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Solve $x$, where $x\in[0,9]\cap\mathbb{Z}:$

$$13^{11^7}\equiv x\pmod{10}$$

Here is my thoughts:

Since $13\equiv3\pmod{10}$

Implies $13^{11^7}\equiv 3^{11^7}\pmod{10}$

$$3^0=1,3^1=3,3^2=9,3^3=27,3^4=81\cdots$$

For mod $10$, have reminder repeat between:

(they have to repeat, but i will skip the induction here) $$R_1=\{1,3,9,7\}$$

Then solve

$$11^7+1\equiv x\pmod{|R_1|}$$

$$11^0+1=2,11^1+1=12,11^2+1=122,11^3+1=1332$$

For mod $4$, have reminder index repeat between:

$$R_2=\{2,0\}$$

In another word, the reminder repeat between second and last term in $R_1$

Then solve $$7+1\equiv x\pmod{|R_2|},\text{ that }x=0$$

$x=0$, means the reminder is the last term in $R_1$ which is $7$, that implies

$$13^{11^7}\equiv7\pmod{10}$$

Therefore $x=7$

My qeustion: is there any theorem to apply so this calculation could be simpler$?$

Any help would be appreciated.

Community
  • 1
Ethan
  • 5,291
  • Familiar with little fermat or euler ? – AgentS Nov 05 '19 at 02:56
  • There is a theorem called Euler theorem, using Euler totient function we got $$13^{\varphi(10)} = 13^4 \equiv 1 \pmod {10}$$ meaning that you need modulo 4 in the exponent – Azlif Nov 05 '19 at 02:56

3 Answers3

2

HINT:

Using Euler's theorem,
$\phi(10) = 4 \Rightarrow 13^4\equiv 1 \pmod{10}$
and
$11^7 \equiv (-1)^7\pmod{4}$

AgentS
  • 12,195
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$3^4=81\equiv1\mod 10$

$11^7\equiv(-1)^7=-1\equiv3\mod 4$

Therefore, $13^{11^7}\equiv3^{11^7}\equiv3^3=27\equiv7\mod 10$

J. W. Tanner
  • 60,406
1

Hint: Note that $13^n~(\text{mod}~10)=7$ for integers congruent to $3~(\text{mod}~4)$.

And any number congruent to $3~(\text{mod}~4)$ raised to the power of an odd number will always be congruent to $3~(\text{mod}~4)$.