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My best offhand 960, achieved as $10 \times 8 \times 4 \times 3 = (7+9) \times 6 \times 5 \times 2 \times 1$. Mostly from trial and error: assuming that generally you get the biggest numbers thro...
#3: Post edited
by
Karl Knechtel
·
2026-03-11T23:44:58Z (6 months ago)
correct terminology
- <details><summary>My best offhand</summary>
- **960**, achieved as $10 \times 8 \times 4 \times 3 = (7+9) \times 6 \times 5 \times 2 \times 1$.
- Mostly from trial and error: assuming that generally you get the biggest numbers through multiplication.
- I started with $10 \times 9 \times 8 \times 7$ versus $6!$ (which is obviously off by a factor of 7), and noticed that I could shuffle the 7 across to make it work out: $10 \times 9 \times 8 = (7-1) \times 6 \times 5 \times 4 \times (3-2)$. The idea is that 6 is much easier to work with than 7 here; the added factor of 6 replaces the 3 and 2, which can in turn be subtracted to remove them.
- From there, by swapping the 9 with 3 and 4, the left-hand side increases, and the 2 can be used for multiplication; then with 9, 7 and 1 I had to make 16 which is straightforward.
- This was just the first of several ideas for improving on 720 that worked out. I suspect that the true maximum does not involve multiplying by 1, since that wastes the number entirely (and it could trivially be put on either side).
- </details>
<details><summary>But since the reference includes a solution,</summary>- The above can easily enough be modified to get to the maximum cited result of **1920**. As I suspected, adding 1 instead of multiplying by 1 is useful: specifically, $7 \times 9 + 1$ is 64, exactly 4 times the $7+9$ I used. Swapping the 2 across to the left-hand side corrects that imbalance at double the value. The resulting equation is $10 \times 8 \times 4 \times 3 \times 2 = (7 \times 9 + 1) \times 6 \times 5$, equivalent to the cited solution (just with the $10 \times 3$ and $6 \times 5$ parts swapped).
- </details>
- <details><summary>Can we prove that result?</summary>
- I can only offer an informal argument and hand-waving here.
- From the citation,
- > When k=1, Bryce Herdt pointed out the optimal expression is $n·(n–1)·. . .·(2+1) = (3/2)n!$
- The reasoning behind this is straightforward. It also follows that for $k=2$, the result is upper-bounded by the square root of this total, since otherwise we could improve the result for $k=1$ by taking the product of the left-hand and right-hand sides in our $k=2$ result. In our case, the highest integer less than that square root is 2333, so there is not a lot of search space. We can also be sure that our result is fairly "smooth" (has a lot of factors) because of the need to multiply to get larger results. We can also be fairly sure that addition is used at most once in the optimum result, and multiplication everywhere else.
- But for example, we can fairly easily rule out 2310, because this would require us to make two factors of 7 and also two factors of 11 (one for each side of the equation). It seems reasonable that we could write an equation with $7$ on one side and $6+1$ on the other, but the 11s would require addition (or even subtraction).
- Ideally we would like to add our 1 to a smaller number, so that it has more impact. Again, adding it to 6 makes sense; then we've easily dealt with the awkward 7. But the remaining numbers multiply to 86400, which is not a perfect square; so we cannot possibly use pure multiplication with them. Similarly, if we start with $7 + 1 = 8$, or with $2 \times 7+1 = 3 \times 5$, the remaining numbers don't multiply to a perfect square, never being able to split them into two equal products.
- A search based on principles like this should disprove higher results fairly easily.
- </details>
- <details><summary>My best offhand</summary>
- **960**, achieved as $10 \times 8 \times 4 \times 3 = (7+9) \times 6 \times 5 \times 2 \times 1$.
- Mostly from trial and error: assuming that generally you get the biggest numbers through multiplication.
- I started with $10 \times 9 \times 8 \times 7$ versus $6!$ (which is obviously off by a factor of 7), and noticed that I could shuffle the 7 across to make it work out: $10 \times 9 \times 8 = (7-1) \times 6 \times 5 \times 4 \times (3-2)$. The idea is that 6 is much easier to work with than 7 here; the added factor of 6 replaces the 3 and 2, which can in turn be subtracted to remove them.
