Question

If \(\color{blue}{g = 10^{100}}\) and \(\color{green}{H = 10^{g}}\) then in which \(\text{Interval}\) does \(\color{red}{g! = 1 \cdot 2 \cdot 3 .......... 10^{100}}\) lie?

a) \(10H < g! < H \) b) \(H < g! <10^{H}\)
c) \(10H < g! < 10^{H}\) d) \(10^{H} < g!\)

Answer 1

x=?

Answer 2

What?

Answer 3

Wait, a minute I have to edit question more..

Answer 4

Isnt x given?

Answer 5

Not at all..

Answer 6

If x=1 then option D is correct

Answer 7

x=1000000000000000000000000000000000000000000000000000000000000000
then none of the option is correct

Answer 8

What is logic behind it @sauravshakya

Answer 9

Sorry, I misread, that is not x that is g there..

Answer 10

option a) isn't right ,

Answer 11

Hello Uncle Rocks, how are you??

Happy New Year to you and to your Family, First of all..

What is logic behind this question can you teach me??

Answer 12

Sorry, I was helping someone else, that is why I am late as I usually do..

Answer 13

hey im alright

a) isnt right because \[10H\not <H\]

Answer 14

Yep..

Answer 15

And that is H = 10 raised to the power g and not 9..

Answer 16

\[\large H = 10^{\text{g}}\]

Answer 17

I think this is an excellent question for using
Stirling's Approximation for large factorials

\[\boxed{\log n!=n\log n-n+\mathcal O(\log n)}\]

take the log_10 for each side of inequality

Answer 18

Now, what is Stirling's Approximation?

I Have solution for this at the back of book, but I don't want to see it..

Answer 19

Stirling's Approximation is the thing in the box ^^

Answer 20

What is O like character there??

Answer 21

that big O indicates the order of the error in the approximation

Answer 22

You think differently @UnkleRhaukus

For your way of answering this question, I must buy A Mentos..

Answer 23

the error term is smaller (compared to the approximation) the larger n is , in these examples you can neglect the \(\mathcal O(\cdot)\) term completely

Answer 24

10H<g!
10*10^g < 1*2*3*...*10^100
10^g < 1*2*3*...*(10^100 -1)*10^(100-1)
10*10*10*...*10 < 1*2*3*...*(10^100 -1) *10^(100-1)
Which is TRUE as 0nly first 9 terms of RHS are less than that of LHS
So, 10H<g!

Answer 25

So, it cannot be option B

Answer 26

g!<10^H
log(g!) < log10^H
log(1*2*3*...*10^100) < H
log1 + log2 +log3 +... +log10^100 < 10^g
log1+log2+10g3+...+100 <10*10*10*...*10
This above inequality is TRUE as
log1+log2+log3+...+100<100+100+100+...+100<10*10*10*...*10
So, g!<10^H

Answer 27

Thus, 10H<g!<10^H

Answer 28

NOTE: