$2^{100}$

[sm.joty]

Yup. RGS rocks
But unfortunately now I haven't much time to go there.

[Phlembac Adib Hasan]

Ok,I'm giving my solve.I needed to use e.But I didn't use it in the formal way so that juniors can realize my solve easily.

Notice that \[2^{10}=1024\]\[\therefore 10^3<2^{10}<10^{4}\]
\[\Rightarrow \left (10^3 \right )^{10} <\left (2^{10} \right )^{10}<\left (10^4 \right )^{10} \]
\[\Rightarrow 10^{30}<2^{100}<10^{40}\]
It implies $2^{100}$ has at least $30$ digits and at most $39$ digits.Now we have to find the least power of ten greater than $2^{100}$.I'll just compere all of $10^{31},10^{32},...,10^{39}$ one by one with $2^{100}$.First I'll take $10^{31}$.\[10^{31}\; ?\; 2^{100}\]\[\Rightarrow 2^{31}5^{31}\; ?\; 2^{100}\]\[\Rightarrow 5^{31}\; ?\; 2^{69}\]
\[\Rightarrow 5^{31}\; ?\; 4^{31}2^{7}\]\[\Rightarrow \left (\frac{5}{4}\right )^{31}\; ?\; 2^7\]
Notice that \[625>512\]\[\therefore 5^4>2^9\]\[\Rightarrow 5^4>4^4.2\]\[\Rightarrow \left (\frac {5}{4}\right )^4>2\]\[\Rightarrow \left (\frac {5}{4}\right )^{28}>2^7\]\[\therefore \left (\frac{5}{4}\right )^{31}> \left (\frac {5}{4}\right )^{28}>2^7\]Hence I have to put $'>'$ instead of $'?'$.So $10^{31}>2^{100}$.So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $31$ digits.

[nafistiham]

Ok,I'm giving my solve.I needed to use e.But I didn't use it in the formal way so that juniors can realize my solve easily.

Notice that \[2^{10}=1024\]\[\therefore 10^3<2^{10}<10^{4}\]
\[\Rightarrow \left (10^3 \right )^{10} <\left (2^{10} \right )^{10}<\left (10^4 \right )^{10} \]
\[\Rightarrow 10^{30}<2^{100}<10^{40}\]
It implies $2^{100}$ has at least $30$ digits and at most $39$ digits.Now we have to find the least power of ten greater than $2^{100}$.I'll just compere all of $10^{31},10^{32},...,10^{39}$ one by one with $2^{100}$.First I'll take $10^{31}$.\[10^{31}\; ?\; 2^{100}\]\[\Rightarrow 2^{31}5^{31}\; ?\; 2^{100}\]\[\Rightarrow 5^{31}\; ?\; 2^{69}\]
\[\Rightarrow 5^{31}\; ?\; 4^{31}2^{7}\]\[\Rightarrow \left (\frac{5}{4}\right )^{31}\; ?\; 2^7\]
Notice that \[625>512\]\[\therefore 5^4>2^9\]\[\Rightarrow 5^4>4^4.2\]\[\Rightarrow \left (\frac {5}{4}\right )^4>2\]\[\Rightarrow \left (\frac {5}{4}\right )^{28}>2^7\]\[\therefore \left (\frac{5}{4}\right )^{31}> \left (\frac {5}{4}\right )^{28}>2^7\]Hence I have to put $'>'$ instead of $'?'$.So $10^{31}>2^{100}$.So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $30$ digits.

but, you see,
\[2^{100}=1,267,650,600,228,229,401,496,703,205,376\]
$31$ digits.
try http://www.wolframalpha.com/input/?i=2^100


[something is going wrong.i can't give urls ]

[nafistiham]

1.It was about two and a half years ago, when I was in class VIII.
2. Very lame typo, fixed.
And yes,there is Ramanujan Gonit Songho in Dhaka.

thanks.i thought there is another PMS in Bangladesh.

[sourav das]

So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $30$ digits.

Actually everything is ok. $10^{n}>a\geq 10^{n-1}$ then $a$ has $n$ digits . So, not $30$, it'll be $31$

[*Mahi*]

Ok,I'm giving my solve.I needed to use e.But I didn't use it in the formal way so that juniors can realize my solve easily.

Notice that \[2^{10}=1024\]\[\therefore 10^3<2^{10}<10^{4}\]
\[\Rightarrow \left (10^3 \right )^{10} <\left (2^{10} \right )^{10}<\left (10^4 \right )^{10} \]
\[\Rightarrow 10^{30}<2^{100}<10^{40}\]
It implies $2^{100}$ has at least $30$ digits and at most $39$ digits.Now we have to find the least power of ten greater than $2^{100}$.I'll just compere all of $10^{31},10^{32},...,10^{39}$ one by one with $2^{100}$.First I'll take $10^{31}$.\[10^{31}\; ?\; 2^{100}\]\[\Rightarrow 2^{31}5^{31}\; ?\; 2^{100}\]\[\Rightarrow 5^{31}\; ?\; 2^{69}\]
\[\Rightarrow 5^{31}\; ?\; 4^{31}2^{7}\]\[\Rightarrow \left (\frac{5}{4}\right )^{31}\; ?\; 2^7\]
Notice that \[625>512\]\[\therefore 5^4>2^9\]\[\Rightarrow 5^4>4^4.2\]\[\Rightarrow \left (\frac {5}{4}\right )^4>2\]\[\Rightarrow \left (\frac {5}{4}\right )^{28}>2^7\]\[\therefore \left (\frac{5}{4}\right )^{31}> \left (\frac {5}{4}\right )^{28}>2^7\]Hence I have to put $'>'$ instead of $'?'$.So $10^{31}>2^{100}$.So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $30$ digits.

