I need a calendar. No, it won't work (Part 2)

If today is $\text{Saturday}$ , $7$ days from now would also be a $\text{Saturday}$ , and $7$ days from then would also be a $\text{Saturday}$ .

So, to find what day it would be ${10}^{20}$ days later, we need to find the remainder obtained on dividing ${10}^{20}$ by $7$ .

If we get remainder as $0$ , then the day is $\text{Saturday}$ . If we get the remainder as $1$ , we add one day to $\text{Saturday}$ , i.e, we get $\text{Sunday}$ . If we get the remainder as $2$ , we add two days to $\text{Saturday}$ , i.e, we get $\text{Monday}$ . If we get the remainder as $3$ , we add three days to $\text{Saturday}$ , i.e, we get $\text{Tuesday}$ and so on.

Now,

$10\equiv 3(mod\quad 7)\\ \Rightarrow { 10 }^{ 20 }\equiv { 3 }^{ 20 }(mod7)\\ \\ { 3 }^{ 20 }={ 9 }^{ 10 }\\ \\ 9\equiv 2(mod7)\\ \Rightarrow { 9 }^{ 10 }\equiv { 2 }^{ 10 }(mod7)\\ \\ { 2 }^{ 10 }={ 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 3 }\times { 2 }^{ 1 }\\ \\ { 2 }^{ 3 }\equiv 1(mod7)$

Thus, ${2}^{10} \text{ mod } 7 = 1 \times 1 \times 1 \times 2$

Hence, the remainder is $2$ and the day is $\huge {\text {Monday}}$

$10^{20} = 10^{18} \times 10^2 = (10^3)^6 \times 10^2 = 1000^6 \times 100$ . Now an interesting point is that 1001 is divisible by 7 so 1000 = -1 (mod 7).

So we have $1000^6 \times 100 = (-1)^6 \times 100 (mod 7) = 1 \times 100 (mod 7) = 2 (mod 7)$

10/7 has remainder 3.

100/7 has remainder 2.

1,000/7 has remainder 6.

10,000/7 has remainder 4.

100,000/7 has remainder 5.

1,000,000/7 has remainder 1.

10,000,000/7 has remainder 3 again.

Then the remainders repeat.

The repeating pattern of remainders has 6 elements.

10^18 divided by 7 has remainder 1.

10^19 divided by 7 has remainder 3.

10^20 divided by 7 has remainder 2.

So, the day is 2 days after a Saturday - Monday.

$10^{20}\equiv3^{20}\equiv3^2\equiv2 \pmod7,$ so, many weeks and two days will pass, and it will be $\boxed{Monday}$ (if the universe still exists). We have used $3^6\equiv1 \pmod7$ , by Fermat.

Saturday to Friday is 7 days. Three days after that is Monday. That makes 10 days. No matter what power you raise ten to, Monday will still be three days after Friday.

10^20 = (7 + 3)^(3 X 6 + 2) = 3^2 (mod 7) = 2 (mod 7)

So the answer is Monday

if the remainder of "a number of days" divided by 7 is 0, then "a number of days" after saturday is also saturday

by using fermat's little theorem, 10^{6} mod 7 = 1 mod 7

so, 10^{20} mod 7 = 10^{(6 x 3) + 2} mod 7 = (10^{6 x 3} x 10^{2}) mod 7 = 10^{2} mod 7 = 2

we get 2 days after saturday which is monday

10^20 mod 7 days. With euler function, we know that 10^20 mod 7 = 2 mod 7... So, the day is Saturday + 2 days = Monday