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What's the formula for calculating large exponents? 
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#1
Nov107, 12:27 PM

P: 76

Hi! I came across the following sum and was wondering if there was some kind of formula for solving it:
16^198 Is it some kind of binomial expansion? Do you calculate the last digits first, advancing to the previous ones later? I can't use a calculator. Please, please help. Sums like this might come in an assessment exam I'm sitting. Thank you very much! 


#2
Nov207, 02:06 AM

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P: 3,348

It wants you to evaluate 16^(198) without a calculator?
Its approximately [tex]2.60469314 \cdot 10^{238}[/tex], even with an efficient method it would take an enormous amount of time to evaluate, you sure you don't mean approximate, or find the last 5 digits of it? 


#3
Nov207, 03:28 AM

Sci Advisor
P: 2,751

Yeah it's not actually solving it but I can make a very rough approximation because I happen to remember that log_10(2) is pretty close to 0.3. Keeping this approx figure in your head lets you do quick and nasty power of 2 to power of 10 (or visa versa) conversions without a calculator.
So in this case [tex]16^{198} = 2^{4*198} \simeq 10^{0.3 * 4 * 198}[/tex], which gets you an order of magnetude calcuation of 10^238. 


#4
Nov207, 04:32 AM

P: 76

What's the formula for calculating large exponents?
I can do the approximations, but is there any way to find the last 6 digits? The question asked for both the approximation and the last 6 digits, so I thought perhaps the method would be similar and that there would be a formula.
Does this have something to do with modular mathematics? Thank you for your help. :) 


#5
Nov207, 08:31 AM

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P: 1,123

26046931378436930758124421057504913270096712196546516251547882077203270 46022512527938059453465450894821456996325559859549175313146140376984516 93595794.... (Edit I've removed the last load of digits not to make it too easy for you). Takes fractions of seconds on modern computers. And yeah, doing it via modular arithmetic is even easier. Because if you want to know the last 6 digits you can just take it mod 1'000'000, so you can just keep knocking off any digits larger than the first 6. 


#6
Nov207, 08:33 AM

P: 367




#7
Nov207, 09:21 AM

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P: 1,123

But yeah, it looks like you have to do some clever modular arithmetic, it'd pretty easy mod 10, mod 100 or perhaps even mod 1'000 by hand, but it's a bit of a pain mod 1'000'000. I'll think on a bit and see if I come up something (certainly nothing that immediately came to mind worked). 


#8
Nov207, 10:52 AM

P: 2,157

I would suggest to give the answer in the binary number system



#9
Nov207, 11:44 AM

P: 76

Thank you all for trying. Forget 6 digits. My exam's tomorrow afternoon, so if it'd be easier to find the last two digits without using a calculator, could you guys show me how? I'd just like to have an idea even if I can't answer.



#10
Nov207, 12:24 PM

P: 2,157

16^198 = 16^(2*3^2 * 11)
All computations mod 100 below: 16^2 = 256 = 56 16^(2*3) = 56^3 = 16 16^(2*3^2) = 16^3 mod(100) = 96 = 4 16^(2*3^2*11) = (4)^11 =2^22 now 2^10 = 1024 = 24 so 2^20 = 24^2 = 576 = 76 2^22 = 2^20 * 4 = 76*4 = 4 > 16^198 = 4 = 96 


#11
Nov207, 09:57 PM

P: 76




#12
Nov407, 02:50 PM

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P: 1,123

16^198 = 16^(2*3^2 * 11) 16^(2*3^2*11) = (4)^11 =2^22 2^22 = 4 16^(198) = 4 4 is the same as 96, mod 100. 


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