I am a beginner in Number Theory so just starting. I tried this problem the following way, I tried using Euler Theorem on the problem and got $7^4 \equiv 1 \text{(mod 10})$. Then I focused on the exponent $7^7$ and reduced it by mod 10 as $7^4 \times 7^3$ to $7^3 \equiv (-3)^3 \equiv (-1)(3)^3 \equiv (-1)(-3) \equiv 3$ and again repeated on the base and got $3$ as the answer, Is this correct? If I am wrong please correct me. And If you have any alternate solutions then greatly appreciated. Thanks for your time.
Asked
Active
Viewed 156 times
-1
-
@Ganit $7^7$ is not 49 so it will not be true I think. – Beena Jay Oct 04 '21 at 06:01
1 Answers
0
The answer is $3$. Observe the following:
$7^{1} \equiv 7 \pmod{10}$
$7^{2} \equiv 9 \pmod{10}$
$7^{3} \equiv 3 \pmod{10}$
$7^{4} \equiv 1 \pmod{10}$.
This pattern repeats. That is, powers of $7$ modulo $10$ repeat in cycles of $4$.
This means that if we can compute $x = 7^{7} \pmod{4}$, then we can easily use the above observation to compute $7^{x} \pmod{10}$. To find $x$, we once again note that $7^{2} \equiv 1 \pmod{4}$, which means that
$$7^{7} \equiv 7 \times (7^{2})^{3} \equiv 7 \times 1 \pmod{4} \equiv 3.$$
Thus, $7^{7^{7}} \equiv 7^{3} \equiv 3 \pmod{10}$.
Ekesh Kumar
- 3,500
-
-
I might be wrong but when I checked this in Wolfram Alpha it's giving $7$ as the answer. Check here https://www.wolframalpha.com/input/?i=remainder+of+%287%5E7%29%5E7%2F10 – Ganit Oct 04 '21 at 06:08
-
1@Ganit You have misplaced the brackets. It should be $7^\left(7^7\right)$, but you're using $\left(7^7\right)^7$ instead with Wolfram Alpha. – John Omielan Oct 04 '21 at 06:11
-
@Ganit https://www.wolframalpha.com/input/?i=remainder+of+7%5E7%5E7%2F10 I think this – Beena Jay Oct 04 '21 at 06:13
-
@Ganit - you had parentheses wrong - it should be remainder of 7^(7^7)/10 – Mike O'Connor Oct 04 '21 at 06:18