Adib,Adib,Adib.
This can be done so much easily... you just went too far.
Lets see...

$2^{10}>10^3$
So $2^{100}>10^{30}$
Again, $2^{13}=8192<10^4$
So $2^{100}<10^{\frac {400}{13}}=10^{30.769...}$
So,$10^{30}<2^{100}<10^{31}$ and thus $2^{100}$ has $31$ digits.

[nafistiham]

Ok,I'm giving my solve.I needed to use e.But I didn't use it in the formal way so that juniors can realize my solve easily.

Notice that \[2^{10}=1024\]\[\therefore 10^3<2^{10}<10^{4}\]
\[\Rightarrow \left (10^3 \right )^{10} <\left (2^{10} \right )^{10}<\left (10^4 \right )^{10} \]
\[\Rightarrow 10^{30}<2^{100}<10^{40}\]
It implies $2^{100}$ has at least $30$ digits and at most $39$ digits.Now we have to find the least power of ten greater than $2^{100}$.I'll just compere all of $10^{31},10^{32},...,10^{39}$ one by one with $2^{100}$.First I'll take $10^{31}$.\[10^{31}\; ?\; 2^{100}\]\[\Rightarrow 2^{31}5^{31}\; ?\; 2^{100}\]\[\Rightarrow 5^{31}\; ?\; 2^{69}\]
\[\Rightarrow 5^{31}\; ?\; 4^{31}2^{7}\]\[\Rightarrow \left (\frac{5}{4}\right )^{31}\; ?\; 2^7\]
Notice that \[625>512\]\[\therefore 5^4>2^9\]\[\Rightarrow 5^4>4^4.2\]\[\Rightarrow \left (\frac {5}{4}\right )^4>2\]\[\Rightarrow \left (\frac {5}{4}\right )^{28}>2^7\]\[\therefore \left (\frac{5}{4}\right )^{31}> \left (\frac {5}{4}\right )^{28}>2^7\]Hence I have to put $'>'$ instead of $'?'$.So $10^{31}>2^{100}$.So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $30$ digits.

Adib,Adib,Adib.
This can be done so much easily... you just went too far.
Lets see...

$2^{10}>10^3$
So $2^{100}>10^{30}$
Again, $2^{13}=8192<10^4$
So $2^{100}<10^{\frac {400}{13}}=10^{30.769...}$
So,$10^{30}<2^{100}<10^{31}$ and thus $2^{100}$ has $31$ digits.

so easy!I used binomial with a bunch of calculation.

but, I'll request adib to show the $e$ way too.

[Phlembac Adib Hasan]

Ok,I'm giving my solve.I needed to use e.But I didn't use it in the formal way so that juniors can realize my solve easily.

Notice that \[2^{10}=1024\]\[\therefore 10^3<2^{10}<10^{4}\]
\[\Rightarrow \left (10^3 \right )^{10} <\left (2^{10} \right )^{10}<\left (10^4 \right )^{10} \]
\[\Rightarrow 10^{30}<2^{100}<10^{40}\]
It implies $2^{100}$ has at least $30$ digits and at most $39$ digits.Now we have to find the least power of ten greater than $2^{100}$.I'll just compere all of $10^{31},10^{32},...,10^{39}$ one by one with $2^{100}$.First I'll take $10^{31}$.\[10^{31}\; ?\; 2^{100}\]\[\Rightarrow 2^{31}5^{31}\; ?\; 2^{100}\]\[\Rightarrow 5^{31}\; ?\; 2^{69}\]
\[\Rightarrow 5^{31}\; ?\; 4^{31}2^{7}\]\[\Rightarrow \left (\frac{5}{4}\right )^{31}\; ?\; 2^7\]
Notice that \[625>512\]\[\therefore 5^4>2^9\]\[\Rightarrow 5^4>4^4.2\]\[\Rightarrow \left (\frac {5}{4}\right )^4>2\]\[\Rightarrow \left (\frac {5}{4}\right )^{28}>2^7\]\[\therefore \left (\frac{5}{4}\right )^{31}> \left (\frac {5}{4}\right )^{28}>2^7\]Hence I have to put $'>'$ instead of $'?'$.So $10^{31}>2^{100}$.So the first inequality becomes \[10^{31}>2^{100}>10^{30}\]Which implies $2^{100}$ has exactly $30$ digits.

Adib,Adib,Adib.
This can be done so much easily... you just went too far.
Lets see...

$2^{10}>10^3$
So $2^{100}>10^{30}$
Again, $2^{13}=8192<10^4$
So $2^{100}<10^{\frac {400}{13}}=10^{30.769...}$
So,$10^{30}<2^{100}<10^{31}$ and thus $2^{100}$ has $31$ digits.

so easy!I used binomial with a bunch of calculation.

but, I'll request adib to show the $e$ way too.

Ok, anyone can ask how I realized I have to compere $512$ and $625$.Actually
\[\left ( \frac {5}{4}\right )^4=\left (1+\frac {1}{4}\right )^4\approx e>2\]And here I used $e$..I am very sorry for that mistake in my proof.