MMM2024 Solutions_0.pdf
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… MMM Competition 2024 Problem 1. (a) 10 = 1⇥ 10 = 2⇥ 5 ) 4 divisors, so white 64 = 1⇥ 64 = 2⇥ 32 = 4⇥ 16 = 8⇥ 8 ) 7 divisors, so black 123 = 1⇥ 123 = 3⇥ 41 ) 4 divisors, so white (b) Paintings with a square number are black and all other paintings white. So 11 black paintings: 1,4,9,16,25,36,49,64,81,100,121. (c) We make us of prime factorizations. 98 = 2 1 · 72. So there are 2 · 3 = 6 divisors, of which 3 are odd (the exponent related to 2 … 3 Problem 4. Say that the common ratio in the geometric progression is r and that a, b and c are the initial number of peanuts of Alice, Bob and Charlie respectively. Then: a+b+c = 1397 and b = ar and c = br = ar2. This gives a·(1+r+r2) = 1397, which can be rewritten as 1 + r + r2 = 1397 a , or (r + 1) 2 = r + 1397 a Now let d be the common di↵erence in the arithmetic progression. Then we have that d = (b � 9) � (a � 5) = (c � 24) � (b � 9) (with b = ar and c = ar2). This gives: (d =) ar � 9 � (a � … are two ways you can go about it. (1) The hard way, by systematically analyzing in which way the outcome three can be obtained from such expressions, step by step. Not too difficult, but easy to make mistakes. For instance, in the following manner: * When using 0 operations, there is 1 way to get the outcome 3 from a single number. * When using 1 operation, there are 21 ways to arrive at the outcome 3: from 24-21, 23-20, 22-19, …, 4-1. * When using 2 operations, there are 42 ways to arrive at the …