2
$\begingroup$

$Given$

The six Interrelated Equations A, B, C, D, E, F

All equal to 3.

Allowed Digits.. 1 to 9.

All the missing 9 are distinct...9?s...9 Different Digits

Fillin all the digits that satisfy all these Relations.

 Six Ernst.with 3

$\endgroup$
  • $\begingroup$ On the ordering of operations, do we assume multiplication and division are performed before addition and subtraction? Also, is the bottom row and rightmost column "(?-?)+?" or "?-(?+?)" $\endgroup$ – hexomino Jun 13 at 9:30
  • $\begingroup$ Is there a unique solution and are all the values unique? $\endgroup$ – Bee Jun 13 at 9:31
  • $\begingroup$ Order of operations ..A, B, C ...3 Horizontal. D, E, F...3 Vertical...Bottom right 4 squares...blank..like other black cells. $\endgroup$ – Uvc Jun 13 at 9:36
  • $\begingroup$ Based on what I could figure out, it is unique $\endgroup$ – Uvc Jun 13 at 9:38
  • 1
    $\begingroup$ Order of operations as given would be..3+3=6,,,6x2=12....follow the operations in the order given in both horizontal and vertical sets. $\endgroup$ – Uvc Jun 13 at 9:50
3
$\begingroup$

I've found a solution, which is unique given the constraints.

 1 * 9 - 6  = 3
 *   -   -
 7 + 8 / 5  = 3
 -   *   +
 4 - 3 + 2  = 3
 =   =   =
 3   3   3

Proof

The digit in the bottom middle must be either $1$ or $3$ since the product of the second column is $3$.
If it is $1$ then the other entries in the bottom row are either $2,2$ or $1,3$ but we are not allowed to repeat digits. Hence, it must be $3$.

This means the other digits in the bottom row will be $2,4$ or $1,5$.
Now, suppose $1$ is on the left in the bottom row. This means the first product on the left-hand column is $4$ which cannot be constructed without repeating digits.
Alternatively, suppose $5$ is on the left in the bottom. Then the other digits in the left-hand column must be $2$ and $4$ since $1$ has already been used. This means the last entry in the second row is at least $6$ which is not permissible since the sum cannot be greater than $17$.
Hence, the other two entries in the bottom row must be $2$ and $4$.

Now suppose $2$ is on the left. This means that the other entries in the left-hand column are $1$ and $5$ which forces the last entry in the second row to be at least $6$, not allowed. Hence the bottom row must be $4,3,2$.

From there, we know that the other entries in the left-hand column are $1$ and $7$.
This means the last entry in the second row must be $5$ since it is the only option available that could possibly work. Hence the sum of the first two digits in the second row is $15$ which makes the first entry $7$ and the second $8$.
This means the top left-hand corner must be $1$ and we are left with $9$ and $6$ to complete the grid.

$\endgroup$
  • $\begingroup$ This morning, my mind is not sharp..forgot to mention that there are no repeating digits...will edit to reflect that.. $\endgroup$ – Uvc Jun 13 at 9:55
  • $\begingroup$ Got it !!...sorry about all those edits $\endgroup$ – Uvc Jun 13 at 10:04
  • $\begingroup$ @Uvc No, that's fine. Nice puzzle! $\endgroup$ – hexomino Jun 13 at 10:10
2
$\begingroup$

Here's a solution:

 1 * 3 - 0  = 3
 *   -   -
 3 + 0 / 1  = 3
 -   *   +
 0 - 1 + 4  = 3
 =   =   =
 3   3   3

$\endgroup$
  • $\begingroup$ I forgot to mention..that only 1 to 9 are allowed...will edit it now..sorry about that $\endgroup$ – Uvc Jun 13 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.