White Christmas

Since the Hulk is tearing down the entire stack into $2$ , then $4$ , then $8$ , and so on,

$2^{x} = 1, 000, 000, 000,$ where $x$ is the number of times Hulk has to tear the stack into $2$ .

$2^{x} = 1000^{3}$

Since the power of $2$ closest to $1000$ is $1024$ ,

$2^{x}$ is about $1024^{3}$

$2^{x}$ is about $(2^{10})^{3}$

$2^{x}$ is about $2^{30}$

Therefore, the smallest whole number $x$ that gives $2^{x} > 1, 000, 000, 000$ is $30$

as we keep dividing by 2 it forms a progression 2^n > 10^9

taking log on both sides n * log2 = 9

n = 9/log 2

the corresponding integer value is 30

For $x$ tears, the Incredible Hulk can produce $2^x$ pieces of paper.

The problem wants us to find the smallest natural value of $x$ such that the inequality $2^x \geq 10^9$ holds.

Applying $\log_2$ to both sides, we get that $x \geq 9 + 9 \times \log_25$ . If I know that $\log_{10}2 \approx 0.3$ , we can say that $\log_25 \Rightarrow \frac{log_{10}5}{log_{10}2} \approx \frac{0.7}{0.3} = 2. \overline{3}$ . Thus, $x \geq 29. \overline{9} \Rightarrow \boxed{x = 30.}$

Every time the paper is divided into 2.

so the equation is ${ \quad 2 }^{ x }={ 1000 }^{ 3 }$

${ 32 }^{ 2 }=1024$

32 happens to be a power of 2

${ 2 }^{ 10 }=1024$

${ 2 }^{ 30 }\quad$ is obviously about $1,000,000,000$

Therefore the paper must be halved 30 times.

the sequence we have is : (1.2.4.8......) "Geometric sequence"

then: a=1

r= 2/1 = 4/2 = 8/4 .....=2

Sum = 1,000,000,000

sum = a(1-r^n / 1-r)

1,000,000,000 = 1x ( 1-2^n / -1) ------> by solving, n = 29.897 = 30

Our equation is two to the x power is greater than 1000000000. First we will solve for what value of x actually gets 1000000000. Let's use our logarithmic properties: x will be equal to log(1000000000) divided by log(2). Well, log(1000000000) just equals 9, so divided by log(2) equals 29.897. The next integer above that is 30.

1 piece of paper when torn will give u 2 pieces..... hence 1=2^1 2=2^2 3=2^3 ... ... ... ... n=2^n ...given here is total pieces obtained.and we have to find torn we made hence
2^n=1,000,000,000 taking log on both sides n* log2=log(1,000,000,000) .... n * 0.3=9 ..... n=9/0.3 n=30

On the first tear, it yields two pieces, on the second it yields 4, one the third it yields 8, and so on. So 2^{n} = (number of pieces)

2^{34} = 1x10^{9}

log {2}(2^{n}) = log {2}1x10^{9}

n = log_{2}1x10^{9}

n ≈ 30

We have to find $k$ such that $2^k>1,000,000,000$ but $2^{k-1}<1,000,000,000$ . So $k=30$ as $\log_2(1,000,000,000)\sim29$ .

Since 2^29<1000000000<2^30

Answer is 30