At least one of these is a Nice puzzle.

2 ^ 100 --- What is it mod 2016 ?

2 ^ 2022 --- What are the least 2 digits?
(or ... What is it mod 100 ?)

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"henh...@gmail.com" <henhanna@gmail.com> writes:
> At least one of these is a Nice puzzle.

Different shortcuts make them very easy to do manually.

> 2 ^ 100 --- What is it mod 2016 ?

2016 = 2*1008 = 4*504 = 8*252 = 16*126 = 32*63
so we only need to calculate 2^100 mod 63, the mod 32 bit's easy
2^6=64 == 1, so 2^96 == 1
2^100 == 2^4 * 2^96 == 2^4 == 16 (mod 63)
16 != 0 mod 32 so add a multiple of 63 to correct that
63 == -1 (mod 32), so we want 16+16*63 = 16*64 = 1024
2^100 == 1024 (mod 2016)

> 2 ^ 2022 --- What are the least 2 digits?
> (or ... What is it mod 100 ?)

powers of 2 mod 25 go:
1, 2, 4, 8, 16, 7, 14, 3, 6, 12, -1, ... it's a primitive root, period 20
2022 = 101*20+2 so we want 2^2 == 4 (mod 25)
conveniently, 4 == 0 mod 4
so 2^2022 == 4 mod (100)

GP verification once the manual graft's done:
? Mod(2,2016)^100
Mod(1024, 2016)
? Mod(2,100)^2022
Mod(4, 100)

Phil
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In article <2993bdc7-2af6-4bdc-be83-bf4eae529600n@googlegroups.com>,
henh...@gmail.com <henhanna@gmail.com> wrote:

> 2 ^ 100 --- What is it mod 2016 ?

Use that 2016 = 2^5 x 63.

And 2^6 == 1 mod 63, so 2^90 == 1 mod 63, so 2^100 == 2^10 mod 2016.

Slightly differently, notice that 2048 = 2016 + 32. So 2^11 == 2^5 mod
2016, so we can throw away 2s in sixes as long as we have at least
eleven left. I.e., 2^100 == (2^6)^15 x 2^10 == 2^10 == 1024 mod 2016.

> 2 ^ 2022 --- What are the least 2 digits?

phi(25) = 20, so 2^2020 == 1 mod 25, so 2^2022 == 04 mod 100.

Gareth

On Thursday, June 16, 2022 at 6:18:58 AM UTC-7, Phil Carmody wrote:
> "henh...@gmail.com" <henh...@gmail.com> writes:
> > At least one of these is a Nice puzzle.
> Different shortcuts make them very easy to do manually.
> > 2 ^ 100 --- What is it mod 2016 ?
> 2016 = 2*1008 = 4*504 = 8*252 = 16*126 = 32*63
> so we only need to calculate 2^100 mod 63, the mod 32 bit's easy
> 2^6=64 == 1, so 2^96 == 1
> 2^100 == 2^4 * 2^96 == 2^4 == 16 (mod 63)
> 16 != 0 mod 32 so add a multiple of 63 to correct that
> 63 == -1 (mod 32), so we want 16+16*63 = 16*64 = 1024
> 2^100 == 1024 (mod 2016)
> > 2 ^ 2022 --- What are the least 2 digits?
> > (or ... What is it mod 100 ?)
> powers of 2 mod 25 go:
> 1, 2, 4, 8, 16, 7, 14, 3, 6, 12, -1, ... it's a primitive root, period 20
> 2022 = 101*20+2 so we want 2^2 == 4 (mod 25)
> conveniently, 4 == 0 mod 4
> so 2^2022 == 4 mod (100)
>
> GP verification once the manual graft's done:
> ? Mod(2,2016)^100
> Mod(1024, 2016)
> ? Mod(2,100)^2022
> Mod(4, 100)
>
> Phil

thanks! is GP a computer language? i can't find it on the Net
https://en.wikipedia.org/wiki/GP

Are there some well-known (or simpler) problems of this kind ?

> > 2 ^ 100 --- What is it mod 2016 ?

> > 2 ^ 2022 --- What are the least 2 digits?
> > (or ... What is it mod 100 ?)