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Let's convert the equation into base 10

\(AAA_7\) = \(56A+1\)

\(BBB_7\) = \(56B+1\)

And \(666_7\) = \(337\)

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So we have : \(56A+56B=335\)

OK.... Now we need to find how many (A,B) there is.

To find the maximum and minimum one of the variables can be (A or B) we do \(\lfloor\frac{335}{56}\rfloor\)

So maximum can be 5

The minimum is obviously 1.

So we have to substitute 1, 2, 3, 4, 5, for A or B to see if there are solutions.

There are no integer solutions.

So there are zero ordered pairs (A,B)

In place of ABC, you have 0s and 1s. Since they can occupy 3 spots, then you have the following possible placements:2^3 = 8 possible placements as follows:
000, 001, 010, 011, 100, 101, 110, 111. Using a short computer code to convert them, you get the following results:
B=2;a=(1,0,1,1,0,1,0,1,0,1,0,1,0,1,0,1,0,1,1,0 ); s=0;d=1;i=0; j=1; m=B^(count a-d)*a[0];s=s+m;cycle:m=B^(count a - (d+1))*a[i++];s=s+m;j++;d++; if(j

000 =  742726
001 =  742734
010 =  742742 - Only this one is a multiple of 7
011 =  742750
100 =  742758
101 =  742766
110 =  742774
111 =  742782

2)  What is the average of all positive integers that have four digits when written in base 3, but two digits when written in base 6? Write your answer in base 10.

The integers in base 3 can range from

1000₃ to  2222₃   =   [27  to 80] base10

The integers in base 6 can range from

10₆  to  55₆  = [6 to 35] base 10

So  the common integers are      27, 28, 29 , 30 , 31, 32, 33 , 34 and 35

Their average  =   [ 4(62) + 31]  / 9  =  31

1): We can use the Chinese Remainder Theorem.

I can post a solution if you want.

Thank you CPhill, I do appreciate it all!

50(20)  = 1000 

When each data point is doubled we have 1000 * 2  =  2000

So

[ 1000 + 2000] / 50  =

3000 / 100  =  

30 =  new average

I think it's supposed to be    sec^2 x - 1

To see why.....the left hand side  can be factored as

tan^2x  (sin^2 x + cos^2 x)  =

tan^2 x ( 1)  =

tan^2 x

But 

tan^2x =  sec^2 x - 1      add 1 to both sides

tan^2 x + 1  =  sec^2 x      which is an identity

x^3 + ax^2 + 11x + 6 = 0  (1)

x^3 + bx^2 + 14x + 8  = 0   (2)

By the Rational Roots Theorem, the possible  zeroes  for the first polynomial are ±  [ 1, 2 , 3 , 6]

And for the second polynomial they are ±  [ 1,2,3,4] 

Subtract (1) from (2)  and we get that

(b - a) x^2 + 3x + 2  = 0    

Two possible shared roots  of  -1 and - 2  can be found if we let  b - a  = 1

So we have that

x^2 + 3x+ 2  = 0

(x + 1) ( x + 2)  = 0

x = - 1  and x = -2

Now   let x  = -1   as a shared root

Then  (-1)^3 + a(-1)^2 + 11(-1) + 6  = 0

-1 + a - 11 + 6  = 0

And a =6

Also

(-1)^3 + b(-1)^2 + 14(-1) +8 = 0

-1 + b -14 + 8  = 0

And b = 7

So testing (-2) as a root we have that

(-2)^3 + 6(-2)^2 + 11(-2) + 6  =

-8 + 24 - 22 + 6  =

0

And 

(-2)^3 + 7(-2)^2 + 14(-2)  + 8 =

-8 + 28 -28 + 8  =

0

So

(a,b ) =  (6, 7)

Here is the graph to show that this is true :https://www.desmos.com/calculator/y69o2yts2n

Melody, no, I am really stuck...please help me...

Dearest Melody, it's juriemagic here...working from my phone....uhm, I made an error..it's supposed be \(1+sec^2x\)

LHS 

factorise and it will fall out to     tan^2 (x )    in just 2 lines

RHS

\(1-sec^2x\\ =1-\frac{1}{cos^2x}\\ =\frac{cos^2x-1}{cos^2x}\\ =\)

You should be able to finish if from there.  

Thanks Alan,

I have not finished looking at your answer but your start is similar to mine.

I did not bother with expanding

I used  this

if:

\(ax^3+bx^2+cx+d=0\\ \text{and the roots are u,v and t}\\ then\\ u+v+t=-\frac{b}{a}\\ uv+ut+vt=+\frac{c}{a}\\ uvt=-\frac{d}{a}\)

The relevance is that I think uvt=-8

I got this far myself but then I got into a mess. 

I am eager to examine the rest of your answer.   

Here is my attempt:

.

I'd also like to see someone answer this.

Babada, you do not need people to do all your homework for you. (Unless you are just bone lazy that is)

Just ask one question, learn from the answer and try doing the others for yourself!

Yes 900   3 digit numbers

1000/11 = 90.9

round down to 90

9 of those are less than 100

so 81 are between 100 and 999

so that is 

P(that a 3 digit number is divisable by 11 )  =  81/900  = 9/100 = 9%

Just as guest said.   

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You rounded up instead of down so

instead of ending with the las multiple of 11 LOWER than 1000,

you ended with the first multiple of 11 AFTER 1000.

Oh, I do not like your 900/11 bit either. That could also have put the answer out by a little bit.

In this case it makes no answer difference but sometimes it would.  Another thing for you to think about.  

I suggest you think very hard about THIS  answer/question

You should not just accept what some answer written down somewhere is. 

How do you know there is not a negative answer?  Answer: Someone told you so.

Well that is not you knowing, that is at best someone else knowing.

If you think about it purely as a logic exercise, you will work out why there cannot be a negative answer.

I took me a while to get my head around it so it might take you a while too but the reason is not very complex.   

There are: 999 - 100 + 1 =900 3-digit numbers.

There are: [990 - 110] / 11 + 1 =81 numbers that are multiples of 11 between 100 and 999

So, the probability is: 81 / 900 = 9%

To denote portions of integers.

This is wrong, why do you round up, why can't you round down?

3) How many positive integers less than 1000 have at least one digit of 0?

10 *9  + 9  = 99   have an ending digit of 0

9*9  = 81  have a middle digit of 0  [ but not an ending digit of 0 ]

So

99 + 81   =   180 have at least one 0