Welcome to DU!
The truly grassroots left-of-center political community where regular people, not algorithms, drive the discussions and set the standards.
Join the community:
Create a free account
Support DU (and get rid of ads!):
Become a Star Member
Latest Breaking News
Editorials & Other Articles
General Discussion
The DU Lounge
All Forums
Issue Forums
Culture Forums
Alliance Forums
Region Forums
Support Forums
Help & Search
General Discussion
In reply to the discussion: Solve this! [View all]Lancero
(3,276 posts)92. 43. My brain hurts now.
Edit
1. 3 pairs of shoes = 30, 10 per pair so 5 per shoe.
2. 2 men and a pair of shoes, remove the pair and you're left with 10/2, so 5 per man.
3. 2 sets of ribbions and a man, remove 5 and you're left with 8/2, 4 per set, so 2 per ribbon.
4. And the difficult one, a man wearing shoes with ribbons on his hands. I'm assuming that everything he's wearing would be treated as in parentheses so thats 5, plus a pair of shoes, plus a set of ribbons for 19, times a ribbon to 38 plus a leftover shoe to get 43.
Though ignoring oop (But still assuming that the 'clothed' man would still count as one value) you'd get 48.
Edit history
Please sign in to view edit histories.
Recommendations
0 members have recommended this reply (displayed in chronological order):
94 replies
= new reply since forum marked as read
Highlight:
NoneDon't highlight anything
5 newestHighlight 5 most recent replies
RecommendedHighlight replies with 5 or more recommendations