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I decided to make another one of these, because they are fun and this one is rather different.

Can you place all numbers from 1 to 16 into cells, such that the following 8 equations hold? Note that the operator "/" only works for non-remainder division, i.e. you can have "8 / 4" but not "8 / 3". As usual multiplication and division are performed before addition and subtraction. Good luck!

enter image description here

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    $\begingroup$ Wow, this one feels much harder to me. $\endgroup$ – Bubbler Oct 8 at 1:29
  • $\begingroup$ Hehe. I didn't plan to make it harder. I thought it would have similar complexity, but require different constraints. $\endgroup$ – Dmitry Kamenetsky Oct 8 at 1:48
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    $\begingroup$ Is $6 \times 1 / 2 = 3$ a valid substitute for the third line, even though $1/2$ is not an integer? $\endgroup$ – Jeremy Dover Oct 8 at 21:11
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    $\begingroup$ No. You can have 1 X 6/2 = 3 though. Thanks for bringing this up. $\endgroup$ – Dmitry Kamenetsky Oct 8 at 21:28
  • $\begingroup$ That does make it a lot easier! $\endgroup$ – Paul Panzer Oct 8 at 21:36
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Solution:

$\begin{matrix}1&+&9&+&5&=&15\\\times&&-&&+&&-\\10&+&12&/&2&=&16\\+&&/&&\times&&/\\4&\times&6&/&3&=&8\\=&&=&&=&&=\\14&/&7&+&11&=&13\end{matrix}$

Reasoning:

$\begin{matrix}\diamond\ast&+&\diamond&+&\diamond&=& \\\times&&-&&+&&-\\\ast&+& &/&\bigcirc&=& \\+&&/&&\times&&/\\\bigcirc&\times&\bigcirc&/&\bigcirc&=&\bigcirc\\=&&=&&=&&=\\ &/&\bigcirc&+& &=& \end{matrix}$

We note that all denominators must be $\le8$ ("small", marked $\bigcirc$) as must be a2 (in checkerboard coordinates) because $b2/c2 \ge 2$ and at least two of those marked $\diamond$ and at least one of those marked $\ast$. Because only 8 small numbers are available a4 must be small and one of b4,c4. All others must be $>8$ ("large", marked $\Box$):

$\begin{matrix}\bigcirc&+&\Box\bigcirc&+&\Box\bigcirc&=&\Box\\\times&&-&&+&&-\\\Box&+&\Box&/&\bigcirc&=&\Box\\+&&/&&\times&&/\\\bigcirc&\times&\bigcirc&/&\bigcirc&=&\bigcirc\\=&&=&&=&&=\\\Box&/&\bigcirc&+&\Box&=&\Box\end{matrix}$


and we see that a4 must be $1$

Observe that $d4 = d3/d2 + a1/b1 + c2\times c3 + c4 \ge 2 + 2 + 2\times 3 + c4$ Therefore c4 must be small and b4 large.

$\begin{matrix}1&+&\Box&+&\bigcirc&=&13\ldots 16\\\times&&-&&+&&-\\\Box&+&\Box&/&\bigcirc&=&\Box\\+&&/&&\times&&/\\\bigcirc&\times&\bigcirc&/&\bigcirc&=&\bigcirc\\=&&=&&=&&=\\\Box&/&\bigcirc&+&9\ldots 12&=&11\ldots 14\end{matrix}$

Next, we see that $2$ must be in either c2 or c3:

$\begin{matrix}1&+&9\ldots 11&+&3\ldots 6&=&14\ldots 16\\\times&&-&&+&&-\\\Box&+&\Box&/&2\ldots 4&=&\Box\\+&&/&&\times&&/\\\bigcirc&\times&\bigcirc&/&2\ldots 4&=&\bigcirc\\=&&=&&=&&=\\\Box&/&\bigcirc&+&10\ldots 12&=&12\ldots 14\end{matrix}$

We are almost done! Observe that in the products $a2\times b2 = c2\times d2$ the primes $5$ and $7$ cannot occur because they are available only once.

$\begin{matrix}1&+&9\ldots 10&+&5&=&15\ldots 16\\\times&&-&&+&&-\\\Box&+&\Box&/&2\ldots 3&=&\Box\\+&&/&&\times&&/\\\bigcirc&\times&\bigcirc&/&2\ldots 3&=&\bigcirc\\=&&=&&=&&=\\14&/&7&+&11&=&13\end{matrix}$

We can finish by filling in column a,

$\begin{matrix}1&+&9&+&5&=&15\\\times&&-&&+&&-\\10&+&\Box&/&2\ldots 3&=&\Box\\+&&/&&\times&&/\\4&\times&\bigcirc&/&2\ldots 3&=&\bigcirc\\=&&=&&=&&=\\14&/&7&+&11&=&13\end{matrix}$

row 3 and, finally, row 2.

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  • $\begingroup$ Very well done and great explanation! I was getting worried that no one will solve this... $\endgroup$ – Dmitry Kamenetsky Oct 10 at 0:46
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Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left).

Define a low digit = 1-8, high digit = 9-16. The cells, b1,a2,b2,c2,d2,c3 are all low. The top row contains two low numbers otherwise d4 > 16. This accounts for all eight low numbers and therefore a3 is high. Ergo, a4=1 to avoid a1>16. Now consider the chain c4 < c1 < d1 < d4. The minimum differences are c1-c4 >= 6, d1-c1 >= 2, d4-d1 >= 2. We soon find a2,c2,c3 must be some permutation of 2,3,4. Therefore c4 is at least 5 and d4 is at least 15.

Since 13 is prime, it must go in a3 or d1. But if a3 = 13 then a1,d3,d4 are all 15 or greater, contradiction. Therefore d1=13, c1=11, c4=5. We have a 2,3 pair in c2,c3 so a2 = 4. Therefore d2=8 and d3=16. Since b2/c2 = 2 we have 4*6/3=8 in row 3 and the rest is easy

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  • $\begingroup$ You got it my friend! $\endgroup$ – Dmitry Kamenetsky Oct 10 at 0:44

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