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Sample text

Then, sh (0) = 0, sh (1) = d, sr (0) = d, sr (1) = 0. With sh and sr also δ(t) = sr (t) − sh (t) is a continous function of t. Because of the different signs of δ(0) = d and δ(1) = −d the function δ(t) has at least one zero to in the intervall [0, 1]. At time to we have sr (to ) = sh (to ). Under which conditions are there more than one such point? Solutions to Problems 45 Problem 6(a) There are eight equations with nine unknowns. And, the solutions have to consist of the natural numbers 1,. .

B needed tB TU for the 100m. Hence his speed is vB = 100/tB = 90/tA . 9. The speed of C is vC = 90/tB = x/tA . 9 = 81m. Thus, the first runner A beats C by 19m. Solutions to Problems 41 Problem 3(a) 36 = 22 · 32 . If one considers also the one year olds, then there are the following combinations: 3. 1 1 1 1 1 2 2 3 2. 1 2 3 4 6 2 3 3 1. 36 18 12 9 6 9 6 4 38 21 16 14 13 13 11 10 Only in case of sum 13 another hint was necessary. But there is an oldest child only if the family has two years old twins and a six years old child.

A one–to–one function f : IN ⊃ D → W ⊂ IN, so that f (x) is easily and f inv (y) is extremely hard to compute. 2. Now, say Alice starts and chooses an odd or even x ∈ D. Now she sends y with y = f (x) to Bob without offenbaren x. 3. Bob receives y and bets whether x was odd or even. ) 4. e. to let Bob compare f (x) with the y he initially received. Solutions to Problems 51 Problem 11(a) They need three locks with two keys each. If each person owns keys according to the following scheme, ✑ lock 1 €€ ✑✑ € ✑€€€ lock 2 Bob €€ ✑✑ €€ ✑ €€ ✑ €€ € lock 3 Claire ✑ Alice €€ € then only at least two persons together have keys for all three locks of the treasure box.

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