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Twenty pairs of integers are formed using each of the integers 1, 2, 3, ..., 40 once. The positive difference between the integers in each pair is 1 or 3. (For example, 5 can be paired with 2, 4, 6 or 8.) If the resulting differences are added together, the greatest possible sum is

 Jun 10, 2020
 #1
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Here is my attempt:

 

14, 25, 36, 710, 811, 912, 1316, 1417, 1518, 1922, 2023, 2124, 2528, 2629, 2730, 3134, 3235, 3336, 3740, 3839

 

The 3-digit numbers should be read as: 7 and 10 or 8 and 11.....etc. The 4-digit numbers as 13 and 16 or 14 and 17...etc.

 

There is an "absolute" difference of 3 between all the pairs, except the last one. Therefore, the maximum sum I get is:

 

19 x 3 + 1 =58

 Jun 10, 2020
 #2
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I do not know if I understand the question but if the biggest difference between the pairs is 3 then the pairs could be

1,4

2,5

3,6

 

7,10

8,11

9,12

 

13,16

14,17

15,18

 

19,22

20,23

21,24

 

25,28

26,29

27,30

 

31,34

32,35

33,36

 

37,40

38,39

 

19*3+1 = 58

 

So I get the same 'guess' answer as guest does.   laugh

 Jun 10, 2020
edited by Melody  Jun 10, 2020

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