Here is a challenging problem from the Polish Mathematical Olympiads published in 1960.
“22. Prove that the polynomial
x44 + x33 + x22 + x11 + 1
is divisible by the polynomial
x4 + x3 + x2 + x + 1.”
See the Polynomial Division Problem
(Update 8/23/2021) The idea expressed in this post that mathematicians are “lazy” and seek short-cuts to solving questions and problems, as I did in this one, was recently the subject of a Numberphile post by Marcus du Sautoy: “Mathematics is all about SHORTCUTS“.
Here is another Brainteaser from the Quantum magazine.
“Mr. R. A. Scall, president of the Pyramid Bank, lives in a suburb rather far from his office. Every weekday a car from the bank comes to his house, always at the same time, so that he arrives at work precisely when the bank opens. One morning his driver called very early to tell him he would probably be late because of mechanical problems. So Mr. Scall left home one hour early and started walking to his office. The driver managed to fix the car quickly, however, and left the garage on time. He met the banker on the road and brought him to the bank. They arrived 20 minutes earlier than usual. How much time did Mr. Scall walk? (The car’s speed is constant, and the time needed to turn around is zero.) (I. Sharygin)”
I struggled with some of the ambiguities in the problem and made my own assumptions. But it turned out there was a reason they were ambiguous.
See the Walking Banker Problem for solutions.
This is another simple problem from Five Hundred Mathematical Challenges:
“Problem 57. Let X be any point between B and C on the side BC of the convex quadrilateral ABCD (as in the Figure). A line is drawn through B parallel to AX and another line is drawn through C parallel to DX. These two lines intersect at P. Prove that the area of the triangle APD is equal to the area of the quadrilateral ABCD.”
See the Triangle Quadrangle Puzzle
This problem comes from the Scottish Mathematical Council (SMC) Senior Mathematical Challenge of 2008:
“S2. In Tiffany’s, a world famous jewellery store, there is a string necklace of 33 pearls. The middle one is the largest and most valuable. The pearls are arranged so that starting from one end, each pearl is worth $100 more than the preceding one, up to [and including] the middle one; and starting from the other end, each pearl is worth $150 more than the preceding one, up to [and including] the middle one. If the total value of the necklace is $65,000 what is the value of the largest pearl?”
I included the words in brackets to erase any ambiguity.
See the Pearl Necklace Problem for solutions.
Again we have a puzzle from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).
“On one occasion, Holmes and I were asked to, solve the robbery of a number of dresses from the workshop of a recently deceased ladies’ tailor to the upper echelons of society. Holmes took a short look at the particulars of the case, and sent them all back to the gown-maker’s son with a scribbled note to the effect that it could only be one particular seamstress, with the help of her husband.
However, glancing through my observations some period later, I observed certain facts about the robbery which led me to an interesting little exercise. The stock at the workshop had been very recently valued at the princely sum of £1,800, and when examined after the theft, comprised of precisely 100 completed dresses in a range of styles, but of equal valuation. However, there was no remaining record of how many dresses had been there beforehand. The son did recall his father stating, of the valuation, that if he’d had thirty. dresses more, then a valuation of £1,800 would have meant £3 less per dress.
Are you able to calculate how many dresses were stolen?”
See the Fashion Puzzle for solutions.
Ian Stewart has a nice logic problem in his Casebook of Mathematical Mysteries, which includes a pastiche of Sherlock Holmes in the form of Herlock Soames and Dr. Watsup, along with brother Spycraft and nemesis Dr. Mogiarty.
“An important document was accidentally mislaid, and then stolen,” Spycraft said. “It is essential to the security of the British Empire that it be recovered without delay. If it gets into the hands of our enemies, careers will be ruined and parts of the Empire may fall. Fortunately, a local constable caught a glimpse of the thief, enough to narrow it down to precisely one of four men.”
“Petty thieves?”
“No, all four are gentlemen of high repute. Admiral Arbuthnot, Bishop Burlington, Captain Charlesworth, and Doctor Dashingham.”