- From there, by swapping the 9 with 3 and 4, the left-hand side increases, and the 2 can be used for multiplication; then with 9, 7 and 1 I had to make 16 which is straightforward.
- This was just the first of several ideas for improving on 720 that worked out. I suspect that the true maximum does not involve multiplying by 1, since that wastes the number entirely (and it could trivially be put on either side).
- </details>
- <details><summary>But since the reference includes a "best known answer",</summary>
- The above can easily enough be modified to get to the maximum cited result of **1920**. As I suspected, adding 1 instead of multiplying by 1 is useful: specifically, $7 \times 9 + 1$ is 64, exactly 4 times the $7+9$ I used. Swapping the 2 across to the left-hand side corrects that imbalance at double the value. The resulting equation is $10 \times 8 \times 4 \times 3 \times 2 = (7 \times 9 + 1) \times 6 \times 5$, equivalent to the cited solution (just with the $10 \times 3$ and $6 \times 5$ parts swapped).
- </details>
- <details><summary>Can we prove that result?</summary>
- I can only offer an informal argument and hand-waving here.
- From the citation,
- > When k=1, Bryce Herdt pointed out the optimal expression is $n·(n–1)·. . .·(2+1) = (3/2)n!$
- The reasoning behind this is straightforward. It also follows that for $k=2$, the result is upper-bounded by the square root of this total, since otherwise we could improve the result for $k=1$ by taking the product of the left-hand and right-hand sides in our $k=2$ result. In our case, the highest integer less than that square root is 2333, so there is not a lot of search space. We can also be sure that our result is fairly "smooth" (has a lot of factors) because of the need to multiply to get larger results. We can also be fairly sure that addition is used at most once in the optimum result, and multiplication everywhere else.
- But for example, we can fairly easily rule out 2310, because this would require us to make two factors of 7 and also two factors of 11 (one for each side of the equation). It seems reasonable that we could write an equation with $7$ on one side and $6+1$ on the other, but the 11s would require addition (or even subtraction).
- Ideally we would like to add our 1 to a smaller number, so that it has more impact. Again, adding it to 6 makes sense; then we've easily dealt with the awkward 7. But the remaining numbers multiply to 86400, which is not a perfect square; so we cannot possibly use pure multiplication with them. Similarly, if we start with $7 + 1 = 8$, or with $2 \times 7+1 = 3 \times 5$, the remaining numbers don't multiply to a perfect square, never being able to split them into two equal products.
- A search based on principles like this should disprove higher results fairly easily.
- </details>
#2: Post edited
by
Karl Knechtel
·
2026-03-11T23:44:00Z (6 months ago)
Sort factors; add derivation of the cited answer from my attempt; sketch of proof
- <details><summary>My best offhand</summary>
**960**, achieved as $10 \times 3 \times 8 \times 4 = (7+9) \times 6 \times 5 \times 2 \times 1$.- Mostly from trial and error: assuming that generally you get the biggest numbers through multiplication.
- I started with $10 \times 9 \times 8 \times 7$ versus $6!$ (which is obviously off by a factor of 7), and noticed that I could shuffle the 7 across to make it work out: $10 \times 9 \times 8 = (7-1) \times 6 \times 5 \times 4 \times (3-2)$. The idea is that 6 is much easier to work with than 7 here; the added factor of 6 replaces the 3 and 2, which can in turn be subtracted to remove them.
- From there, by swapping the 9 with 3 and 4, the left-hand side increases, and the 2 can be used for multiplication; then with 9, 7 and 1 I had to make 16 which is straightforward.
- This was just the first of several ideas for improving on 720 that worked out. I suspect that the true maximum does not involve multiplying by 1, since that wastes the number entirely (and it could trivially be put on either side).
</detail>
- <details><summary>My best offhand</summary>
- **960**, achieved as $10 \times 8 \times 4 \times 3 = (7+9) \times 6 \times 5 \times 2 \times 1$.
- Mostly from trial and error: assuming that generally you get the biggest numbers through multiplication.
