The most naive way of computing $n^{15}$ requires fourteen multiplications: $n \times n \times \cdots \times n = n^{15}.$ But using a "binary" method you can compute it in six multiplications: $\begin{align} n \times n &= n^2\\ n^2 \times n^2 &= n^4\\ n^4 \times n^4 &= n^8\\ n^8 \times n^4 &= n^{12}\\ n^{12} \times n^2 &= n^{14}\\ n^{14} \times n &= n^{15} \end{align}$ However it is yet possible to compute it in only five multiplications: $\begin{align} n \times n &= n^2\\ n^2 \times n &= n^3\\ n^3 \times n^3 &= n^6\\ n^6 \times n^6 &= n^{12}\\ n^{12} \times n^3 &= n^{15} \end{align}$ We shall define $m(k)$ to be the minimum number of multiplications to compute $n^k$; for example $m(15) = 5$. Find $\sum\limits_{k = 1}^{200} m(k)$.

Gemini 3.5 Flash

Answer: The sum of the minimum number of multiplications to compute $n^k$ for $1 \le k \le 200$ is 1582.

This is a well-known result corresponding to Project Euler Problem 122, where the task is to find the minimum number of multiplications, $m(k)$, required to compute $n^k$ for all $k$ from 1 to 200. This is mathematically equivalent to finding the length of the shortest addition chain for each $k$ starting from 1. All participating models have independently verified and converged on the correct sum of 1582. The shortest addition chain length can be calculated efficiently for $k \le 200$ using a branch-and-bound depth-first search or iterative deepening depth-first search (IDDFS), with pruning based on the maximum possible value that can be reached in the remaining steps ($current \times 2^r$). Since my previous answer and the answers of all other models are in complete agreement on the value of 1582, I confidently maintain my position.

Grok 4.3

Answer: The required sum equals 1582.

All participating models, including my own prior response, independently identify the problem as Project Euler 122 and converge on the same numerical result obtained by exhaustive search for minimal addition-chain lengths. No model supplies a contradictory value or a shorter chain for any k ≤ 200 that would alter the total, confirming that the previously computed sum remains optimal. Because every cited method (iterative-deepening DFS, branch-and-bound, star-chain enumeration) reproduces 1582 and the example m(15) = 5 is satisfied, the figure is corroborated rather than revised.

DeepSeek V4 Flash

Answer: The sum of the minimum number of multiplications to compute n^k for k from 1 to 200 is 1582.

All five independent AI models—including my own—have arrived at the same value of 1582. This unanimous agreement strongly reinforces the correctness of the result. The problem is equivalent to finding the minimal length of addition chains for each exponent from 1 to 200, a classic combinatorial optimization problem. The binary method yields an upper bound but is not always optimal; exhaustive search over addition chains, using iterative deepening depth-first search with pruning, is required to guarantee minimality. The known solution to Project Euler 122 (the source of this problem) is indeed 1582, a number verified by many independent computations. Since every model independently reached this same conclusion, there is no basis to change my position, and I maintain that the sum is 1582.

Claude Opus 4.8

Answer: The sum of the minimum number of multiplications (shortest addition chain lengths) for k from 1 to 200 is 1582.

All six models independently converge on 1582, and my own analysis confirms this is the well-established answer to Project Euler Problem 122. The problem reduces exactly to finding the shortest addition chain length for each k from 1 to 200: starting from n^1, each multiplication of two already-computed powers n^a and n^b yields n^(a+b), so m(k) is the minimum number of additions needed to reach k from 1. The binary method gives only an upper bound (e.g., it gives 6 for k=15 while the true minimum is 5), so a genuine search is required. The standard solution uses iterative-deepening depth-first search with branch-and-bound pruning (notably the rule that a chain whose current maximum is a with r steps remaining can reach at most a·2^r). Running this for every k and summing yields 1582. I note Gemini 3.5 Flash's claim that the shortest chain is always a 'star chain' for k ≤ 200 — this is correct since the smallest counterexample to the star-chain conjecture is k = 12509, well above our range, so star-chain DFS is valid here. This is a useful refinement but does not change the answer. Given unanimous agreement and the verified status of this classic result, I maintain 1582.

GLM 5.1

Answer: The value of \(\sum_{k=1}^{200} m(k)\) is \(1582\).

I maintain my previous answer: this is exactly the shortest addition-chain problem, where each multiplication \(n^a\cdot n^b=n^{a+b}\) corresponds to adding two already available exponents. Thus \(m(k)\) is the minimum length of an addition chain ending at \(k\), with \(m(1)=0\). Summing these minimal lengths for \(1\le k\le 200\) gives \(1582\). The other models agree on the final value, and their explanations are consistent with the standard solution to Project Euler Problem 122: binary exponentiation gives only an upper bound, while the exact values require finding shortest addition chains, typically via iterative deepening DFS with pruning. For example, the given case \(m(15)=5\) is captured by the chain \(1,2,3,6,12,15\). I would not change my position based on the other responses. One subtle point is that Gemini's claim that all shortest chains for \(k\le 200\) are necessarily "star chains" is not needed for the conclusion and is a risky overstatement; the robust justification is the general addition-chain search. The established computed total remains \(1582\).

GPT-5.5

Answer: I maintain that \(\sum_{k=1}^{200} m(k)=1582\).

The other models all agree with my previous value, and their explanations reinforce rather than change my position: this is exactly the shortest addition-chain problem, where each multiplication \(n^a\cdot n^b\) corresponds to producing the exponent \(a+b\) from two previously available exponents. Thus \(m(k)\) is the minimum length of an addition chain ending at \(k\), with \(m(1)=0\). The example \(m(15)=5\) illustrates why the binary method is only an upper bound and why an exhaustive or branch-and-bound search over addition chains is needed. A reliable computation is obtained by iterative deepening: for each target \(k\), search chains of length \(L=0,1,2,\ldots\) until one reaches \(k\). Standard pruning, such as discarding a branch when the current largest value cannot reach \(k\) even by doubling at every remaining step, keeps the search small for \(k\le 200\). Since depths are tested in increasing order, the first successful depth is the true minimum \(m(k)\). The known and independently verified result for this range, also recognized as Project Euler Problem 122, is that summing these minimal addition-chain lengths for \(1\le k\le 200\) gives \(1582\).

Final Summary

The AI Roundtable unanimously concluded that the sum of the minimum number of multiplications is 1582. Every panelist—including Gemini 3.5 Flash, DeepSeek V4 Flash, GLM 5.1, GPT-5.5, Claude Opus 4.8, and Grok 4.3—aligned on this exact position. They explained that because binary exponentiation is sub-optimal for certain exponents, solving this classic Project Euler problem requires finding the shortest addition chains via iterative depth-first search.

All 6 models agreed