Mally,

I posted the table in direct response to Xyzzy's query. I was not answering Wacky's original question. That had been adequately answered already.

What happened to "Its better to give than to receive"?

Fusion

[QUOTE=mfgoode]:huh:
Evidently you have not read my post No. 13 where I clearly mention the cycle of digits repeating much before Fusion Power used brute force when it was not necessary. It appears to me that FP merely amplified my rule after reading it
Hey Richard please give credit when it is deserved
[/QUOTE]

sorry: I did read it and I did appreciate it. But in post 13 you did not give the correct answer!
[QUOTE=me, mainly]
[ "you can read off everything you need to find the solution"] AMBIGUOUS!
[/QUOTE]
the details of the method of solution had been given before so I didn't go through it again: in order to identify the last digit of the original number you need the exponent mod 4 so you need the last two digits of the exponent. And you pointed this out too, thank you.

[QUOTE=Mally]From Fusion Power's ready table I observe that the digits 0 and 4 do not always alternate but its always one of them
[/QUOTE]

I used Fusion Power's table because it was convenient and available. And from there you can see that the pattern for two digits is 07,49,43,01, so the penultimate digit goes 0,0,4,4.

If you extend that table, you will see that the third last digit has a period of 20, the fourth last 100, and -I was sensing a pattern by this time- the fifth last 500.

I don't know how exceptional 7 is in this respect, I think some digits have maximal periods.


Richard

PR4#32
 

[QUOTE=Fusion_power]Mally,

I posted the table in direct response to Xyzzy's query. I was not answering Wacky's original question. That had been adequately answered already.

What happened to "Its better to give than to receive"?

Fusion[/QUOTE]
:smile:
Fusion: Yes I certainly believe that phrase and practice it too
Bu here it is slightly out of context though.
So here lets say "Forgive us our trespasses as we forgive them that trespass against us"
Regards,
Mally:coffee:

Pattern
 

K[QUOTE=Richard Cameron]sorry: I did read it and I did appreciate it. But in post 13 you did not give the correct answer!

the details of the method of solution had been given before so I didn't go through it again: in order to identify the last digit of the original number you need the exponent mod 4 so you need the last two digits of the exponent. And you pointed this out too, thank you.



I used Fusion Power's table because it was convenient and available. And from there you can see that the pattern for two digits is 07,49,43,01, so the penultimate digit goes 0,0,4,4.

If you extend that table, you will see that the third last digit has a period of 20, the fourth last 100, and -I was sensing a pattern by this time- the fifth last 500.

I don't know how exceptional 7 is in this respect, I think some digits have maximal periods.


Richard[/QUOTE]
:smile: Excellent observation Richard.
The reciprical of 7 is 142857 repeating indefinitely.
You may try 1/13 ,1/17, 1/19.
See if you can crack out a pattern for the bigger primes

Mally:coffee:

[QUOTE=Richard Cameron]

If you extend that table, you will see that the third last digit has a period of 20, the fourth last 100, and -I was sensing a pattern by this time- the fifth last 500.

I don't know how exceptional 7 is in this respect, I think some digits have maximal periods.
[/QUOTE]

Do you feel sad by knowing that the sixth last has a period of 5000?

[QUOTE=fetofs]Do you feel sad by knowing that the sixth last has a period of 5000?[/QUOTE]

yes, devastated. My great mathematical theory is shown to be flawed.
Oh well, back to proving the Riemann Hypothesis for me.

Richard

PR4#32
 

[QUOTE=Richard Cameron]!
#~
the details of the method of solution had been given before so I didn't go through it again: in order to identify the last digit of the original number you need the exponent mod 4 so you need the last two digits of the exponent. And you pointed this out too, thank you.

Richard[/QUOTE]

Sorry Richard but I cant understand your terminology.
Kindly be so kind as to explain this to me?

mally :coffee:

[QUOTE=mfgoode]Sorry Richard but I cant understand your terminology.
Kindly be so kind as to explain this to me?

mally :coffee:[/QUOTE]

by the original number I meant 7^7^7; by the exponent i meant 7^7. I hope the rest makes sense.



I've thought about this a little more and if you look back at R Gerbicz's elegently concise solution and modify it a little, an even shorter solution becomes apparent:

[QUOTE=R Gerbicz]
Yes, the answer is 3.

7^7=823543=4*k+3, so
7^(7^7)=7^823543=7^(4*k+3)=343*2401^k==3*1==3 mod 10.[/QUOTE]

If you start from
7^7 = 7^3 * 7^4 = 343 * 2401
Its obvious without actually multiplying this out that the last two digits of 7^7 are 43 and so 7^(7^7) will be 7^(4k+3) == 343 mod 10.


Richard

[QUOTE=Richard Cameron]


I've thought about this a little more and if you look back at R Gerbicz's elegently concise solution and modify it a little, an even shorter solution becomes apparent:



Richard[/QUOTE]

There is even a shorter solution. I will give a hint: 7 = -1 mod 4.

Try 7^(7^(7^(7^(7^(7^7))))) mod 10 ..........

[QUOTE=R.D. Silverman]There is even a shorter solution. I will give a hint: 7 = -1 mod 4.

Try 7^(7^(7^(7^(7^(7^7))))) mod 10 ..........[/QUOTE]

[spoiler]Since we want the solution x mod 10, we have to work both mod 2 and mod 5.

The first case (mod 2) is simple because we have:

x = 7^(7^(7^(7^(7^(7^7))))) = 1^(...) = 1 (mod 2)

The second case (mod 5) can be worked as follows:

From Fermat's Little theorem we have:

x = 7^(7^(7^(7^(7^(7^7))))) = 2^a (mod 5)

where a = 7^(7^(7^(7^(7^7)))) (mod 4)

so a = 3^(7^(7^(7^(7^7)))) (mod 4)

a = 3^odd = 3 (mod 4)

So x = 2^3 = 8 = 3 (mod 5)

Since x = 1 (mod 2) and x = 3 (mod 5), we must have [b]x = 3 (mod 10)[/b]

Notice that it does not matter the number of 7s.[/spoiler]

[QUOTE=alpertron][spoiler]Since we want the solution x mod 10, we have to work both mod 2 and mod 5.

The first case (mod 2) is simple because we have:

x = 7^(7^(7^(7^(7^(7^7))))) = 1^(...) = 1 (mod 2)

The second case (mod 5) can be worked as follows:

From Fermat's Little theorem we have:

x = 7^(7^(7^(7^(7^(7^7))))) = 2^a (mod 5)

where a = 7^(7^(7^(7^(7^7)))) (mod 4)

so a = 3^(7^(7^(7^(7^7)))) (mod 4)

a = 3^odd = 3 (mod 4)

So x = 2^3 = 8 = 3 (mod 5)

Since x = 1 (mod 2) and x = 3 (mod 5), we must have [b]x = 3 (mod 10)[/b]

Notice that it does not matter the number of 7s.[/spoiler][/QUOTE]


While you got the right answer, you ignored the hint. There is a
MUCH easier way to do this.