[Continuous] Samster's Puzzle of the Day - Join in at any time! page 7

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Because my good computer is still out of bounds, no fancy visuals today. :P


Puzzle 18:

The son of a rich businessman left home on the death of his father. All he had with him was a gold chain that consisted of 114 links. He rented a flat, in which he was required to pay every week one link of the gold chain as rent for the flat. The landlord told him that he wanted one link of the gold chain at the end of one week, two gold links at the end of two weeks, three gold links at the end of three weeks and so on. The son realised that he had to cut the links of the gold chain to pay the weekly rent. If the son wished to rent the place for 114 weeks, what would be the minimum number of links he would need to cut?

 
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…not 57?

 
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Originally posted by LouWeed:

…not 57?

Nope.

 
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56?

EDIT: Nononono wait

 
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7 cuts

 
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Originally posted by WiiPlayer113:

7 cuts

Nope.

 
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6 cuts, I meant. 7 segments of chain.

Segments of chain length 1, 2, 4, 8, 16, 32, and 51

 
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Originally posted by WiiPlayer113:

6 cuts, I meant. 7 segments of chain.

Segments of chain length 1, 2, 4, 8, 16, 32, and 51

Still incorrect. :P

 
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I guess I’ll just throw 5 out as a guess, then.

 
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4 Cuts. 4,1,8,1,16,1,32,1,50.

 
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I hate everything.

 
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Originally posted by hok0003:

4 Cuts. 4,1,8,1,16,1,32,1,50.

Correct! Sorry Wii :P

 
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At least I had the right idea, haha.

 
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Wait, so the landlord will actually give you back pieces of the chain? That doesn’t sound like any landlord I know.

 
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Originally posted by LouWeed:

Wait, so the landlord will actually give you back pieces of the chain? That doesn’t sound like any landlord I know.

Yup :L

 
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Puzzle 19:

As contributed by the wonderful Bluji :D

 
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Please forgive my horrible writing—it’s really hard to make suit shapes on a trackpad.

 
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That is a correct solution — though, you could flip 3 ♠ and 9 ♠ as well as 4 ♦ and 8 ♦. Well done!

 
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Originally posted by racefan12:

Please forgive my horrible writing—it’s really hard to make suit shapes on a trackpad.

I find it difficult even with a mouse.

 
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From now on, if anyone submits to me (through PMs) a good puzzle that I use, they will earn 1 point however will obviously not be able to answer it.

 
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A festive one today :D

Puzzle 20:

Below are 12 anagrams, each of which when arranged correctly display a Christmas phrase. Answer all twelve to get your point.

1: Travel and Dance

2: Rich Star Meets

3: Smashing tricks cost

4: Wean again Mary

5: Vicars Themes

6: The magic swiftness heart throb

7: Spice in em

8: Modestly switch after shave

9: He’s Fat Smart Rich

10: Scathing mud drips

11: Red Ernie

12: Let in Things

 
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Advent Calendar
Christmas Tree
Christmas Stockings
Away in a Manger
Christmas Eve
Twas the night before Christmas
Mince Pies
The Twelve Days of Christmas
Father Christmas
Christmas Pudding
Reindeer
Silent Night

 
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Originally posted by WiiPlayer113:

Advent Calendar
Christmas Tree
Christmas Stockings
Away in a Manger
Christmas Eve
Twas the night before Christmas
Mince Pies
The Twelve Days of Christmas
Father Christmas
Christmas Pudding
Reindeer
Silent Night

Correct!

 
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Sorry guys, no puzzle today. Super busy. Two tomorrow though ;)

 
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Sorry they’re a little late.

Puzzle 21:

Once upon a time, and old lady went to sell her vast quantity of cabbages at the local market.

When asked how many she had, she replied:

Son, I can’t count past 100 but I know that.

If you divide the number of cabbages by 2 there will be one cabbage left.
If you divide the number of cabbages by 3 there will be one cabbage left.
If you divide the number of cabbages by 4 there will be one cabbage left.
If you divide the number of cabbages by 5 there will be one cabbage left.
If you divide the number of cabbages by 6 there will be one cabbage left.
If you divide the number of cabbages by 7 there will be one cabbage left.
If you divide the number of cabbages by 8 there will be one cabbage left.
If you divide the number of cabbages by 9 there will be one cabbage left.
If you divide the number of cabbages by 10 there will be one cabbage left.

Finally, if you divide the number of cabbages by 11 there will be NO CABBAGES left!

Puzzle 22:

It was the hey-day of the late 1970s when the archbishop went to open the charity collection box. It contained 21 coins, totalling £2.21. Each of the normally current coins (50p, 10p, 5p, 2p, 1p, ½p) was represented, but each of these six denominations was represented by a different number of coins. Exactly how many coins of each denomination were in the box? – Thanks to LouWeed for this puzzle!