- I started with $10 \times 9 \times 8 \times 7$ versus $6!$ (which is obviously off by a factor of 7), and noticed that I could shuffle the 7 across to make it work out: $10 \times 9 \times 8 = (7-1) \times 6 \times 5 \times 4 \times (3-2)$. The idea is that 6 is much easier to work with than 7 here; the added factor of 6 replaces the 3 and 2, which can in turn be subtracted to remove them.
- From there, by swapping the 9 with 3 and 4, the left-hand side increases, and the 2 can be used for multiplication; then with 9, 7 and 1 I had to make 16 which is straightforward.
- This was just the first of several ideas for improving on 720 that worked out. I suspect that the true maximum does not involve multiplying by 1, since that wastes the number entirely (and it could trivially be put on either side).
- </details>
- <details><summary>But since the reference includes a solution,</summary>
- The above can easily enough be modified to get to the maximum cited result of **1920**. As I suspected, adding 1 instead of multiplying by 1 is useful: specifically, $7 \times 9 + 1$ is 64, exactly 4 times the $7+9$ I used. Swapping the 2 across to the left-hand side corrects that imbalance at double the value. The resulting equation is $10 \times 8 \times 4 \times 3 \times 2 = (7 \times 9 + 1) \times 6 \times 5$, equivalent to the cited solution (just with the $10 \times 3$ and $6 \times 5$ parts swapped).
- </details>
- <details><summary>Can we prove that result?</summary>
- I can only offer an informal argument and hand-waving here.
- From the citation,
- > When k=1, Bryce Herdt pointed out the optimal expression is $n·(n–1)·. . .·(2+1) = (3/2)n!$
- The reasoning behind this is straightforward. It also follows that for $k=2$, the result is upper-bounded by the square root of this total, since otherwise we could improve the result for $k=1$ by taking the product of the left-hand and right-hand sides in our $k=2$ result. In our case, the highest integer less than that square root is 2333, so there is not a lot of search space. We can also be sure that our result is fairly "smooth" (has a lot of factors) because of the need to multiply to get larger results. We can also be fairly sure that addition is used at most once in the optimum result, and multiplication everywhere else.
- But for example, we can fairly easily rule out 2310, because this would require us to make two factors of 7 and also two factors of 11 (one for each side of the equation). It seems reasonable that we could write an equation with $7$ on one side and $6+1$ on the other, but the 11s would require addition (or even subtraction).
- Ideally we would like to add our 1 to a smaller number, so that it has more impact. Again, adding it to 6 makes sense; then we've easily dealt with the awkward 7. But the remaining numbers multiply to 86400, which is not a perfect square; so we cannot possibly use pure multiplication with them. Similarly, if we start with $7 + 1 = 8$, or with $2 \times 7+1 = 3 \times 5$, the remaining numbers don't multiply to a perfect square, never being able to split them into two equal products.
- A search based on principles like this should disprove higher results fairly easily.
- </details>
#1: Initial revision
by
Karl Knechtel
·
2026-03-11T23:13:42Z (6 months ago)
<details><summary>My best offhand</summary> **960**, achieved as $10 \times 3 \times 8 \times 4 = (7+9) \times 6 \times 5 \times 2 \times 1$. Mostly from trial and error: assuming that generally you get the biggest numbers through multiplication. I started with $10 \times 9 \times 8 \times 7$ versus $6!$ (which is obviously off by a factor of 7), and noticed that I could shuffle the 7 across to make it work out: $10 \times 9 \times 8 = (7-1) \times 6 \times 5 \times 4 \times (3-2)$. The idea is that 6 is much easier to work with than 7 here; the added factor of 6 replaces the 3 and 2, which can in turn be subtracted to remove them. From there, by swapping the 9 with 3 and 4, the left-hand side increases, and the 2 can be used for multiplication; then with 9, 7 and 1 I had to make 16 which is straightforward. This was just the first of several ideas for improving on 720 that worked out. I suspect that the true maximum does not involve multiplying by 1, since that wastes the number entirely (and it could trivially be put on either side). </detail>