What is the value of `sum _(i+j = odd 1 le | lt | le 10) (i +j) - sum_(i + j = even 1 le | lt | le 10)(i+j)` ?

Answer

To solve the problem, we need to find the value of the expression:

\[ \text{sum}_{i+j \text{ is odd}} (i+j) - \text{sum}_{i+j \text{ is even}} (i+j) \]

where \(1 \leq i, j \leq 10\).

### Step 1: Understanding the sums

First, we need to understand what it means for \(i + j\) to be odd or even.

- \(i + j\) is odd when one of \(i\) or \(j\) is odd and the other is even.
- \(i + j\) is even when both \(i\) and \(j\) are either odd or even.

### Step 2: Calculate the total sums

We will calculate the total sum of \(i + j\) for all pairs \((i, j)\) where \(1 \leq i, j \leq 10\).

The total number of pairs \((i, j)\) is \(10 \times 10 = 100\).

The sum \(S\) of all \(i + j\) can be calculated as follows:

\[ S = \sum_{i=1}^{10} \sum_{j=1}^{10} (i + j) = \sum_{i=1}^{10} \sum_{j=1}^{10} i + \sum_{i=1}^{10} \sum_{j=1}^{10} j \]

Since both sums are equal:

\[ S = 10 \sum_{i=1}^{10} i + 10 \sum_{j=1}^{10} j = 10 \cdot 55 + 10 \cdot 55 = 1100 \]

### Step 3: Count odd and even sums

Next, we need to count how many pairs produce odd and even sums.

1. **Odd sums**:
- \(i\) is odd and \(j\) is even or \(i\) is even and \(j\) is odd.
- There are 5 odd numbers (1, 3, 5, 7, 9) and 5 even numbers (2, 4, 6, 8, 10).
- The number of pairs that give odd sums is \(5 \times 5 + 5 \times 5 = 50\).

2. **Even sums**:
- Both \(i\) and \(j\) are odd or both are even.
- The number of pairs that give even sums is \(5 \times 5 + 5 \times 5 = 50\).

### Step 4: Calculate the contributions to odd and even sums

- For odd sums, the contribution can be calculated by considering the pairs:
- For odd \(i\) and even \(j\):
- The values of \(i + j\) will range from \(3\) (1 + 2) to \(19\) (9 + 10).
- For even \(i\) and odd \(j\):
- The values will also range from \(3\) to \(19\).

The contributions from odd sums and even sums can be calculated similarly, but since they are symmetric, we can conclude that:

\[ \text{sum}_{i+j \text{ is odd}} (i+j) = \text{sum}_{i+j \text{ is even}} (i+j) \]

### Step 5: Final calculation

Since both sums are equal, we have:

\[ \text{sum}_{i+j \text{ is odd}} (i+j) - \text{sum}_{i+j \text{ is even}} (i+j) = 0 \]

### Conclusion

Thus, the value of the expression is:

\[ \boxed{0} \]