Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$7*7~$ is $~25$
$6*6~$ is $~18$
$9*9~$ is $~41$

When is this true?


Think inside the box.

share|improve this question
When Common Core says it is. – EnragedTanker Mar 3 at 19:34
@crayzeedude Common Core is just a standard for what should be taught at what level - it's pretty unobjectionable. The problem is with parents and teachers who make misinterpretations viral rather than accepting change. – Deusovi Mar 4 at 1:22
@Deusovi And I completely get that. I'm mostly just poking fun at some of the ridiculous stuff that parents of elementary schools post. – EnragedTanker Mar 4 at 1:56
@Deusovi Common core is more than just a standard. It is also a prescription of how exactly something should be learned. Case in point: Math addition. They want to force kids to understand 7 + 7 as 7 + (3 + 4) = 10 + 4. That is not the first step in learning addition. That is a higher level concept that some kids intuitively get, others don't until much much later. – Καrτhικ Mar 4 at 20:57
@Καrτhικ "Force kids to understand" - isn't that just called teaching? You make it sound like knowledge is something bad. (Oh, and it's not the first step - it's taught after addition.) If you want to continue this conversation, we should probably move to chat. – Deusovi Mar 4 at 23:42
up vote 17 down vote accepted

These are true,

if you define $~x*y~$ as the number of black squares on a chessboard with $x$ rows and $y$ columns, where the lower left corner square is black.


define $~x*y=\lceil xy/2 \rceil$, that is, as the product of $x$ and $y$, divided by $2$, and then rounded up to the next integer.

With this,

$7*7 ~=~ \lceil 49/2\rceil ~=~ \lceil 24.5 \rceil ~=~ 25$
$6*6 ~=~ \lceil 36/2\rceil ~=~ \lceil 18~~~\rceil ~=~ 18$
$9*9 ~=~ \lceil 81/2\rceil ~=~ \lceil 40.5 \rceil ~=~ 41$

share|improve this answer
Wouldn't $\lceil\frac{xy}{2}\rceil$ work just as well and be marginally simpler? – frodoskywalker Mar 3 at 20:04
This was my thought too. – Improve Mar 3 at 20:57
@frodoskywalker: yes, the two formulas are equivalent; your formulation is simpler to parse; I'll update my answer. – Gamow Mar 4 at 10:05

Because $*$ means:

$\left\lceil\frac{n\times m}{2}\right\rceil$

share|improve this answer
I'm not understanding this one. Could you please elaborate? – LN6595 Mar 3 at 22:54
@LN6595 ceil means ceiling which is the mathematical term for rounding up. So what he does is multiply the two numbers, divide it by two and round that number up – Ivo Beckers Mar 3 at 23:16

It is

The maximum number of knights on a board of that size (i.e. width is the first number and height is the second number) where none of them can attack another. (Assuming it's free-for-all chess for some reason)
And it's inside the "box" because the box is a chess board.

It's also just how many black squares would be on a chess/checkers board, as @Sleafer pointed out. It turns out I over complicated it.

You would continue the sequence

$$1*1=1\\2*2=4\text{ (or }2\text{ if counting squares)}\\3*3=5\\4*4=8\\5*5=13\\6*6=18\\7*7=25\\8*8=32\\9*9=41\\10*10=50$$

share|improve this answer
Or simply the number of black squares. – Sleafar Mar 3 at 20:26
@Sleafar No, it's more than the number of black squares, if you place it properly. I'll make a .gif showing 1x1 up to 10x10. – Artyer Mar 3 at 20:28
For any n > 2 the number should be the same, at least it is for all you posted. – Sleafar Mar 3 at 20:32
@Sleafar It turns out I was counting an inverted board's squares – Artyer Mar 3 at 20:38
If you put 4 knights on a 2x2 board, none of them attack each other. – Kiran Linsuain Mar 4 at 0:16

My take on it is you would change the counting base from decimal (base 10), so 7 * 7 = 49 in decimal, which is 25 in base 22 ((22 * 2) + 5).

Thus 6 * 6 = 36 in b10, which is 18 in b28, and 9 * 9 = 81 in b10, which is 41 in b20.

share|improve this answer
What's the system for which base to use? – charfellow Mar 4 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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