Soames sat bolt upright. “Mogiarty has a hand in this, then.”
Not following his reasoning, I asked him to explain.
“All four are spies, Watsup. Working for Mogiarty.”
“Then … Spycraft must be engaged in counter-espionage!” I cried.
“Yes.” He glanced at his brother. “But you did not hear that from me.”
“Have these traitors been questioned?” I asked.
Spycraft handed me a dossier, and I read it aloud for Soames’s benefit. “Under interrogation Arbuthnot said ‘Burlington did it.’ Burlington said ‘Arbuthnot is lying.’ Charlesworth said ‘It was not I.’ Dashingham said ‘Arbuthnot did it.’ That is all.”
“Not quite all. We know from another source that exactly one of them was telling the truth.”
“You have an informer in Mogiarty’s inner circle, Spycraft?”
“We had an informer, Hemlock. He was garrotted with his own necktie before he could tell us the actual name. Very sad—it was an Old Etonian tie, totally ruined. However, all is not lost. If we can deduce who was the thief, we can obtain a search warrant and recover the document. All four men are being watched; they will have no opportunity to pass the document to Mogiarty. But our hands are tied; we must stick to the letter of the law. Moreover, if we raid the wrong premises, Mogiarty’s lawyers will publicise the mistake and cause irreparable damage.”
Which man was the thief?
See the Purloined Papers Puzzle for solutions.
This seemingly magical result from Futility Closet defies proof at first. Go to the Wolfram demo by Jay Warendorff and then …
“Grab point B above and drag it to a new location. Surprisingly, M, the midpoint of RS, doesn’t move.
This works for any triangle — draw squares on two of its sides, note their common vertex, and draw a line that connects the vertices of the respective squares that lie opposite that point. Now changing the location of the common vertex does not change the location of the midpoint of the line.
It was discovered by Dutch mathematician Oene Bottema.”
As we shall see, Bottema’s Theorem has shown up in other guises as well.
Here is surprising problem from the 1875 The Analyst
“81. By G. W. Hill, Nyack Turnpike, N. Y. — Prove that, identically,
”
By “identically” the proposer means for all n = 1, 2, 3, ….
See the Surprising Identity
(Update 8/20/2021) James Propp at his website has an informative, extensive article on mathematical induction and its variations.
This is another simple problem from H. E. Dudeney.
“103. PAINTING THE LAMP-POSTS.
Tim Murphy and Pat Donovan were engaged by the local authorities to paint the lamp-posts in a certain street. Tim, who was an early riser, arrived first on the job, and had painted three on the south side when Pat turned up and pointed out that Tim’s contract was for the north side. So Tim started afresh on the north side and Pat continued on the south. When Pat had finished his side he went across the street and painted six posts for Tim, and then the job was finished. As there was an equal number of lamp-posts on each side of the street, the simple question is: Which man painted the more lamp-posts, and just how many more?”
See Painting Lampposts for a solution.
This is a nice puzzle from Alex Bellos’s Monday Puzzle column in the Guardian.
“My cultural highlight of recent weeks has been the brilliant BBC documentary Gods of Snooker, about the time in the 1980s when the sport was a national obsession. Today’s puzzle describes a shot to malfunction the Romford Robot … and put the Whirlwind … in a spin.
Baize theorem
A square snooker table has three corner pockets, as [shown]. A ball is placed at the remaining corner (bottom left). Show that there is no way you can hit the ball so that it returns to its starting position.
The arrows represent one possible shot and how it would rebound around the table.
The table is a mathematical one, which means friction, damping, spin and napping do not exist. In other words, when the ball is hit, it moves in a straight line. The ball changes direction when it bounces off a cushion, with the outgoing angle equal to the incoming angle. The ball and the pockets are infinitely small (i.e. are points), and the ball does not lose momentum, so that its path can include any number of cushion bounces.
Thanks to Dr Pierre Chardaire, associate professor of computing science at the University of East Anglia, who devised today’s puzzle.”
See the Snooker Puzzle
