Math Is Fun Forum

lashko

Member

Registered: 2009-08-13

Posts: 8

2^3687

What are the three last digits of 2^3687??? can anyone explain the way how to solve this kind of problem?
Thanks a lot

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: 2^3687

Here's a hint: Adding 100 to a power of two (excluding 0, 1 and 2) doesn't change the last three digits of the result.

eg. The last three digits of 2^29 are 912. But the last three digits of 2^129 are also 912.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: 2^3687

Hi;

one way is

2^3687 = 2 * 2^2 * 2^4 * 2^32 * 2^64 * 2^512 * 2^1024 * 2^2048

Take each sub problem mod 1000

2 = 2

2^2  = 4

2^4  = 16

2^32 = (2 ^ 4)^8 = 16 ^ 8 = 296

2^64 = (2^32)^2 = 296 ^2 = 616

2^512 = (2^64)^8 = 616^8 = 96

2^1024 = (2^512)^2 = 96^2 = 216

2^2048 = (2^1024)^2 = 216^2 = 656

656 *216 = 696

696 * 96 = 816

816 * 616 = 656

656 * 296 = 176

176 * 16 = 816

816 * 4 = 264

264 * 2 = 528

so

So the last 3 digits are 528

Using mathsypersons hint you would use the same process but you only have to do it to 2^87.

Last edited by bobbym (2009-09-18 08:33:40)

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In mathematics, you don't understand things. You just get used to them.
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