Question #11493

what is the remainder when 7^7^7^upto infinite times(7 to the power 7 to the power 7 to the power ....infinite times) is divided by 13

Expert's answer

If we have 7^7^7^7^... then it is obvious an infinity.

So we must have

7^7^7^...^7 only some n-times.

Euler function of 13 is 12. Then 7^12=1(mod

13). 7^7=7(mod 12).

So, starting from tail of 7^7^7^...^7 we can replace

7^7^7 by 7 . So, finally we get the remainder 7 or 7^7=6(mod 13).

So we must have

7^7^7^...^7 only some n-times.

Euler function of 13 is 12. Then 7^12=1(mod

13). 7^7=7(mod 12).

So, starting from tail of 7^7^7^...^7 we can replace

7^7^7 by 7 . So, finally we get the remainder 7 or 7^7=6(mod 13).

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