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Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

Powers? A non-calculator question...?

What is 2 (to the power of) 2009?

The question explained that we should work through patterns, and predict the last digit, which I got 2. Now I just need to check if the answer's right, but my scientific calculator says 'Maths Error' and no calculators on the internet have this function.

If anyone would please do the question, or do it on a calculator that works, or give me a link to an internet calculator that works?

Thanks!!

2 Answers

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  • 1 decade ago
    Favorite Answer

    You are correct.

    The way to do this is like you did, looking at powers. There are more advanced techniques like a theorem called "Fermat's Little Theorem" that can help you figure this out:

    http://en.wikipedia.org/wiki/Fermat%27s_little_the...

    You could also realize that the last digit is all you need to look at in any step. For example, if you have 18342 × 1842056 and you want the 1s digit, just look at the 1s digit of what you have. You get 12, or 2, which would be the 1s digit of the full product of those big numbers.

    So for your problem, you can use that trick by realizing:

    2^5 = 32

    Any time you see a 2^5, you can replace it with 2^1. So rewrite:

    2^2009 = (2^5)^401 2^4 = 2^401 2^4 = 2^405

    = (2^5)^81 = 2^81 = (2^5)^16 2= 2^16 2 = 2^17

    = (2^5)^3 2^2 = 2^3 2^2 = 2^5 = 2

    Note nothing here is actually equal "=" is just standing for equality in the 1s digit, not all digits. The actual product is

    2^2009 =

    58784291598041831640721059900297317

    58194266634694119426445530812547923

    25832893600694609656994051210198244

    33389516158094000492490796188432969

    00768543573264309203455444239988736

    03526549238989029741716106189125049

    57328187117386950842341026317332718

    77323310335823777914819017965035807

    91355645625160816488103328482144814

    00042754868418296221651998157278605

    56821964939095379242522726816370497

    60213817691562584097786856429660810

    35151287502869585844829824788935390

    15787106332413838519791208404996196

    20949148583707547778988677199505145

    78646749211908564621201347904089822

    99074602129549865879831232623864378

    8303040512

    That's all one number, but look at the last digit. Good!

  • Anonymous
    1 decade ago

    to find out the last digit we need not know the actual number. we basically require here the remainder when divided by 10.

    we can see that for every five 2s we get a number that is ending with two. so 2^2005 will end with 2. now it is easy to see that the last digit

    of 2^2009 will be 2. so your answer was correct.

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