(1) Suppose .

Case

Case

Case

Case

(2) Since there are more 2’s than 5’s in 2017! and , all we need to do is to count the number of 5’s in the prime factorization of 2017!.

(3) Suppose that .

, for some

By completing the square,

Case

Case

Case

Case

If we define zero being a perfect square, then . Otherwise .

(4)

constant term:

In particular,

In general, for any integer *n,*

(5) Let be the radius of the smallest semi-circle.

Note that the perimeter of semi-circle is ,

(6) Let the side length of B, C, D, E, F, G, H, and I be b, c, d, e, f, g, h, and i respectively.

(7) By Fermat’s little theorem,

(8) By Pythagoras theorem,

(9) By Euclidean Algorithm,

hence

(10) Divide the numbers from 1 to 100 into congruence classes mod 11.

[1] : 1, 12, … , 100; There are 10 members so we can choose at most 5 members, since we don’t want any pair of numbers with difference of 11. (Any adjacent members in the class have difference of 11)

[2] : 2, 13, … , 90; There are 9 members so we can choose at most 5 members.

[3] : 3, 14, … , 91; There are 9 members so we can choose at most 5 members.

… … , similarly for classes such as [4] to [11], each of them have 9 members. Therefore we can at most pick numbers without any pair in the set having difference of 11.

### Like this:

Like Loading...

*Related*