SOLUTION: In the following addition problem, Y, O, and U are different digits. Solve the following problem: YOU + YOU + YOU = UUU

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Question 176031: In the following addition problem, Y, O, and U are different digits. Solve the following problem:
YOU + YOU + YOU = UUU
Found 2 solutions by josmiceli, Edwin McCravy:
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
First of all, Y cannot be 0 because it is
a leading digit, and U cannot be 0 because
That would make the sum 000, which it can't be
-------
So, Y and U must be one of the digits 1-9
Also, UUU must be less than 999 since one
more would be 4 digits, not 3.
If UUU is less than 999, that means
3 x YOU must be less than 333. It can't be
333, because each digit must be different
-------
When I add up the units column which is U + U + U,
I still get U in the sum, UUU, but I must have a
carry which would be 10, 20,30,
40 up to 90.
First I'll say the carry is 10
U + U + U = U + 10
3U = U + 10
2U = 10
U = 5 and
UUU = 555
Now I'll say the carry is 20
U + U + U = U + 20
3U = U + 20
2U = 20
U = 10
This is 2 digits, and it needs to be 1
Also, any higher carrys, 30,40, etc
would also be 2 digits
So, U = 5 and the carry is 10
I have 3 x YO5 = 555
I'll divide both sides by 3
YO5 = 555/3
YOU = 185
185 + 185 + 185 = 555 answer

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!

In the following addition problem, Y, O, and U are different digits. Solve the following problem:

I'm going to change the letter "O" to the letter "Q" 
because the letter "O" looks too much like zero.


YOU + YOU + YOU = UUU

Sorry there can't be a solution. Why?

 Y Q U
 Y Q U
 Y Q U
 -----
 U U U

Adding the rightmost column

U + U + U

If there was 0 (nothing) to carry, then

U + U + U = U
       3U = U
       2U = 0
        U = 0

If U were 0, then the next column would be

Q+Q+Q = 0 or Q+Q+Q = 10
   3Q = 0       3Q = 10
    Q = 0        Q = 3%261%2F3  

But Q can't be 0 because U = 0, and it can't be
3%261%2F3 so there had to be 1 to carry.  
Therefore

U + U + U = 10 + U
       3U = 10
        U = 5

So therefore we have:

   1
 Y Q 5
 Y Q 5
 Y Q 5
 -----
 5 5 5

Adding up the middle column with the carry,
we have

1 + Q + Q + Q = 5 or 15
  
1 + 3Q = 5   or   1 + 3Q = 15 
    3Q = 4   or       3Q = 14
     Q = 3%2F4     Q = 4%262%2F3

But Q can't be a fraction, so there
can be no solution.  Sorry.

Edwin