[Is this conjecture regarding primes true?]

For every prime number (p) greater than 3, there exists a natural number (n) such that

p^2 = 12(n)-11. Can you provide a counterexample? Else, can you prove it?

All prime nos. >3 can be written as 6k+1 or 6K-1.

So p^2=36k^2+1+12K or -12K. on comparing with 12n-11 or 12(n-1)+1, we get

12(n-1)=12(3k^2 +or-k) hence n-1 = 3k^2+k or 3k^2-k, where k is > or equal to 1, so it shows, its true for all primes.

[a simple pattern.?!]

(n-3)^4 + 2(n-3)^3 - (n-3)^2 + 6

(Note: word 'obvious' is used as a joke.)

How did you get that? Awesome!

[My friend and I play in a chess tournament]

either 7 or 14 eighth graders!!

[Sharing Circles]

Though Prime has already solved it.

[Sharing Circles]

7 players, with 2$,3$.....8$.

In the end the one with 2$ ends up with 20$, and the neighbour (with 8$) ends with 5$.

[A child is drowning.]

The calculation part is a headache..........

[Modifying two standard dice]

Let me rephrase the op. the probability of rolling a 2 with two normal dice is 1/36. The probability of rolling a 3 is 2/36. Can you relabel the dice so that you still get these probabilities and all others using different dice configurations

How about (1,2,3,4,5,6) and (a,a,a,a,a,a) where a is any +ve integer?

[Cabbage Patch Partitions]

^same here.

[A game]

A and B played a game with some amount. They agreed that whoever will lose have to give half his amount to the winner.

They played 100 games and a result was that A won First game B won next two games A won next three games and so on

A B B A A A B B B B A A A A A...............

Who will have graeter amount after the 100th game?

Whosoever wins the latest game will have more money, so B.

[Who am I?]

2 possiblities giving same answer

all lying - 7-bill 8-susan

1 true rest lying 7- bill, 8-susan

[A formal proof for x^2 = 2^x]

roots are easy to find and 2 roots may be somehow proved. 3rd is difficult to prove.

Using log base e,

2 lnx = x ln 2
x/lnx = 2/ln2, By comparison, x=2 and also LHS= 4/2lnx, again compare, x=4.

[My first sequence]

11, 12, 13, 22, 15, ?, 17, 32, 23, 111, ?

Fill in the ?'s

are you sure 111 will follow 23 and nothing is between then?

as per me,

16 in 1st blank

211 in 2nd blank

My apologies. The last question mark should be before the 111:

11, 12, 13, 22, 15, ?, 17, 32, 23, ?, 111

16

101

Nope, try again.

ok then can you tell me this?

what will come after 111?

211? or something else?

[Number game 2]

Look at this pair

42 24 ( digits of 42 are interchanged i.e =24)

x 48 x 84 (digits of 48 are interchanged i.e = 84)

-------- ----------

2016 2016

When the digit of multiplicand and multiplier are interchanged then also the result is same.

How many such pair can you find?

If you are asking for 2 digit numbers, then

lets say the numbers are AB and CD, then this property is true if A*C=B*D. eg. 14*82, 24*63 and so on....

[My first sequence]

11, 12, 13, 22, 15, ?, 17, 32, 23, 111, ?

Fill in the ?'s

are you sure 111 will follow 23 and nothing is between then?

as per me,

16 in 1st blank

211 in 2nd blank

My apologies. The last question mark should be before the 111:

11, 12, 13, 22, 15, ?, 17, 32, 23, ?, 111

16

101

[My first sequence]

11, 12, 13, 22, 15, ?, 17, 32, 23, 111, ?

Fill in the ?'s

are you sure 111 will follow 23 and nothing is between then?

as per me,

16 in 1st blank

211 in 2nd blank

[Magic Square!]

2 circles represent 00, which is the square of 0 HAHAHA

[A chessboard]

Pretty sure it cant be done. trying to pinpoint the problem in it.

[What is ....]

What is the value of

(m-a)(m-b).............................(m-y)(m-z)

if i am right in thinking what you are thinking, then

(m-m ) will make it 0

[How many tickets?]

Suppose n tickets numbered 1-n are in a box, and you draw one of them

You don't know how many tickets are in the box, but you are asked to estimate how many there are.

Your ticket has the number p on it.

What estimate of n has the highest likelihood of being correct?

It must be p only because then the Probabability of picking out p = 1/p, for n>p, the probability decreases.

[Concentric circles on steroids]

If i got the question right, then

R is approx. 125.54

[Very easy]

I have 3 friends,right?So then, including me, there are 4 people in total.So i can give 4 chocolates each to my friends and keep 4 chocolates with me!

(This is the very first answer i am giving.I suppose i am the latest member here!)

Use spoilers buddy! The 'S' option!!

[4 x 4 cards]

HA SK DJ CQ

CJ DQ SA HK
SQ HJ CK DA
DK CA HQ SJ

[Very easy]

4 each

the man is dead, his widow's .....

so he cant

[A few logical cases]

i did some reasearch on non smoking car in trains and found out that

there is a disease called small pox and its cure is to be in a "tobacco in the air" sort of atmosphere, so he shud hv chosen smoking car.