PR 4 # 32

[Wacky]

What is the rightmost digit of 7^7[sup]7[/sup] ?

[xilman]

Originally Posted by Wacky

What is the rightmost digit of 7^7[sup]7[/sup] ?

1

Proof:

7^2 = 49 == -1 mod 10
Therefore, 7^6 = (7^2)^3 == (-1)^3 == -1 mod 10

Therefore 7^7 == -3 == 4 mod 10.

Therefore 7^(7^7) == 7^4 == (7^2)^2 == (-1)^2 == 1 mod 10

[Citrix]

3. To get answer, all you have to do is calculate mod 10.

I was going to be the first to answer this, the 300 sec posting time limit let Xilman answer first.

[axn]

Whoa there!

Order(7,10) = 4

Therefor 7^(7^7) = 7^(Mod(7^7,4)) (mod 10).

[Wacky]

Paul,

"No cigar". Try again.

[fetofs]

Originally Posted by Wacky

What is the rightmost digit of 7^7[sup]7[/sup] ?

7

So, we have 7*7*7*7 (49 times) . Since a*b (mod 10) = (a mod 10)*(b mod 10), we can reduce the multipliers as we go through. This is a sequence, but we must find it first. We can know for sure it's smaller than 11! :)
Exponent = 2 mod 4 7*7 = 9 (mod 10)
Exponent = 3 mod 4 9*7 = 3 (mod 10).
Exponent = 0 mod 4 3 * 7 = 1 (mod 10)
Exponent = 1 mod 4 1 * 7 = 7 (mod 10).
We have returned to our original point. Therefore every exponent that is -2 mod 4 equals 9 mod 10, and so on. Since 49 == 1 (mod 4), the result is congruent to 7 (mod 10)

[Xyzzy]

3?

[Wacky]

I have ten possibilities:
0, 1, 2, 3, 4, 5, 6, 7, 8, and 9

Please justify why you choose a particular one / (exclude some of them)

[Xyzzy]

Originally Posted by Wacky

Please justify why you choose a particular one / (exclude some of them)

$ echo '7^7^7' | bc | tail --bytes=2
3

[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.

[Richard Cameron]

since 7^4 = 2401 and so ==1 mod 100, the penultimate digit must be 4.