Here is a fairly straight-forward problem from 500 Mathematical Challenges.
“Problem 256. Let n be a positive integer. Show that (x – 1)2 is a factor of xn – n(x – 1) – 1.”
Here is a fairly straight-forward problem from 500 Mathematical Challenges.
“Problem 256. Let n be a positive integer. Show that (x – 1)2 is a factor of xn – n(x – 1) – 1.”
This is another thoughtful puzzle from the imaginative mind of James Tanton (with slight edits).
“Three poles of height 1183 feet, 182 feet, 637 feet stand in the ground. Pick a pole and saw off all the taller poles at that height. Plant those tops in the ground too. Repeat until no more such saw cuts can be made. Despite choices made along the way, what final result is sure to occur? [Four poles, heights a, b, c, d ft?]”
See the Pole Leveling Puzzle for a solution.
James Tanton has come up with another imaginative concrete problem harboring a mathematical pattern.
“60 trees in a row. Their stars are yellow, orange, blue, Y, O, B, Y, O, B, … Their pots are orange, yellow, pink, blue, O, Y, P, B, O, Y, P, B, … Their baubles are mauve, pink, yellow, blue, orange, M, P, Y, B, O, M, P, Y, B, O, … Must there be an all yellow tree? All B? One with star = O, pot = O, baubles = M?”
See the Christmas Tree Puzzle for a solution.
Here is another delightful problem from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).
“In Sussex, Holmes and I ran into a pair of woodcutters named Doug and Dave. There was an air of the unreliable about them—not helped by a clearly discernable aroma of scrumpy—but they nevertheless proved extremely helpful in guiding us to a particular hilltop clearing some distance outside of the town of Arundel. A shadowy group had been counterfeiting sorceries of a positively medieval kind, and all sorts of nastiness had ensued.
The Adventure of the Black Alchemist is not one that I would feel comfortable recounting, and if my life never drags me back to Chanctonbury Ring I shall be a happy man. But there is still some instructive material here. Whilst we were ascending our hill, Doug and Dave made conversation by telling us about their trade. According to these worthies, working together they were able to saw 600 cubic feet of wood into large logs over the course of a day, or split as much as 900 cubic feet of logs into chunks of firewood.
Holmes immediately suggested that they saw as much wood in the first part of the day as they would need in order to finish splitting it at the end of the day. It naturally fell to me to calculate precisely how much wood that would be.
Can you find the answer?”
See the Loggers Problem for solutions.
Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos). This one is quite a bit more challenging, at least for me.
“When Holmes and I met with Wiggins one afternoon, he was accompanied by a rather scrappy-looking mutt, who eyed me with evident suspicion.
‘This is Rufus,’ Wiggins said. ‘He’s a friend.’
‘Charmed,’ I said.
‘He’s very energetic,’ Wiggins told us. ‘Just this morning, he and I set out for a little walk.’
At the word ‘walk’, the dog barked happily.
‘When we set out, he immediately dashed off to the end of the road, then turned round and bounded back to me. He did this four times in total, in fact. After that, he settled down to match my speed, and we walked the remaining 81 feet to the end of the road at my pace. But it seems to me that if I tell you the distance from where we started to the end of the road, which is 625 feet, and that I was walking at four miles an hour, you ought to be able to work out how fast Rufus goes when he’s running.’
‘Indeed we should,’ said Holmes, and turned to look at me expectantly.
What’s the dog’s running speed?”
See the Rufus Puzzle for solutions.
Here is a collection of puzzles from the great logic puzzle master Raymond Smullyan in a “Brain Bogglers” column for the 1996 Discover magazine.
See Family Values for solutions.
Here is another problem (slightly edited) from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).
“Holmes and I were walking along a sleepy lane in Hookland, making our way back to the inn at which we had secured lodgings after scouting out the estates of the supposed major, C. L. Nolan. Up ahead, a team of horses were slowly pulling a chained tree trunk along the lane. Fortunately it had been trimmed of its branches, but it was still an imposing sight.
When we’d overtaken the thing, Holmes surprised me by turning sharply on his heel and walking back along the trunk. I stopped where I was to watch him. He continued at a steady pace until he’d passed the last of it, then reversed himself once more, and walked back to me.
‘Come along, old chap,’ he said as he walked past. Shaking my head, I duly followed.
‘It took me 140 paces to walk from the back of the tree to the front, and just twenty to walk from the front to the back,’ he declared.
‘Well of course,’ I said. ‘The tree was moving, after all.’
‘Precisely,’ he said. ‘My pace is one yard in length, so how long is that tree-trunk?’
Can you find the answer?”
See the Tree Trunk Puzzle for solutions.
This is another problem from the 2020 Math Calendar.
“Find the difference between the highest and lowest roots of
f(x) = x3 – 54x2 + 969x – 5780”
See Root Difference for a solution.
The craziness of manipulating radicals strikes again. This 2006 four-star problem from Colin Hughes at Maths Challenge is really astonishing, though it takes the right key to unlock it.
“Problem Consider the following sequence:
For which values of [positive integer] n is S(n) rational?”
See Amazing Radical Sum for a solution.
Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).
“Our pursuit of the dubious Alan Grey, whom we encountered during The Adventure of the Third Carriage, led Holmes and myself to a circular running track where, as the sun fell, we witnessed a race using bicycles. There was some sort of substantial wager involved in the matter, as I recall, and the track had been closed off specially for the occasion. This was insufficient to prevent our ingress, obviously.
One of the competitors was wearing red, and the other blue. We never did discover their names. As the race started, red immediately pulled ahead. A few moments later, Holmes observed that if they maintained their pace, red would complete a lap in four minutes, whilst blue would complete one in seven.
Having made that pronouncement, he turned to me. ‘How long would it be before red passed blue if they kept those rates up, old chap?’
Whilst I wrestled with the answer, Holmes went back to watching the proceedings. Can you find the solution?”
See the Track Problem for a solution.