I'm stumped on this one too:

s e e d
+ i c e d
----------
s p i c e

c= ___ d=___ e=___ i= ___ p= ___ s=___
Hint: Solve for s and i. Then is e is even or odd?

Thanks for your help.

c=2

d=3
e=6
i=9
p=0
s=1

To solve this puzzle, let's start step-by-step:

1. Start by writing out the equation:
```
s e e d
+ i c e d
---------
s p i c e
```

2. From the left-most column, we can see that `s + i = s`. This means that `i` must be equal to zero since any number plus zero equals itself. So, we can fill in the value for `i = 0`:
```
s e e d
+ 0 c e d
---------
s p 0 c e
```

3. Now, let's focus on the carry value. In the right-most column, we have `d + d = e`. Since `d` is a single digit number (0-9), the sum `d + d` must result in a two-digit number that ends with `e`. Knowing this, it's clear that `d` must be equal to 5, which would make `e = 0`:
```
s e e 5
+ 0 c 5
---------
s p 0 c 0
```

4. Now, we can move on to the next column from right to left. We have `e + e + c = 10 + i`. Since `i = 0`, we can simplify this to `e + e + c = 10`. As `e` can only be a single digit number (0-9) and we already have e = 0, we know that `c` must be equal to 1:
```
s e e 5
+ 0 1 5
---------
s p 0 1 0
```

5. Continuing to the left-most column, we have `0 + 0 + 1 = p`. Therefore, `p = 1`:
```
s e e 5
+ 0 1 5
---------
s 1 0 1 0
```

6. Now let's solve for `s`. Looking at the equation, `s + 0 + 0 = s`. This tells us that `s` must be equal to 1:
```
1 e e 5
+ 0 1 5
---------
1 1 0 1 0
```

7. Finally, we can determine the value of `e`. Since `e + e = 10`, it means that `e` must be an even number (0, 2, 4, 6, or 8).

So, the solution to the puzzle is:
```
1 e e 5
+ 0 1 5
---------
1 1 0 1 0
```

Since `e` must be an even number, the possible values for `e` are 0, 2, 4, 6, or 8.

To solve this type of puzzle, we need to assign each letter a unique digit from 0 to 9.

Let's start by considering the ones place in the sum. We can see that there is no carryover in the ones place, which means that d + e must be equal to e (since e + d cannot exceed 9). Therefore, d = 0.

Next, let's consider the hundreds place. Since d = 0, there is no carryover from the tens place. This means that i + c must be equal to c. The only way this can occur is if i = 0 as well, because any other digit added to 0 will result in a different number.

We now have d = 0 and i = 0. Let's substitute these values into the puzzle:

s e e 0
+ 0 c e d
----------
s p i c e

From here, we can make some observations. Since d = 0, the digit 0 is used in the sum, which means that s + c must be greater than or equal to 10 in order to have a carryover. Therefore, s must be 1, and the value of c will be greater than or equal to 9.

So far we have:

1 e e 0
+ 0 c e 0
----------
1 p i c e

Now let's consider the tens place. We can see that there is a carryover from the ones place, which means that the sum of e + e + 0 must be greater than or equal to 10. This implies that e must be greater than or equal to 5 (since 5 + 5 + 0 = 10).

We now have:

1 e e 0
+ 0 c e 0
----------
1 p i c e

-e-e- The value of e is 5 or greater.

Lastly, let's consider the thousands place. We have a carryover from the hundreds place, which means that 1 + e + c must be greater than or equal to 10. We know that e is 5 or greater, and c is 9 or greater. Therefore, the sum of 1 + e + c will always be greater than or equal to 15. This means that the value of p must be at least 5.

We now have:

1 e e 0
+ 0 c e 0
----------
1 p i c e

Here are the possible values for each letter:

c= 9 or more
d= 0
e= 5 or more
i= 0
p= 5 or more
s= 1

To determine if e is even or odd, we need to know the value of c, which could be any number higher than or equal to 9. Without more information, we cannot determine the parity of e.