Find the remainder when 1!+2!+...+299!+300! is divided by 21

Answer 1

a) Find the last two digits(units and tens digit) in 1829^(1829)

b) Find the units digit in 23^(7777)

Answer 2

To find large powers modulo some r, you can repeatedly square and/or multply numbers and reduce it modulo r at each step. E.g. if we do all coputations Mod 10 in the following, we can write:

1829 = -1

The 1829 th power of both sides is thus:

1829^1829 = (-1)^1829 = -1

Therefore the last digit of 1829^1829 is 9.

Answer 3

You can compute this Mod 100 using the Chinese Remainder Theorem, by ecaluating it separately Mod 25 and
Mod 4. Mod 4 you have:

1829 = 1, so

1829^1829 = 1

Mod 25 you have:

1829 = 4

Now 4^5 = 2^10 = 1024 = -1

So, 4^10 = 1 and thus

4^1829 = 4^9 = 4^(-1)

4*6 = 24 = -1, so the inverse of 4 is -6 = 19

We thus have that:

1829^1829 = 1 Mod 4

1829^1829 = -6 Mod 25

We can then write down the answer Mod 100 as follows. If we denote the inverse of x mod y is denoted as
[x^(-1)]_y, we can write down the solution as:

1 * 25 * [25^(-1)]_4 +

(-6) * 4 * [4^(-1)]_25

Modulo 25 the first term is zero as it is multiple of 25, ad the second term is -6 as the 4 and the inverse of 4 mod 25 cancel. Modulo 4 the last term is zero as that isnow a multiple of 4 while the first term is 1 as 25 times the inverse of 25 mod 4 now cancel.

We have:

1 * 25 * [25^(-1)]_4 = 25

(-6) * 4 * [4^(-1)]_25 = (-6)*4*(-6) = 24*6 = 44

Note that in these computations we can reduce Mod 100.

So, 1829^1829 Mod 100 = 25 + 44 = 69

Answer 4

23^(7777) Mod 10 =

3^7777 Mod 10

Mod 10 we have:

3^2 = 9 = -1 --->

3^4 = 1

7776 is a multiple of 4, so

3^7777 = 3

The last digit of 23^7777 is thus 3.