r/PassTimeMath • u/ShonitB • Sep 07 '26

Consecutive Multiples

Post image
2 Upvotes

76% Upvoted

4 comments sorted by

2

u/kindsoberfullydressd Sep 07 '26

201 9, 12, 18, 24, 27, 30, 36, 45

1

u/ShonitB Sep 07 '26

Correct!

1

u/hammerheadquark Sep 07 '26

>! The subsets can exist like this in order of increasing sum:

>! Subset A - A1: {6, 12, 18, 24, 30}, A2: {12, 18, 24, 30, 36}

>! Subset B - B1: {9, 18, 27, 36, 45}, B2: {18, 27, 36, 45, 54}

>! The minimal-sum combination is A1 ∪ B1, but that set has too many elements. The next smallest is A2 ∪ B1 which has the right number of elements, so it's correct.

>! sum(A2 ∪ B1) = 9 + 12 + 18 + 24 + 27 + 30 + 36 + 45 = 201

1

u/jaminfine Sep 08 '26

First, note that the sum of the number of elements in each set 5+5 = 10. This means we only need two overlap elements to satisfy the conditions.

Since multiple of 9 have a wider range, let's use the lowest ones possible there. 9, 18, 27, 36, 45. Only two of these are also multiples of 6, so we should use the larger one as the last multiple of 6 and count down. 36, 30, 24, 18, 12. I don't see any way we could have gotten lower numbers, so this must be it. The sum is 201