PR 4 # 32

[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

[Richard Cameron]

Originally Posted by mfgoode

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

sorry: I did read it and I did appreciate it. But in post 13 you did not give the correct answer!

Originally Posted by me, mainly

[ "you can read off everything you need to find the solution"] AMBIGUOUS!

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.

Originally Posted by 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

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

[mfgoode]

Originally Posted by 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

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

[mfgoode]

K

Originally Posted by 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

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

[fetofs]

Originally Posted by 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.

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

[Richard Cameron]

Originally Posted by fetofs

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

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

Richard

[mfgoode]

Originally Posted by 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

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

mally

[Richard Cameron]

Originally Posted by mfgoode

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

mally

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:

Originally Posted by 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.

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

[R.D. Silverman]

Originally Posted by 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

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 ..........

[alpertron]

Originally Posted by 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 ..........

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.

[R.D. Silverman]

Originally Posted by alpertron

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.

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