Puzzles with numbers and things

[abextra]

Round 279 - here's my pic, 25 dots and 8 lines...

[snookersfun]

oh, how cute!!!!

...and one more option, staying 'within the dots'
Skittle Pool Scoreboard.JPG

[davis_greatest]

Round 279 - congratulations abextra, Monique and snookersfun, for the 8 lines!

Round 280 - I've had answers from Monique and snookersfun. Please answer on the thread...

[Monique]

Charlie had 25 bananas ...

we have C^2=4*c^2 and c=f^2
where C is the number of rows in the big square, c the number od rows in a small sqare and f the number of fruits (bananas)
also C^2=n*b-1 and n=b+100
where n is the number of buckets and b the number of balls in a bucket

so we have ; b^2+100*b-1=4*f^4

the discriminant of this equation must be a square (because the solution is integer)

so 10000 + 4*(1+4*f^4) is a square
4*(2501 +4*f^4) is a square and
2501 + 4*f^4 is a square say r^2

then 2501=(r-2*f^2)*(r+2*f^2)
2501 is 1*2501 or 41*61
so we have either
r-2*f^2=41 and r+2*f^2=61
which yields only a solution where f=sqrt(5) and is not integer
or
r-2*f^2=1 and r+2*f^2=2501
or 1 + 4*f^2=2501
giving 2f=50 or f=25

[davis_greatest]

Charlie had 25 bananas ...

we have C^2=4*c^2 and c=f^2
where C is the number of rows in the big square, c the number od rows in a small sqare and f the number of fruits (bananas)
also C^2=n*b-1 and n=b+100
where n is the number of buckets and b the number of balls in a bucket

Thank you Monique! 25 is correct! snookersfun did it a different way - snookersfun, you can put it up here!

Here is yet another way, using the same notation as Monique:

If there are f bananas, then each small square has f² rows and (f²)² = f^4 balls, so the Big Square has 4f^4 balls and the number of balls delivered to Barry was 4f^4+1.

We also know that the number of balls delivered was
buckets x number of balls per bucket = n.b = n(n-100) (Expression 1)

Since Charlie was able to work out the difference between n and b, expression 1 must be the only way of factorising the total number of balls delivered.

From the above,
Number of balls delivered = 4f^4+1 = (2f²+1)² - (2f)² = (2f²+1+2f)(2f²+1-2f) = z(z-4f) (Expression 2)

where z = (2f²+1+2f)

Comparing expressions 1 and 2 and making them equal, we see that we must have n=z and 4f=100.

So Charlie had f=25 bananas.

[snookersfun]

Thank you Monique! 25 is correct! snookersfun did it a different way - snookersfun, you can put it up here!

ok, guess I can add to the equation mess. Why not? Though nice explanations up there

so here is yet another way, not using the same notation as Monique or d_g:
let's keep the part, where the number of balls in that big square was 4b^4 with b=# of bananas.

now the buckets/balls expression I rather wrote in the form:
(n+50)(n-50)-1= 4b^4
and was looking for a case where left hand side would result in a square.

(written as is, that square can't be n², as (n+50)(n-50)+50²=n², so I have to write n in the form m+p)

(m+p+50)(m+p-50)-1= m² + 2p*m-(50+p)(50-p)-1

these equal m² for m=((50+p)(50-p)+1)/2p
in the simplest case p=1 (and one is bound to find a solution for this case, as 50 is even, p odd, thus numerator divisible by 2)

and m=2500/2
also 1250²=4b^4, so b turns out to be 25.

so m=1250, the mean n=m+1 = 1251, number of balls = 1201, number of buckets =1301.

(there is a possible additional square in this case and that is m+p=51 (buckets 100, balls 1each), m=10 or m^2=100 but not of the form 4b^4)

[davis_greatest]

Round 280 Square balls

Charlie goes to the warehouse at the back of Barry The Baboon's Ball Shop, carrying a sack of bananas, and sees a nice arrangement of balls that Barry has put on display. The balls are arranged in a solid Big Square, which is divided into four equal Small Squares of balls: one brown, one blue, one pink and one black.

"That's nice," says Charlie. "When did you get all those balls?"

"This morning," replied Barry. "The balls arrived in big buckets, each containing an equal number of balls banging together. Once all the buckets had been delivered, I discovered that I could make this nice arrangement, and had exactly one ball left over!"

"How many buckets were there?" Charlie asked.

"I don't remember," replied Barry, "but I do remember noticing that the number of buckets was more than the number of balls in each bucket."

Charlie examined the balls more closely, and noticed that the number of rows of balls in each Small Square was equal to the square of the number of bananas in his sack!

"In that case," said Charlie, "the number of buckets must have been exactly 100 more than the number of balls in each bucket!"

How many bananas in Charlie's sack?

Answers initially by Private Message please...

Round 281 - Less square balls

Charlie has just realised that he had completely miscounted the number of rows of balls in each Small Square in round 280 – on recounting, he saw that the number of rows was equal to the square of the number of plums, not bananas, in his sack.

Meanwhile, Barry has just remembered that he hadn’t originally had enough balls in the buckets to make that Big Square after all, as he had thought, but had had to fetch some additional balls from his shop in order to make the pretty design.

The additional balls that he had fetched had originally been laid out in his shop in a solid Big Rectangle, which had been divided into four equally sized Small Rectangles of balls (one brown, one blue, one pink and one black). Each Small Rectangle was one ball taller than it was wide – and the number of balls across its width equalled the number of plums in Charlie’s sack!

Moreover, Barry had not had one ball over as he had previously thought, but had been one ball short, and had had to borrow it from Lenny the Lemur!

On hearing all this, Charlie announced to Barry that the number of buckets must have been exactly 398 more than the number of balls in each bucket, not 100 more as he had said before!

How many plums in Charlie's sack?

Answers initially by Private Message please...

[davis_greatest]

First answer to round 281 came quite quickly from snookersfun.... congratulations! Round 281 will remain open for a little while.

But here is...

Round 282 - Twice Nightly Snooker Whiteley

Here are some snooker balls. You can add them, divide, multiply or subtract, and need to make the Target or as close as you can get to it.

You don't have to use all the balls, and can't use any more than once.

You cannot combine digits - so, for example, with yellow and green, you could make 2+3 = 5, or 3-2=1, or 2x3 = 6, or even 3/2=1½, but you can't make 23 or 32.

Answer on the thread - winner / joint winners for each round will be decided according to how quickly the answer is given and how close it is to the Target. Your answer must be a whole number, and you must state how you use the balls to get it.

The first Target will be ... 902

[Monique]

first bid 905 = (7*7+1)*(3*6)+5

[snookersfun]

well 900: (7x7+1)x3x6

[Monique]

second bid 902: ((5-3)^7)*7+6

[snookersfun]

903: ((6x7)+1)x7x3

[davis_greatest]

second bid 902: ((5-3)^7)*7+6

Nice idea but powers aren't allowed. Only add, subtract, multiply, divide.

So closest bid so far stands at 903...

[Semih_Sayginer]

((6x7)-1) x ((5x3)+7)

=

(42-1) x (15+7)

=

41 x 22

=

902

[Stella147]

902 : ((5x3)+7) x ((7x6)-1)