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Firlakuza [10]
3 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]3 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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d =  \frac{4 - 3c}{c + 1}

Step-by-step explanation:

To make d the subject of formula, we need to rearrange the equation such that we arrive at d= _____.

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<em>Expand</em><em>:</em>

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<em>Bring</em><em> </em><em>all</em><em> </em><em>the</em><em> </em><em>d</em><em> </em><em>terms</em><em> </em><em>to</em><em> </em><em>one</em><em> </em><em>side</em><em> </em><em>and</em><em> </em><em>move</em><em> </em><em>the</em><em> </em><em>others</em><em> </em><em>to</em><em> </em><em>the</em><em> </em><em>other</em><em> </em><em>side</em><em> </em><em>of</em><em> </em><em>the</em><em> </em><em>equation</em><em>:</em>

cd + d = 4 - 3c

<em>Factorise</em><em> </em><em>d</em><em> </em><em>out</em><em>:</em>

<em>d(c + 1) = 4 - 3c</em>

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<em>d =  \frac{4 - 3c}{c + 1}</em>

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