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This is the Fibonacci sequence and every 6th term is a multiple of 4.

Therefore: Floor[1000 / 6] = 166 such integers that are divisible by 4

Im confused, which one is correct? i feel like the second one but im not sure...

You can rewrite this as:

(n−3)(n+3)(n−2)(n+2)

Consecutive integers are always coprime, so there are no common factors of n−2 and n−3 or n+2 and n+3 - so the question boils down to how many positive integers from 1 to 2016 have a common factor for n−2 and n+3 or n−3 and n+2 - and since the difference is 5, the only possible common factor can be 5.

In other words, we have a solution whenever n=2 or 3(mod5) - with the exception of n=2 which gives a division by 0. We start counting at n=3, since gcd(0,5)=5.

Let’s check n=7 and n=8 to gut-check the answer: gcd(7^2−9,7^2−4)=5 and gcd(8^2−9,8^2−4)=5 - all looks good! All solutions other than 3 are of the form 5k+2 or 5k+3 for 1≤k≤402 giving 805 possible solutions.

13(3x - 2) =26 (mod 8)..............(1) 

Divide both sides by 13

(3x - 2) =2 (mod 8)

3x =4  (mod 8) - this is the same as (1) above

x =8m  +  4,  where m=0, 1, 2, 3........etc. [This is the full solution]

So, when m=0,  x =4

And when m=1, x =(8*1  +  4) =12

Alright, the first step is very simple

So basically subtract 6ab^2 from the first equation and you will get the second equation so 6ab^2 = 6

\(6ab^2 = 6\)

\(ab^2 = 1\)

now, you are asking for a - b

\(b^2 = 1/a\)

\(\cdot 1/b\) on both sides

\(b = 1/ab\)

im just doing every possible thing i can do bc this is my first time seeing this so im doing it as well as you sorry

\(ab = 1/b\) ok that might be useful

oh......

\(a^3 + 3 = 679\)

\(a^3 = 676\)

\(a \approx 8.77639\)

so b is 1/ whatever that is. Square rooted.....

its approximately 8.43883.... probably not correct, looking at the numbers. Maybe its like the square root of something?

 its like \(\sqrt{71.212}\) not exacly but close

\((1+ax)^n=1+nax+\frac{n(n-1)}{2!}a^2x^2+\frac{n(n-1)(n-2)}{3!}a^3x^3+...\)

Let  \(na=10\)

and    \(\frac{n(n-1)}{2!}a^2=150\)

and solve for a and n, then find c from \(c=\frac{n(n-1)(n-2)}{3!}a^3\)

That's correct, thank you!

(sqrt(2)  +  sqrt(3))^2002 =3854913 6553434375 7852540977........... 4542810952 6835265508 6915760009.9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999999999 9999997405 908

So, the first 10 digits after the decimal point are:9,999,999,999

69 is 23(2+0!)

No, NOT the FIRST 10 digits, but the first 10 digits AFTER THE DECINAL POINT !!.

\(\frac{\left(2z-1\right)^4}{\left(z\sqrt{z}+1\right)^4}\)this 

is what you get okay now for the expansion.......

\(\frac{16z^4-32z^3+24z^2-8z+1}{z^6+4z^{\frac{9}{2}}+6z^3+4z\sqrt{z}+1}\)

i agree, the numbers are huge but u have your answer. multiply it out and you will get 1 as your constant term because of the +1 in your original equation

\([(\sqrt (3) + \sqrt(2)]^{2002}\)

alright so the first ten digits of the decimal expansion.

just plug it in to a calc and u should be able to get it

First of all, what? This doesnot make any sense. I am assuming you meant

\(7/100\cdot 2/10\)

that will be \(14/1000 = 7/500\)

so we can get 2 equations here, please ask if u want to know how i got these.

\(4a + 4b = 16\)

\(a^2 + 2ab + b^2 = 16\)

so now we combine them

\(4a + 4b = a^2 + 2ab + b^2\)

\(4a + 4b - 2ab - 2b^2 = (a+b)(a-b)\)

more on how that works if you ask me.

\(4(a+b) - 2b(a+b) = (a+b)(a-b)\)

\(4-2b = a-b\)

i think thats correct..

\(a+b = 4\)

oh well.....................

thats depressing oofoof

okok for b) its just 4, literally

and for c i don't think any integers work therfore no real numbers work becuase the max number for a or b is 1, becuase 2^4 is already 16 and negatives^4 are always postive for i think c is none.

and for a) when i did the failed equations u see a 2ab when i squared the first equation,try to single that out and u will get the answer. for a

HINT. for a, do \(2ab = a^4+b^4-a^2 - b^2\)

that will be equal to \(2ab = (a^2)(a^2 - 1) + (b^2)(b^2 - 1)\)

then use the second equation to substiture the a^2 and b^2 and u will get somewhere...

We are working on it with students. We got so far to alomst 50.

I only have a few

1 = 2*0+3-2

2=3*2*0+2

3=2*2*0+3

4=3*0+2+2

5=2*0+3+2

6=(2*0)+(3*2)

7=3+2+2+0

8=(2^3)+(2*0)

9=3^2+(2*0)

10=3^2+2^0

11=3^2+2+0

12=3*2*2+0

13=3!*2+2^0

14=20-(3*2)

15=20-(3+2)

The 114th term of :     (2u - 3v + u^2 - v^2)^9 ==  - 577368 u^2 v^9

What calculator are you using?

Sorry, that is not correct. But thanks.