Puzzles with numbers and things

[Monique]

That's because you know only the pink variety

[davis_greatest]

No points to snookersfun or Monique. The elephants don't look elephantish enough.

[Monique]

Here it is ...
svs.jpg

And 2 more 15, one with a length 5 line
doc1.PNG

[davis_greatest]

Monique has bid a 16 too...

[snookersfun]

I've now just seen snookersfun's pictures for scores of 15 and 16, which had been sent to me earlier - both very nice!

...and here they are

[davis_greatest]

OK - pictures on the thread please... and keep the bids coming!

I've now just seen snookersfun's pictures for scores of 15 and 16, which had been sent to me earlier - both very nice!

[davis_greatest]

OK - pictures on the thread please... and keep the bids coming!

[Monique]

15 only for now ...

[snookersfun]

16 for now...

[dantuck_7]

I'll volunteer 15 points.

Dan

[davis_greatest]

Someone please post the answer to snookersfun's round 177! Because we're moving onto:

Round 178 - From basketballs to elephants

Take 10 perfectly spherical elephants shrunk to the size of peas and position them wherever you want on a postage stamp that has been stretched to the size of Outer Mongolia annexed with Sesame Street.

For any line of 3 spherical elephants, score 1 point.

For any line of 4 spherical elephants, score 3 points.

For any line of 5 or more spherical elephants, score 5 points.

The lines must be straight, and can go in any direction. Only the longest line counts. For example, 10 spherical elephants in a line OOOOOOOOOO scores 5 points (as it has 5 or more elephants) - you cannot also count lines of 3 or 4 elephants within that same line!


Post here whatever scores you can find - preferably as high as possible!
You don't have to find the highest theoretically possible - in fact, I don't know what it is!

[snookersfun]

Round 177 was solved speedily by Monique, d_g and dantuck. Next answer on the thread please (and it is really easy)

[Monique]

R176 solution

As requested ...
Black. And all four apes were wearing square bow ties.

We have
sum of (score of "square bow tie" ape)^2 = 8*n +b
where n is the score of one group of apes, b the value of the ball shown by Charlie

The left part of this expression has 1 to 4 members, all squares.
The right part has a "modulo 8" value of b, b being the value of the ball Lois Lane saw is potentially anything from 1 to 7.
Now square numbers "modulo 8" can only yield values of 1, 4 or 0. Those that yield 0 become in a way "invisible" in the rigth part oh the expression.

The only value of b that actually requires 4 "visible" squares is 7 : 1+1+1+4. All other values can be obtained with 3 or less "visible" squares leaving room for at least one additional invisible one.
So a value of 7 is the only possibility that gives Lois Lane the clue to the number of square bow ties.

PS:n,m,k being integer numbers, n modulo m equals k if k is the remainder of n/m

[snookersfun]

answer to Round 176: The black ball was held up and therefore all the 4 bowtie wearers had square ties!

One can prove that using remainders: The total number of points are 8n, while 8n+(1 to 7) has to be reached by addition of 1-4 square numbers. Square numbers are unique in that they all have remainders (if dividing by 8) of 0,1 or 4. The only remainder which can be reached by a unique amount of square numbers is 7, which needs 4 squares with remainders 4,1,1,1.

[davis_greatest]

Monique has joined snookersfun as a solver of round 176! Congratulations!

It is much easier than might appear at first sight, and will remain open until 9.30a.m. BST tomorrow...

Someone please post the answer to round 176 on this thread...