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Firlakuza [10]
4 years ago
14

You randomly select k integers between 1 and 100, inclusive. What is the smallest k that guarantees that at least one pair of th

e selected integers will sum to 101? Prove your answer.
Mathematics
1 answer:
vredina [299]4 years ago
6 0

Answer:

51

Step-by-step explanation:

The possible components to sum up to 101 can only be divided into 2 groups, 1 is larger than 50 and the other is less than or equals to 50

For example

50 + 51 = 101

52 + 49 = 101

60 + 41 = 101

...

99 + 2 = 101

100 + 1 = 101

Therefore, the worst case scenario is to pick all numbers from only 1 group, either all number less than or equal to 50, which there are 50 of them from 1 to 50, or greater than 50, which there are 50 of them from 51 to 100.

So k has to be at least 51 to guarantee that at least one pair of the selected integers will sum to 101

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Let's say the 3 points are A, B, C.

If A, B and C lie on one line then m_{AB} = m_{BC} = m_{AC}

m_{AB} = \frac{y_{B}-y_{A} }{x_{B}-x_{A}} = \frac{2-6}{3-5} = \frac{-4}{-2} = 2\\m_{BC} = \frac{y_{C}-y_{B} }{x_{C}-x_{B}} = \frac{8-2}{6-3} = \frac{6}{3} = 2\\\\m_{AC} = \frac{y_{C}-y_{A} }{x_{C}-x_{A}} = \frac{8-6}{6-5} = \frac{2}{1} = 2\\\\

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You don't need to use a graph paper to prove it.

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