4
$\begingroup$

A three-by-three square grid with a blue background containing: ((four red squares and one yellow square)(four red squares)(three red squares and one yellow square))((three red squares and three yellow squares)(three red squares and two yellow squares)(two red squares and three yellow squares))((one red square and two yellow squares)(one red square and one yellow square)(two yellow squares))

This image represents, in a fundamental manner, a square containing many squares.

Hints:

The tag is important. The image represents a grid of numbers. The correct interpretation will not give the solution directly, you'll need to take it a step further.

Red is the first, yellow the second. Blue is just another primary color.

Everything must be squared!

$\endgroup$
9
  • 6
    $\begingroup$ And what is the question? $\endgroup$
    – Florian F
    Commented Feb 26, 2023 at 20:45
  • 1
    $\begingroup$ @FlorianF I was deliberately vague, not wanting to add hints that were too direct. The tag mathematics is important. The image represents numbers. Perhaps 'fundamental' would have been more accurate than 'primitive'. $\endgroup$ Commented Feb 26, 2023 at 21:51
  • $\begingroup$ "a square containing many squares": 49 squares, actually, if you include the blue ones. 36 if you don't. Those are both square numbers. I'm guessing that's merely a coincidence. $\endgroup$
    – msh210
    Commented Feb 27, 2023 at 5:10
  • $\begingroup$ @FlorianF The numbers are distinct, but at least you are looking in the right direction. $\endgroup$ Commented Feb 27, 2023 at 20:41
  • 2
    $\begingroup$ @DanielMathias No, because I still don't know what the question is. $\endgroup$
    – Florian F
    Commented Mar 5, 2023 at 18:42

1 Answer 1

6
$\begingroup$

First of all, various implicit and explicit hints.

"In a fundamental manner": this is gesturing towards the so-called Fundamental Theorem of Arithmetic, which is the thing that says that integers have basically-unique prime factorizations.
"You'll need to take it a step further ... Everything must be squared!": when we have obtained some numbers in a reasonably straightforward way we will be squaring them.
"Red is the first, yellow the second. Blue is just another primary color." In keeping with the "fundamental" thing, this is clearly pointing at prime numbers.

OK, so what does this mean?

The most obvious thing to do, given the above, is to treat e.g. a square containing four red squares and one yellow square as denoting the number $2^43^1=16\cdot3=48$. This would give us [48, 16, 24 | 216, 72, 108 | 18, 6, 9]. I worried that the reference to blue as a third primary colour might mean that there are some blue squares in the boxes and we just can't see them against the blue background, or something, but fortunately this turns out not to be so.

OK, so

square the numbers, getting [2304 256 576 | 46656 5184 11664 | 324 36 81]. But what now? You might e.g. hope for this to be a magic square, but obviously it isn't since the rows are of very different sizes.

But

something magic-square-like is true: if you take any row, column, or full diagonal then the sum of the numbers in those cells is a square. As is the sum of all the numbers in the 3x3 grid. (There are some other subsets that add up to squares to, but I think these are the specifically-intended ones.)

$\endgroup$
5
  • $\begingroup$ You have the grid correct, but you have overlooked something. The arrangement is definitely not a magic square, but it is also not arbitrary. $\endgroup$ Commented Mar 10, 2023 at 2:02
  • $\begingroup$ I thought maybe there was rot13(Bar oyhr fdhner sbe rnpu 1/9gu bs gur ovt fdhner, znlor ercerfragvat 5?) $\endgroup$
    – Chengarda
    Commented Mar 10, 2023 at 3:03
  • $\begingroup$ Note: the comments above were written when my answer was somewhat different from what it is now. $\endgroup$
    – Gareth McCaughan
    Commented Mar 10, 2023 at 19:55
  • $\begingroup$ The solution grid has the smallest total sum for an all-squares version of this question. $\endgroup$ Commented Mar 10, 2023 at 21:34
  • $\begingroup$ @DanielMathias Neat! $\endgroup$
    – Gareth McCaughan
    Commented Mar 11, 2023 at 16:43

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