Power of 2

Note that $2^{10} = 1024$ and $x^2 - y^2 = (x+y)(x-y)$

$3^{2^{10}} -1$

$= (3^{2^{9}} +1)(3^{2^{9}} -1)$

$= (3^{2^{9}} +1)(3^{2^{8}} +1)(3^{2^{8}} -1)$

$= (3^{2^{9}} +1)(3^{2^{8}} +1) ......... (3^{2^{1}}+1) (3^{2^{0}}+1) (3-1)$

We know $3^{2^n}+1$ is divisible by $2$ , but not divisible by $4$ .

So $(3^{2^{9}} +1)(3^{2^{8}} +1) ......... (3^{2^{1}}+1)$ is divisible by $2^9$ and $(3^{2^{0}}+1) (3-1) = 4*2 = 8$ is divisible by $2^3$ .

Hence, $3^{1024} -1$ is divisible by $2^{9+3} = 2^{12}$ and $n= \boxed {12}$ .

We use Theorem 4 in this article:

$v_2(3^{1024} - 1^{1024}) = v_2(3-1) + v_2(3+1) + v_2(1024) - 1 = 1 + 2 + 10 - 1 = \boxed{12}$

Notice that $3^{1024} - 1 = (3^{512})^2 - 1 = (3^{512} - 1)(3^{512} + 1)$ .

Now we do the same with the new difference.

$(3^{512} - 1)(3^{512} + 1) = (3^{256} - 1)(3^{256} + 1)(3^{512} + 1)$

Continuing this, we get:

$(3 - 1)(3 + 1)(3^2 + 1)(3^4 + 1)(3^8 + 1)(3^{16} + 1) \dots (3^{512} + 1)$

Now we will find the largest power of two in each of these factors. We will first do all of the ones where the exponent on the $3$ is greater than $1$ .

By Euler's Theorem we have that, $3^{2n} + 1 \equiv 1 + 1 \equiv 2 \pmod 4$

And so each factor where the exponent on the $3$ is greater than one is divisible by $2$ but not by $4$ . This gives a factor of $2^9$ thus far, and now we will incorporate the final two terms. $(3 - 1)(3 + 1) = 2\cdot4 = 2^3$ , and so the answer is $3 + 9 = \boxed{12}$

Okay, let's be careful, 'cause this solution is tricky. First of all, observe that decomposing $3^{1024} - 1$ using $a^{2q} - b^{2} = (a^{q} + b) \times (a^{q} - b)$ 10 times, we have

$3^{1024} - 1 = (3^{512} + 1) \times (3^{256} + 1) \times (3^{128} + 1) \times (3^{64} + 1) \times (3^{32} + 1) \times (3^{16} + 1) \times (3^{8} + 1) \times (3^{4} + 1) \times (3^{2} + 1) \times 2^{3}$

But we may write

$3^{x} = (2 + 1)^{x} = \sum_{n=0}^x {x \choose n} 2^{x - n}$

For some natural $x$ . Remember that we want only even $x$ 's so that $x = 2m$ implies that

${2m \choose n} = \frac{(2m!)}{n!(2m - n)!}$

Is always even (unless for $n = 2m$ in which chase the binomial becomes $1$ ). In order to get rid of the only non-even case we may write the sum in this way:

$3^{x} = (2 + 1)^{x} = 1 + \sum_{n=0}^{x-1} {x \choose n} 2^{x - n}$

And, therefore, the sum may be written as $2^{k} \theta$ with $k \geq 2$ , since the binomial is even and $2^{x - n}$ is also even. Then we are able to write

$3^{x} + 1 = 2 + \sum_{n=0}^{x-1} {x \choose n} 2^{x - n} = 2 + 2^{k_{1}} \theta_{1} + \cdots + 2^{k_{x}} \theta_{x} = 2 \times ( 1 + 2^{k_{1} - 1} \theta_{1} + \cdots + 2^{k_{x} - 1} \theta_{x})$

But you may notice that $2^{k_{1} - 1} \theta_{1} + \cdots + 2^{k_{x} - 1} \theta_{x}$ is still an even number, because each $k_{i}$ is $\geq 2$ . Then $1 + 2^{k_{1} - 1} \theta_{1} + \cdots + 2^{k_{x} - 1} \theta_{x}$ is odd, which implies that

$2 \times ( 1 + 2^{k_{1} - 1} \theta_{1} + \cdots + 2^{k_{x} - 1} \theta_{x}) = 3^{x} + 1$

Has only ONE prime factor $2$ . Therefore, by the first decomposition, each of the parentheses have only ONE prime factor $2$ . We have $9$ parentheses and a factor $2^{3}$ multiplying them, so that we have a total of $12$ prime factors $2$ . There is our answer.

