-4
$\begingroup$

You have numbers 1, 2, 3, 4, 5, and 6.

enter image description here

It is required to put the numbers in triangles (one number in a triangle), so that when cutting a hexagon into two parts, the product of all the numbers in the first part is divided by the sum of all the numbers in the second one.

You can cut through one of the three straight lines.

Question. Can you put numbers and cut the hexagon?

$\endgroup$
7
  • 2
    $\begingroup$ By cutting, do you mean cutting through one of the three straight lines? Also, should the property hold for all three lines or only one? $\endgroup$
    – Bubbler
    Nov 10 '20 at 9:08
  • $\begingroup$ @Bubbler, three straight lines are three diametrs. You can cut through two of the six straight lines. $\endgroup$
    – Nick
    Nov 10 '20 at 9:17
  • $\begingroup$ By two of the six straight lines, do you mean that I can choose two lines that do not form a diameter, like the left and right sides of 1 (which makes the puzzle trivial)? $\endgroup$
    – Bubbler
    Nov 10 '20 at 9:20
  • 4
    $\begingroup$ If so, it is really trivial: I can fill in the numbers randomly and cut out the single 1 (or single 3 for that matter) because 1 divides 20 (and 3 divides 18). $\endgroup$
    – Bubbler
    Nov 10 '20 at 9:51
  • 2
    $\begingroup$ This problem seems like something a math teacher might use as a tool for teaching the concept of divisibility at grade school. Certainly there is no puzzle of any kind involved. $\endgroup$
    – Bass
    Nov 10 '20 at 11:19
2
$\begingroup$

Yes, this is easy to do:

One option is to place 1, 2, and 4 in one group, and 3, 5, and 6 in the other. Then the sums are 7 and 14, and 14 evenly divides into 7.

The only actual constraint in finding a solution is that the smaller group must sum to 1, 3, or 7. This is easily satisfied.

$\endgroup$
1
  • $\begingroup$ I am sorry, in my question after your answer, I encountered a bug. Instead of "the sum of all the numbers in the first part" should be "the product of all the numbers in the first part" $\endgroup$
    – Nick
    Nov 10 '20 at 14:18

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