Question

Three Questions for loads of points :-

1. Let $\frac{ 10^{27}+2 }{3} =k$ Find the sum of digits of K.

2. Find the remainder when $9^{1990}$ is divided by 11.

3. Look at the following system of calculations :-
    [tex]3 || 5 = 1 \\ 7 || 14 = 7 \\ 3 || 15 = 15 \\ 15 || 20 = 5 \\ 17 || 34 = 17
[/tex]
    Find 64 || 30

[Where "||" is any calculation including BODMAS or combination of several functions.]

Answer 1

1) Note that working mod 10 yields the sum of the digits.
(10^27 + 2) = 1^27 + 2 = 3 (mod 10).
==> k = (10^27 + 2)/3 = 1 (mod 10).

So, the sum of the digits of k equals 1.
----
2) Working mod 11,
9^1990 = (-2)^1990
............= 2^1990
............= (2^10)^199
............= 1^199, since 2^(11-1) = 1 (mod 11) by Fermat
............= 1.

So, the remainder equals 1.
----
3) It appears to be the greatest common factor of the two numbers...

3 || 5 = 1
7 || 14 = 7
3 || 15 = 15 <-- Hmm... this should be 3?
15 || 20 = 5
17 || 34 = 17

The greatest common factor of 64 and 30 is 2:

64 || 30 = 2

Answer 2

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1) Note that working mod 10 yields the sum of the digits.

(10^27 + 2) = 1^27 + 2 = 3 (mod 10).

==> k = (10^27 + 2)/3 = 1 (mod 10).

So, the sum of the digits of k equals 1.

----

2) Working mod 11,

9^1990 = (-2)^1990

= 2^1990

= (2^10)^199

= 1^199, since 2^(11-1) = 1 (mod 11) by Fermat

= 1.

So, the remainder equals 1.

----

3) It appears to be the greatest common factor of the two numbers...

3 || 5 = 1

7 || 14 = 7

3 || 15 = 15 <-- Hmm... this should be 3?

15 || 20 = 5

17 || 34 = 17

The greatest common factor of 64 and 30 is 2:

64 || 30 = 2