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/wsr/ - Worksafe Requests

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How would you work out the last digit of, say, 38 to the power 38 to the power 38... (38 times)
I know it has something to do with congruences but I'm struggling to get my head around it
>>
>>628261
3838 mod 10
= 8^[3837] mod 10

8^1=8
8^2=64=4 mod 10
8^3=8*4=32=2 mod 10
8^4=8*2=16=6 mod 10
8^5=8*6=48=8 mod 10

So 8 has an cycle of length 4 that sweeps 6->8->4->2 for the modulo 4 values of 0 to 3 respectively

= 8^[38*38*38^(3836-2)] mod 10
= 8^[4*19^2*38^(3836-2)] mod 10
= 8^[4*k] mod 10
= 6
>>
I don't follow how you did the very first bit and got 8^3837 sorry.

Can we do another example so I can see where everything's coming from?
How about 42 to the power 42 to the power 42 (42 times). thank you
>>
bump for interest
>>628503
I don't follow this either, looks like it's the last digit of original number to the power of [original number -1].
Do we use mod 10 because our number system is base 10?
>>
>>628503
Fuck gookmoot added the up arrow symbol to autodelete Unicode list
>>
>>628406
Let ^^ be power tower aka iterated exponentiation ie n^^m = n^n^n^...(m time)...^n
https://en.wikipedia.org/wiki/Tetration

38^^38 mod 10
= (38 mod 10)^[38^^37] mod 10
= 8^[38^^37] mod 10

There are only so many values 8^n mod 10 can take so list the first few till it repeats

8^1=8
8^2=64=4 mod 10
8^3=8*4=32=2 mod 10
8^4=8*2=16=6 mod 10
8^5=8*6=48=8 mod 10

So 8 has an cycle of length 4 that sweeps 6->8->4->2 for the modulo 4 values of 0,1,2,3 respectively

= 8^[38*38*38^(38^^36-2)] mod 10
= 8^[4*19^2*38^(38^^36-2)] mod 10
= 8^[4*k] mod 10
so 4*k is 0 mod 4 therefore it's the first element of the cycle
= 6
>>
>>628503
42^^42 mod 10
=2^(42^^41) mod 10

look at the powers of 2
2->4->8->6->2
so a cycle of length 4 again

2^(42^^41) mod 10
=2^(42^(42^^40)) mod 10
=2^(42*42*42^[(42^^40)-2]) mod 10
=2^(4*21^2*42^[(42^^40)-2]) mod 10
=2^(4*somebigassnumber) mod 10
=6

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