Brainteaser and math puzzle thread

[ibavly]

Spoiler:

so there are 49*4 combos. remove 49 combos of FF, 43 combos of MM, 84 combos of FM, you are left with 6/20=30%

[TheDean1]

Originally Posted by ibavly

also dean,

lame solving riddles you've already heard of

well i didn't know if this solution applied to this riddle

also lmk if this is right for the red/blue coin flipping thing:

Spoiler:

if there are m red coins, move m coins to its own pile and flip all of them over

Originally Posted by iversonian

ah, i remember.

only ONE of them is a boy born on sunday. what are the odds that both are boys.

Spoiler:

i'm guessing that this is 6/13 since there are 14 day/gender pairings, and we know that child #2 is not sunday/male

[ibavly]

correct

[PyramidScheme]

We should also taken into account the % of the time the babies are fraternal twins and in 100% of these cases the other one MUST be a girl, thus raising the probability the other kid is a girl slightly. lol

[Aksdal]

fraternal twins don't have to be boy/girl

[iversonian]

Originally Posted by ibavly

A couple more

You have 5 coins, one of them weighs a different amount. You have a standard electronic scale, and you can use it 3 times. How do you find out a - which coin is different b - how much does it weigh

0, 2, 24, 252, 3120, ... what comes next

9, 10, 11, 12, 13, 14, 21, 100, ...

Prove that N = n(n+1)(n+2)(n+3)+1 is a perfect square for any whole number n

ok i think i got it for real now

Spoiler:

weigh coins a+b

weigh coins b+c+d

if a+b * 3/2 = b+c+d then coin e is odd, weigh that.

otherwise, weigh coins b+c.

if b+c=a+b, coin d is odd. its weight is b+c+d - (a+b).

cleared coins d,e so far.

since d = (b+c+d)-(b+c)

if d*2 = a+b, coin c is odd. its weight is b+c+d - (a+b)

cleared coins c,d,e

if d*2 != a+b, then

if (b+c)*3/2 = (b+c+d), then coin a is odd. its weight is a+b - (b+c)/2

cleared a,c,d,e

if (b+c)*3/2 != (b+c+d), then coin b is odd. its weight is (a+b) - ((a+b)-(b+c))/2

[ibavly]

Looks good. General idea is

Spoiler:

Make sure there is no coin there are no 2 coins that are in all the same equations. Bunch of different ways of doing that

[iversonian]

Prove that N = n(n+1)(n+2)(n+3)+1 is a perfect square for any whole number n

Spoiler:

because n(n+1)(n+2)(n+3)+1 can be rewritten as:

(n^2 + 3n + 1)^2

which is obv a perfect square

i didnt actually factor it. i just looked at the pattern and worked backwards

[metsandfinsfan]

Originally Posted by PyramidScheme

We should also taken into account the % of the time the babies are fraternal twins and in 100% of these cases the other one MUST be a girl, thus raising the probability the other kid is a girl slightly. lol

Wat

[David Lyons]

Originally Posted by Aksdal

fraternal twins don't have to be boy/girl

For example; Mary-Kate and Ashley are fraternal twins of the same gender.

[Sun Tzu]

Originally Posted by pwnsall

What's in my pocket?

not a fair quesssstion, precioussss!

[Sun Tzu]

this question is from a math talent search thing from when I was in high school:

Prove there are no 2 integers, the sum of whose squares is 12,345,678

[PyramidScheme]

Originally Posted by Sun Tzu

this question is from a math talent search thing from when I was in high school:

Prove there are no 2 integers, the sum of whose squares is 12,345,678

I guess solving this algorithmically is against the spirit of the question.