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ok thx

5 consecutive heads can occur as follows

HHHHHTT

THHHHHT

TTHHHHH

6 consecutive heads can occur as follows:

HHHHHHT

THHHHHH

7 consecutive heads can occur as follows :

HHHHHHH

So....the probability  is 

[ 3 + 2 + 1 ] / 2^7  =

6 /128 =

3 / 64

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Ok. Thank you for your answer

I don't think you counted the 11 case. 2511 would also work because 2+5+1+1=9

If A,B  can be the same integers....you are correct, HSM  !!!

[ I assumed that they must be unique ]

Whatever.....our answers only differ by 1, so  it probably depends upon our assumptions  ....LOL!!!

25AB

The sum of the digits must be a multiple of 9......so.....

2 + 5  +  A  + B  =   9M

A + B   =  9M - 7

If  M =  1                    M  =  2                M  = 3

A + B  =  2                A + B  =  11         A + B  = 20 

A , B  =                     A , B =                 No possibilities  for A, B when  M > 2

0, 2                           2, 9

and the reverse        3, 8

                                 4, 7

                                 5, 6

                                 and the reverses

So....by my count.....we have 5(2) =  10 such numbers

For AB I got 11, 02, 20, 29, 38, 47, 56, 65, 74, 83, and 92. I'm not sure if that's all of them.

That sounds like NO FUN, CU  !!!

Yes I learned something! I always thought of that the sum of the digits IS nine, now, from I learned from this problem.

If   3AA1  is divisible by 9, the sum of its digits must be divisible by 9

The means  that 

3 + 1 + 2A  = 9M

4 + 2A  = 9M

2A  = 9M - 4

M must be an even number  > 0

So

If

M  = 2    A  = 7

And 3771 / 9  =  419

Thank you for your answers.

Substitute and simplify

2x² + (k+1)x + 2

There is exactly one solution when the discriminate is equal to 0.

That is my hint and I will let you solve.

The rule for divisibility for 9 is that all the digits add up to 9 or a multiply of 9, like 18.

So we can make an equation 

3 + 2a + 1 = 9

4 + 2a = 9

Notice how a will NOT be an integer, so we try 18.

3 + 2a + 1 = 18

4 + 2a = 18

14 = 2a

a = 7

So A must be 7

Trial and error shows a = 0 1 2 3 4 5 6 do not work....but a =7 does work

(3/2)x^2 + 11x + c  = 0

Using the discriminant, we have that

(-11)^2 -  4(3/2)c   = 7

121 -6c =  7      subtract 121 from each side

-6c  =  -114   divide both sides by -6

c =  -114/-6

c = 19

She turns the oven on and the temp rises and she puts the cookies in......shortly after , she opens the door to take the batch out and the temp drops as she puts the next batch in and closes the door....the temp rises until the batch is done and the sequence repeats for a total of 5 batches....then she turns the oven off and the temp drops below baking temp

I assume the question to mean:

N mod 3 = 1

N mod 5 = 2

N mod 7 = 3

N mod 11 =4

N mod 13 =5

N mod 17 =6

N mod 19 =7

I have used the "Chinese Remainder Theorem + Modular multiplicative Inverse", which are incorporated in this computer code to solve your congruences:

i=0;j=0;m=0;t=0;a=(3, 5, 7, 11, 13, 17, 19);r= (1, 2, 3, 4, 5, 6, 7);c=lcm(a); d=c / a[i];n=d % a[i] ;loop1:m++; if(n*m % a[i] ==1, goto loop, goto loop1);loop:s=(c/a[i]*r[j]*m);i++;j++;t=t+s;m=0;if(i< count a, goto4,m=m);printc,"m + ",t % c;return

N = 4,849,845 m +  4,383,592, where m =0, 1, 2, 3.........etc.

Let x  = the number of bees  and y the number of flowers

In the first equation....we have one more bee than the number of flowers  so

x - 1  = y    (1)

In the second equation, x/2  = the number of  flowers occupied by bees.....and if we add 1 to this,we get the total number of flowers....so

x/2 + 1  = y    ⇒   -x/2  - 1  = -y    (2)

Add (1) and (2)   and we have that

(1/2)x - 2  = 0

(1/2)x  = 2

x  = 4

And y  =  4 - 1  =  3

Yeah that was correct

I got 4 and 3 

Oops! Didn't realize it was inccorect!

XD my answer was luck!!

its freezing cold up here in Illinois!

Dang I came back from a race in Cross Country.

The two mile race was fine, except for the freezing weather and that it rained last night so we were swimming in mud instead of running