Just use difference of squares to obtain

$(3-1)(3+1)(3^{2}+1)(3^{4}+1)(3^{8}+1)(3^{16}+1)(3^{32}+1)(3^{64}+1)(3^{128}+1)(3^{256}+1)(3^{512}+1)$

And since $3^{k}+1\equiv2\pmod{4}$ , for $k = 2^{x},x\ge1$ , we cannot have a factor of $2^{y},y\ge2$ ; thus, we have

one $2$ as a factor for $3^{k}+1$ , for $k = 2^{x},1\le x < 10$ . This gives us $2^{9}$

and

one 2 in $3-1$

and

two 2 in $3+1$ The largest value of $n$ must be

$n = 1 + 2 + 9 = \boxed{12}$

This can be solved by induction.

Consider the largest integer n such that $2^{n}$ divides $3^{2^{m}} - 1$ . After trying small cases, we conjecture that n=m+2.

Base case: When m=1, $3^{2^{m}} - 1= 3^{2} - 1 = 8$ , so n=3=1+2. -> Claim true for m=1.

Inductive step: Assume it is true for m=k.

$3^{2^{k+1}} - 1 = (3^{2^{k}} - 1) (3^{2^{k}} + 1)$

We know that $3^{2^{k}} - 1$ is divisible by at most k+2, so all we have to do is show that $(3^{2^{k}} + 1)$ is divisible by at most 2.

Working modulus 4, we see that the powers of 3 have residues 3,1,3,1... . That is, even powers of 3 are all congruent to 1 mod 4. So $(3^{2^{m}} + 1)$ is congruent to 2 mod 4, and so the highest power of 2 that divides it is 2.

Then we have shown that for m=k+1, the largest integer n is k+3. (Assuming the claim is true for m=k.) And since it is true for m=1, by induction, it is true for all positive integers m.

$1024 = 2^{10}$ , so the answer is 10+2 = $\boxed{12}$ .

$3^{1024}-1=(3^{512}-1)(3^{512}+1)=(3^{256}-1)(3^{256}+1)(3^{512}+1)=...=(3-1)(3+1)(3^{2}+1)(3^{4}+1)...(3^{512}+1)$ The largest integer $m$ such that $2^{m}$ divides $3^{2a}-1$ for any positive integer $a$ is 1. This is because $3^{2a}-1$ is even but $3^{2a}+1=9^{n}+1$ is congruent to $1^{a}+1=2$ modulo 4. Also, $(3-1)(3+1)=2^{3}$ . Thus the largest value of n such that $2^{n}$ divides $3^{1024}-1$ is $3+1 \times 9=12$

The result simply follows from a particular case of Lifting the Exponent Lemma: if $$2\mid x-y$$ and n is even, then: $$\upsilon 2(x^n-y^n)=\upsilon 2(x-y)+\upsilon 2(x+y)+\upsilon 2(n)-1$$ In this case, we have: $$\upsilon 2(3^{1024}-1^{1024})=\upsilon 2(2)+\upsilon 2(4)+\upsilon 2(1024)-1=1+2+10-1=12$$ So 12 is the correct result.

I wrote a program...

a = (3**1024)- 1
for n in range (1,10000000):
    b = 2 ** n
    if(a % b == 0):
        print n

We can write $3^{1024} - 1$ as $(3^{512})^{2} - 1^{2}$

Thus, by using $(a^2 - b^2) = (a + b) (a - b)$ ,

$3^{1024} - 1$ becomes $(3^{512} + 1) (3^{512} - 1)$

We can further decompose $(3^{512} - 1)$ as $(3^{256} + 1) (3^{256} - 1)$ .

