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Just received an invitation from a fellow mathematician for a party. The invitation is quite clear: where I need to be at what time is clearly said in the invitation, but at the bottom of the invitation he posted a P.S. as follows:

  • Word 1 (2 letters): $[3*3]+[1*1]+[2*2]+[2*2]$
  • Word 2 (5 letters): $((-[1*1] + [3*3] + [4*4] + [8*8] + [27*27]) \cdot [2*2] + [1*1] + [1*1] + [5*5] \cdot [27*27]) \cdot [2*2]$
  • Word 3 (2 letters): $-[3*3]/2 + [9*9] + [15*15]$
  • Word 4 (2 letters): $[3*3] + [1*1] + [2*2] + [2*2]$
  • Word 5 (6 letters): $-[2*2] - [1*1] + [5*5] + [7*7] + [29*29] + [122*122] + [1791*1791]$

Not sure what he means, the only hint he gave me is as follows:

enter image description here

Can you guys help me out finding what the P.S. says?

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Well, judging from the hint, each $[m\ast n]$ means

the total number of squares (of any size) to be found in an $m\times n$ square grid. The formula for this is $$\sum_{i=0}^{\min(m,n)-1}(m-i)(n-i).$$

In our case, we seem to have only $[n\ast n]$, in which case

the formula is simply $$\sum_{i=1}^{n}i^2=\frac{n(n+1)(2n+1)}{6}.$$

So rewriting all the expressions we've been given:

  • Word 1 (2 letters): $[3*3]+[1*1]+[2*2]+[2*2]$

    $=(14)+(1)+(5)+(5)=25$

  • Word 2 (5 letters): $((-[1*1] + [3*3] + [4*4] + [8*8] + [27*27]) \cdot [2*2] + [1*1] + [1*1] + [5*5] \cdot [27*27]) \cdot [2*2]$

    $=((-(1)+(14)+(30)+(204)+(6930))\cdot(5)+(1)+(1)+(55)\cdot(6930))\cdot(5)=(7177\cdot5+2+381150)\cdot5=2085185$

  • Word 3 (2 letters): $-[3*3]/2 + [9*9] + [15*15]$

    $=-(14)/2+(285)+(1240)=1518$

  • Word 4 (2 letters): $[3*3] + [1*1] + [2*2] + [2*2]$

    $=(14)+(1)+(5)+(5)=25$, same as Word 1.

  • Word 5 (6 letters): $-[2*2] - [1*1] + [5*5] + [7*7] + [29*29] + [122*122] + [1792*1792]$

    Here there is a mistake, as it should actually be $1791$. (We'll see why later.)

    $=-(5)-(1)+(55)+(140)+(8555)+(612745)+(1916589696)=1917211185$

So the overall solution is

$25,2085185,1518,25,1917211185$. Breaking each number up into chunks that represent letters, we get $[2,5],[20,8,5,18,5],[15,18],[2,5],[19,17,21,1,18,5]$

giving

BE THERE OR BE SQUARE

which certainly fits the theme of the puzzle!


I'm guessing this is one of those puzzles that took much longer to create than to solve :-)

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  • $\begingroup$ You're faster. Go for it. $\endgroup$ – MacGyver88 Mar 20 at 14:51
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    $\begingroup$ If you haven't already checked, changing it to 1791 produces the expected result. $\endgroup$ – hagfy Mar 20 at 15:07
  • $\begingroup$ @hagfy Thanks! That makes sense. $\endgroup$ – Rand al'Thor Mar 20 at 15:11
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The answer is most likely

BE THERE OR BE SQUARE

Because

Using deduction based on the first word, the total of which is 25, 2 letters being 2, 5 or BE

Also, the same sequence is used for the 4th word.

So,

BE
(5 letters)
(2 letters)
BE
(6 letters)

The title sold it for me, has to be SQUARE

This is my complete answer/explanation. All math was done completely and faster by Rand al'Thor and would be very redundant and unneeded if I did it as well. :) Plus, I'm at work and my boss gave me other tasks to do.

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  • $\begingroup$ You beat me to it because I did all the calculation before posting :-) +1 anyway, I'm still hunting for a mistake I must have made somewhere ... $\endgroup$ – Rand al'Thor Mar 20 at 14:51
  • $\begingroup$ I did the first one and went, Oh, okay... Sounds like this. :) EDIT: I didn't really want to do all of the necessary math, though. So, yea! Nice work. $\endgroup$ – MacGyver88 Mar 20 at 14:53

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