Engrave numbers on 5 keys on a circular keyring so that the numbers on adjacent groups of keys sum to any value between 1 and 21 inclusively.
For example, 1,1,3,6,6 can sum up to any number between 1 and 17 (1=1, 1+1=2, 3=3, 3+1=4, 3+1+1=5, 6=6, 6+1=7, 6+1+1=8, 6+3=9, 6+3+1=10, etc).
Friday, January 25, 2008
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2, 5, 1, 3, 10
The ring only has 21 combinations. 5 numbers, 5 couples, 5 triples, 5 quad and 1 number from all 5 numbers added, giving 21 combinations total. Hence, every combination must be different.
Using this logic, no numbers can be the same, total sum must be 21. Hence, the number 1 and 2 are automatically one of the 5. The sum of the other 3 numbers is 18.
In order to make the number 4, the number 3 or 4 must be one of the 3 unknown numbers. I personally tried to prove that it can't be 3, but ended up getting the answer while trying.
In case of 1,2,3 being 3 of the 5 numbers, 1 and 2 cannot be next to each other. If 1,2,3 are in a row, "132" (or "231", same thing since its a ring) is the only possible combo, and there is no way to make the number 18 this way as the sum of the leftover is 15, and you can't connect 15 with 3. Hence they can't be in a row.
In case where they aren't in a row.
1 [some_number] 2 3 [another_number];
2 [some_number] 1 3 [another_number];
Are the only possibility (since 1 and 2 can't be connected).
First combination requires a 4 (& 11). 1 4 2 3 11 and 1 11 2 3 4 doesnt work.
Second combination requires a 5(& 10). Solved.
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