Again, we can decompose $(3^{256} - 1)$ as $(3^{128} + 1) (3^{128} - 1)$

Iterating again and again in the same manner, finally we get

$(3^{512} + 1) (3^{256} + 1) (3^{128} + 1) (3^{64} + 1) (3^{32} + 1) (3^{16} + 1) (3^{8} + 1) (3^{4} + 1) (3^{2} + 1) (3^{2} - 1)$

This becomes

$(3^{512} + 1) (3^{256} + 1) (3^{128} + 1) (3^{64} + 1) (3^{32} + 1) (3^{16} + 1) (3^{8} + 1) (3^{4} + 1) (3^{2} + 1) (2^{3})$

Now, a number of the form $(3^b + 1)$ where $b$ is a power of $2$ has only one $2$ as a prime factor. I guessed this by expanding a few terms.

Therefore, there are a total of $12$ two's in the prime factorization of this number, counting the $3$ two's in $2^{3}$ . And as a result, $2^{12}$ will be the highest power of two that will divide $3^{1024} - 1$ .

Thus, our answer is $\boxed{12}$

3^1024 - 1 = (3^512 +1)(3^256 +1)(3^128 +1)(3^64 +1)(3^32 +1)(3^16 +1)(3^8 +1)(3^4 +1)(3^2 +1)(3+1)(3-1) therefore, in each factor except (3+1) we have even number ie.. one two is available and in (3+1),we have two two's =2^2 therefore the answer is (1x10)+(2x1)=10+2=12,the required answer

Trivial by lifting exponent Lemma.

This lemma states that v(x^n - y^n) = v(x-y)+v(x+y)+v(n)-1 if n is even and x,y are odd. Note that v(m) denotes the greatest k such that 2^k | m. Note also that k could be 0 for an odd m.

So applying the lemma we get the answer 1+2+10-1=12.

3^1024-1=(3^512+1)(3^256+1)(3^128+1)(3^64+1(3^32+1)(3^16+1)(3^8+1)(3^4+1)(3^2+1)(3+1)(3-1). The last digits of its factors from 3^512+1 to 3^4+1) are 2 and if they are divided by 2 their last digit will become 1 and it won't get further divided by 2 so 7 is the higest power 2 till factors 3^4+1 and the remaining factors yield 5 as the maximum power of 2. Hence, 12 = n. Note: I factorized the given number by using a^2-b^2 = (a+b)(a-b).

$3^{1024}-1=(3^{512}+1)(3^{512}-1)$ using the identity $a^{2}-b^{2}=(a+b)(a-b)$ So, further simplifying the equation, we get: $3^{1024}-1=(3^{512}+1)(3^{256}+1)(3^{128}+1)(3^{64}+1)(3^{32}+1)(3^{16}+1)(3^{8}+1)(3^{4}+1)(3^{2}+1)(3+1)(3-1)$ Note that, All factors of $3^{1024}-1$ are even numbers, and thus multiples of 2 To check if they are multiples of $2^{2}$ we know, $3=-1(mod 4)$ $3^{even power}=1 (mod 4)$ $3^{even power}+1=1+1=2 (mod 4)$ Again note that the factor $3+1=4$ in which there are two powers of 2 Therefore $n=1+1+1+1+1+1+1+1+1+1+2=12$ note that all factors have even powers of 3 except 3+1

Use Lifting The Exponent Lemma (LTE) that is . Let x and y be two odd integers and let n be an even positive integer. Then $v_{2}(x^{n} - y^{n}) = v_{2}(x - y) + v_{2}(x + y) + v_{2}(n) - 1$ . Substitute the stuff!

Or

Refer to this

First, factor $3^1024-1$ as $(3^512+1)(3^256+1)...(3+1)(3-1)$ . The problem is asking for the number of factors of two in this product. Clearly, $3-1=2$ has one factor of two, and $3+1=4$ has two factors of two. Notice that $3^2+1=10$ is congruent to $2 (mod 4)$ , so it only has one factor of two. The same applies to $3^3+1=28$ . In fact, all the factors from $(3^2+1)$ and above are congruent to $2 (mod 4)$ .

Assume $3^k+1\equiv2 (mod 4)$ . Then: $3^k\equiv1 (mod 4)$ $3^2k\equiv 1 (mod 4)$ $3^2k\equiv 2 (mod 4)$

In total, there are $1+2+9=12$ factors of $2$ , so $n=\fbox{12}$ .