Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$. Consider all possible positive differences of pairs of elements of $S$. Let $N$ be the sum of all of these differences. Find $N.$
When computing $N$, the number $2^x$ will be added $x$ times (for terms $2^x-2^0$, $2^x-2^1$, $\dots,$ $2^x - 2^{x-1}$), and subtracted $10-x$ times. Hence, $N$ can be computed as $$N=10\cdot 2^{10} + 8\cdot 2^9 + 6\cdot 2^8 + \cdots - 8\cdot 2^1 - 10\cdot 2^0.$$Then
\begin{align*}
N & = 10(2^{10}-1) + 8(2^9 - 2^1) + 6(2^8-2^2) + 4(2^7-2^3) + 2(2^6-2^4) \\
& = 10(1023) + 8(510) + 6(252) + 4(120) + 2(48) \\
& = 10(1000+23) + 8(500+10) + 6(250+2) + 480 + 96 \\
&= \boxed{16398}.
\end{align*}
16398
Level 5
Intermediate Algebra
Problem: Let $S = \{2^0,2^1,2^2,\ldots,2^{10}\}$. Consider all possible positive differences of pairs of elements of $S$. Let $N$ be the sum of all of these differences. Find $N.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the sum of all positive differences of pairs of elements from the set S, we first identify the elements of S, which are powers of 2 from 2^0 to 2^10. The elements are: 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, and 1024.

Next, we consider all pairs (a, b) where a and b are elements of S and a > b. The positive difference for each pair is given by a - b. 

For each element 2^i in S, we calculate how many times it will be the larger element in the pairs. Specifically, for each 2^i, it can be paired with all previous elements 2^0, 2^1, ..., 2^(i-1). Therefore, the number of pairs where 2^i is the larger element is i.

The contribution to the sum of differences from each element 2^i can be calculated as follows:
- For each 2^i, the contribution to the sum is the value of 2^i multiplied by the number of times it appears as the larger element, which is i.
- Additionally, we need to account for the contributions from the smaller elements. Each smaller element 2^j (where j < i) contributes negatively to the sum when paired with 2^i. The total contribution from the smaller elements when paired with 2^i is the sum of all smaller elements, which is 2^0 + 2^1 + ... + 2^(i-1) = 2^i - 1.

Thus, the total contribution from each element 2^i can be expressed as:
Contribution from 2^i = i * 2^i - (2^i - 1) = i * 2^i - 2^i + 1 = (i - 1) * 2^i + 1.

Now, we sum this contribution for all i from 1 to 10 (since 2^0 has no larger elements to pair with):
N = Σ from i=1 to 10 of [(i - 1) * 2^i + 1].

Calculating this step-by-step:
- For i = 1: Contribution = 0 * 2^1 + 1 = 1.
- For i = 2: Contribution = 1 * 2^2 + 1 = 4 + 1 = 5.
- For i = 3: Contribution = 2 * 2^3 + 1 = 16 + 1 = 17.
- For i = 4: Contribution = 3 * 2^4 + 1 = 48 + 1 = 49.
- For i = 5: Contribution = 4 * 2^5 + 1 = 128 + 1 = 129.
- For i = 6: Contribution = 5 * 2^6 + 1 = 320 + 1 = 321.
- For i = 7: Contribution = 6 * 2^7 + 1 = 768 + 1 = 769.
- For i = 8: Contribution = 7 * 2^8 + 1 = 1792 + 1 = 1793.
- For i = 9: Contribution = 8 * 2^9 + 1 = 4096 + 1 = 4097.
- For i = 10: Contribution = 9 * 2^10 + 1 = 9216 + 1 = 9217.

Now, summing all contributions:
N = 1 + 5 + 17 + 49 + 129 + 321 + 769 + 1793 + 4097 + 9217 = 16398.

Thus, the final answer is 16398.<|END_OF_LATENT|> The final answer is $16398$.