## MMM #23: Winner!

Kevin Wang is the winner of MMM #23. Congratulations, Kevin!

Here is Kevin's solution:

Each run of heads must be bordered by two tails or a tail and an end. Also, we know that there are only 11 possible positions for the first coin. Thus, we can just list all of the possibilities, number them for reference later, and allow for wildcard places as such:

H H H H H H H H H H T X X X X X X X X X (1)

T H H H H H H H H H H T X X X X X X X X (2)

X T H H H H H H H H H H T X X X X X X X (3)

X X T H H H H H H H H H H T X X X X X X (4)

X X X T H H H H H H H H H H T X X X X X (5)

X X X X T H H H H H H H H H H T X X X X (6)

X X X X X T H H H H H H H H H H T X X X (7)

X X X X X X T H H H H H H H H H H T X X (8)

X X X X X X X T H H H H H H H H H H T X (9)

X X X X X X X X T H H H H H H H H H H T (10)

X X X X X X X X X T H H H H H H H H H H (11)In each scenario, the amount of cases is equal to two to the power of the number of X's, so we can elaborate by case.

Case 1: 2^9 = 512

Case 2: 2^8 = 256

Case 3: 2^8 = 256

Case 4: 2^8 = 256

Case 5: 2^8 = 256

Case 6: 2^8 = 256

Case 7: 2^8 = 256

Case 8: 2^8 = 256

Case 9: 2^8 = 256

Case 10: 2^8 = 256

Case 11: 2^9 = 512Add these all up, and you get 9*(2^8)+2*(2^9), 2304+1024, 3328 total combinations.

Zach Feinstein contributed a Matlab verification of his solution:

function sum = mmm23() sum = 0; for setup = 0:2^20-1 set = dec2bin(setup,20); sum = sum + check(set); end end function result = check(set) consecutive = 0; result = 0; for ii = 1:length(set) if set(ii) == '1' consecutive = consecutive + 1; elseif consecutive == 10 result = 1; return; else consecutive = 0; result = 0; end end if consecutive == 10 result = 1; end end

Stay tuned for another Monday Math Madness contest, at BlinkDagger, on Monday.

LeoFebruary 3rd, 2009 - 19:54

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