+0

+1
100
2

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
+1

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
+1

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. Jun 10, 2020
edited by Melody  Jun 10, 2020