Melody

Maybe I do not understand your question but

COD   COL   RIE      is the same as   COL   COD    RIE     You have counted it twice.

I think 65 is correct.

sin( theta) is positive becasue it is in the second quadrant

Find the opposite side using Pythagoras's Theorem.  Then read of sin (theata)

I would like you to become a member.

We need people who are seriously trying to understand :)

If you get a good reputation your questions will get priority.

I have added my 2c worth on the original thread :)

mmm interesting use of LaTex.

I will try to talk of your solution.    You have made a very good attempt    

1)   CO _    CO_    _ _ _        You got 20 here.  I think you need to divide that by two becasuse

                                            for example COD COL ***      is the same as   COL COD ***

                                             so I think this should be 10

2)   COO  C_ _    _ _ _       You got 10, I agree

3)    CO_   C_ _   O_ _         We agree on 30

4)   C_ _   C_ _   OO_          You got 30 I would halve that for the same  reason as in 1.         So 15 I think.

I think that covers all the cases so I get     10+10+30+15 = 65  ways.

I am not going to guarentee that it is correct though   

ok, I have had a bit of a look, it is not an easy one.

I will look some  more.    

There is already an answer there ??

How many five-digit numbers beginning in a 6 have exactly one pair of identical digits?

6 ** #$

10 posibilities for the double, 4C2 places they can go

9 possibilities for the third number and   8 possibilities for the fourth number

10 * 4C2 * 9 * 8 = 4320 such numbers

I was intending to answer until I noticed your title. 

You obviously are not interested in learning here.

N is a four-digit positive integer. Dividing N by 9, the remainder is 5. Dividing n by 7, the remainder is 2. Dividing n by 5, the remainder is 1. What is the smallest possible value of N?

Thanks guest, it is nice to have the answer

...but I would like to see it done without the omputer.

I have tried to do this  it without a computer program.

Let the number be N where

N=1000a + 100b + 10c + d

since it is 1 (mod5)        d = 1 or 6   (*)  

since it is 5 (mod9)        a+b+c+d (mod9) = 5

so

If    d=1    a+b+c = 4 (mod9)   so  a+b+c = 4  or  13  or  22

if    d=6    a+b+c = 8 (mod9)   so   a+b+c = 8 or 17 or  26

So far we have six possibilities for the sum of a,b and c, we want the smallest number so it will probably be one of the smaller ones but that doesn't help much.

Now  N(mod7) = 2

1000a + 100b + 10c + d (mod7) =2

6a     +    2b   +  3c  + d (mod7) =2

If d=1 then  6a+2b+3c = -a+2b+3c = 1 (mod7)  

so    -a+2b+3c = 1 or 8 or 15 or 22  or 29  or  36  or 43 

If d=6 then  -a+2b+3c = -4 (mod7) 

so       -a+2b+3c = -4 or 3 or 10 or 17 or 24 or 31 or 38 

summarizing what I have 

d =1 or 6

If d=1 then 

 a+b+c = 4  or  13  or  22

-a+2b+3c = 1 or 8 or 15 or 22  or 29  or  36  or 43

If d=6 then

     a+b+c = 8 or 17 or  26

-a+2b+3c = -4 or 3 or 10 or 17 or 24 or 31 or 38

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I want the smallest one so hopefully a will be 1  let me look at that

If d=1 (and a=1) then 

 b+c = 3  or  12

2b+3c = 2 or 9 or 16 or 23  or 30  or  37  or 44

If d=6 (and a=1) then

     b+c = 7 or 16 or  25

2b+3c = -3 or 4 or 11 or 18 or 25 or 32 or 39

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

If d=1 (and a=1) then 

2b+2c+c = 2 or 9 or 16 or 23  or 30  or  37  or 44

 b+c = 3  or  12

2b+2c = 6 or 24  

if 

If 2b+2c=6 then c=3    and we have the number   1031

If d=6 (and a=1) then

2b+2c +c = -3 or 4 or 11 or 18 or 25 or 32 or 39

     b+c = 7 or 16 or  25

    2b+2c = 14 or 32  (50 is too big)

If

2b+2c = 14 then c=4  and so b=3   so the number is  1346

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So   

1031 is the smallest one  (just as guest already determined with his/her computer program. 

That was an exteme effort!    LOL