it possibly be? Yes, 5 just 5.

The answer is ZZZ , which is 555.

Which hundreds could be multiplied by 3 and not exceed 555? Yes, just 1

We now here get X=1 Z=5

What could possibly the other number be?

Since it's 500+, then we should choose number >6

We can try 7 and get 175*3=525 which is incorrect

Next we can try 8 and get 185*3=555 which is correct.

Then we get X,Y,Z are 1,8 and 5 respectively

X * Y * Z= 1 * 8 * 5 = 40

While I got the product 40 for this problem, the product 0 is also mathematically correct because it is not obligatory to assume x, y, and z are unique digits.

XYZ + XYZ + XYZ = ZZZ, 3 * XYZ = ZZZ, 3 * Z = Z + Something * 10 Z = Something * 5 Z = 5 3 * XYZ = 555 XYZ = 185 X * Y * Z = 1 * 8 * 5 = 40

Here,

  1 8 5

  1 8 5

  1 8 5

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  5 5 5

So, 1 * 8 * 5 = 40