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… 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 must be 0) and thus 3 even. 120 = 2 3 · 31 · 51. So there are 4 · 2 · 2 = 16 divisors, of which 2 · 2 = 4 are odd and thus 12 even. 1 Problem 2. (a) p 2 = 1 + 1 1 + p 2 , p 2� 1 = 1 1 + p 2 , ( p 2� 1)(1 + p 2) = 1. (b) p 2 = 1 + 1 1 + p 2 = 1 + 1 2 + 1 1+ p 2 = 1 + 1 2 + 1 2+ 1 1+ p 2 = . . . Hence a0 = … 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 …

… 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 must be 0) and thus 3 even. 120 = 2 3 · 31 · 51. So there are 4 · 2 · 2 = 16 divisors, of which 2 · 2 = 4 are odd and thus 12 even. 1 Problem 2. (a) p 2 = 1 + 1 1 + p 2 , p 2� 1 = 1 1 + p 2 , ( p 2� 1)(1 + p 2) = 1. (b) p 2 = 1 + 1 1 + p 2 = 1 + 1 2 + 1 1+ p 2 = 1 + 1 2 + 1 2+ 1 1+ p 2 = . . . Hence a0 = … 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 …