Alisa T.

asked • 01/13/23

Find the largest natural number n such that |(3^1024-1)|

Mark M.

Gibberish! Review for accuracy.
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01/13/23

Stanton D.

There's no question here, no question about it. Divides evenly, perhaps? If it's "such that 2^n divides evenly", that's traditionally solved by decomposing your "product" (no need for absolute value, is there?) into factors. You will have a bunch of them since (n^2x-1) always has (n^x +1) and (n^x-1) as factors. Then you need to reason just how many multiples of 2 could exist in each of those factors (it's a trivial task, you can do it by just doing a few and making the inference you need.) -- Cheers, --Mr. d.
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01/13/23

Mark M.

Stanton D.: Interesting conjecture, yet powers of 3 always have a units digit of 3, 9, 7, or 1. None of which are multiples of 2, i.e., are "evenly divisible". If my thinking is off, please correct.
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01/14/23

Alisa T.

Apologies the question was mistyped it is meant to say: Find the largest number n such that 2^n |(3^1024-1)|
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01/14/23

Mark M.

Well that does not really clarify. Such that 2^n |(3^1024-1)|does what? Equal something? Why absolute value? Is 2^n a factor?
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01/14/23

Stanton D.

Mark M. -- the factors aren't powers of 3, they are powers of 3 plus 1 in each case. So there is a continued factoring out of terms of successively lower powers of 3 + 1 . Each time you generate a one-lower power of 3 - 1 as a term, right? Alisa T., that terminology for "divides evenly" is not standard except maybe in higher mathematics texts; suggest you simply state verbally next time.
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01/15/23

Stanton D.

Yes, Mark M., she really is looking for the highest value of n such that 2^n is an integer factor of that 3 thing. This is a fairly well-reported problem on the Internet ....
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01/15/23

Mark M.

Stanton D.: Thank you. I failed to read |(3^1024) - 1|
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01/15/23

1 Expert Answer

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Stanton D. answered • 01/15/23

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