MHB How do I convert from this mod number to a regular number?

[SigmaS]

So I was working on solving $(\frac{1}{4900})^{100}$, and I figured the only way to do this neatly is through modular arithmetic.

I found that $4900 \equiv 84 \ (\text {mod 112})$, so I concluded $$\frac{1^{100}}{84^{10}\times84^{10} \ (\text{mod 112})}$$

Which should equal $$\frac{1}{3.06\times10^{38} \ (\text{mod 112})}$$

Now, this is still in mod form. How do I convert that value to a regular number, by mostly hand? When I tried converting it with the equation $n=qm+r$ where n is the number we wish to convert to, q is our quotient, m is the mod we're using, and r is the remainder; solving for n I got 84, but that doesn't sound right at all.

 

[I like Serena]

Hi SigmaS,

What are you 'solving' exactly?
Are you trying to calculate $(4900^{-1})^{100}\pmod{112}$?
Or something else? If something else, where is the $112$ coming from?

 

[SigmaS]

Hi SigmaS,

What are you 'solving' exactly?
Are you trying to calculate $(4900^{-1})^{100}\pmod{112}$?
Or something else? If something else, where is the $112$ coming from?

The 112 was just arbitrarily determined. I wanted to make $4900^{100}$ easy to calculate by hand. And I'm just trying to solve the probability of the problem, but expressed in much simpler terms.

if there's a better way to solve the original problem without modular arithmetic, then that would be great

 

[I like Serena]

How about:
$$4900^{100}\approx (\frac 12\cdot 10^4)^{100}=\frac 1{(2^{10})^{10}}\cdot 10^{400}
=\frac 1{(1024)^{10}}\cdot 10^{400}\approx \frac 1{10^{30}}\cdot 10^{400}=1\cdot10^{370}$$
? (Wondering)

 

[Opalg]

Sixty years ago, this would have been an easy exercise in logarithms. $\log49 \approx 1.690196$, so $\log4900 \approx 3.690196$, and $\log(4900^{100}) \approx 369.0196$. Now take the antilog, to get $4900^{100}\approx 1.046\times 10^{369}.$