Here is a tricky little logarithm problem from the 2021 Math Calendar.

“Find *x*, where

log_{2}(log_{4}(*x*)) = log_{4}(log_{2}(*x*))”

As before, recall that all the answers are integer days of the month.

See Log Jam for a solution.

This is another candle burning problem, presented by Presh Talwalkar.

“Two candles of equal heights but different thicknesses are lit. The first burns off in 8 hours and the second in 10 hours. How long after lighting, in hours, will the first candle be half the height of the second candle? The candles are lit simultaneously and each burns at a constant linear rate.”

See Two Candles for solutions.

Here is a challenging problem from the 2021 Math Calendar.

“Find the remainder from dividing the polynomial

*x*^{20} + *x*^{15} + *x*^{10} + *x*^{5} + *x* + 1

by the polynomial

*x*^{4} + *x*^{3} + *x*^{2} + *x* + 1”

Recall that all the answers are integer days of the month.

See the Remainder Problem for a solution.

This problem comes from the “Problems Drive” section of the *Eureka* magazine published in 1955 by the Archimedeans at Cambridge University, England. (“The problems drive is a competition conducted annually by the Archimedeans. Competitors work in pairs and are allowed five minutes per question ….”)

“There are ten times as many seconds remaining in the hour as there are minutes remaining in the day. There are half as many minutes remaining in the day as there will be hours remaining in the week at the end of the day. What time is it on what day?”

One of the hardest parts of the problem is just being able to translate the statements into mathematical terms. Solvable in 5 minutes?!!!

See the Hard Time Conundrum for a solution.

Here is a challenging problem from the Polish Mathematical Olympiads published in 1960.

**“22.** Prove that the polynomial

*x*^{44} + *x*^{33} + *x*^{22} +* x*^{11} + 1

is divisible by the polynomial

*x*^{4} + *x*^{3} + *x*^{2} +* 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“.

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.

Here is another, more challenging, problem from the Sherlock Holmes puzzle book by Dr. Watson (aka Tim Dedopulos).

“An event that occurred during *The Adventure of the Wandering Bishops *inspired Holmes to devise a particularly tricky little mental exercise for my ongoing improvement. There were times when I thoroughly appreciated and enjoyed his efforts, and times when I found them somewhat unwelcome. I’m afraid that this was one of the latter occasions. It had been a bad week.

‘Picture three farmers,’ Holmes told me. ‘Hooklanders. We’ll call them Ern, Ted, and Hob.’

‘If I must,’ I muttered.

‘It will help,’ Holmes replied. ‘Ern has a horse and cart, with an average speed of eight mph. Ted can walk just one mph, given his bad knee, and Hob is a little better at two mph, thanks to his back.’

‘A fine shower,’ I said. ‘Can’t I imagine them somewhat fitter?’

‘Together, these worthies want to go from Old Hook to Coreham, a journey of 40 miles. So Ern got Ted in his cart, drove him most of the way, and dropped him off to walk the rest. Then he went back to get Hob [who was still walking], and took him into Coreham, arriving exactly as Ted did. How long did the journey take?’

Can you find a solution?”

I added the statement in brackets. I initially thought Hob waited in Old Hook until Ted fetched him. But the solution indicated that was not the case. So I realized Hob had started out at the same time as the others. The solution has some hairy arithmetic. Even knowing the answer it is difficult to do the computations without a mistake.

See the Old Hook Puzzle for solutions.

The “Moving Up” post recalled an unforgettable moment in my past, when I still rode the Washington Metro somewhat sporadically (my youth was spent riding busses, before the advent of the Metro). It was the first time I confronted the escalator at the DuPont Circle stop. I was going to a math talk with a friend and we were busy discussing math when I stepped onto the escalator. Suddenly, I looked up and saw the stairs disappearing 188 feet into the heavens and froze. I have always been afraid of heights, and the escalator brought out all the customary terror. There was of course no turning back. And then people started bolting up the stairs past me, not always avoiding brushing by.

My hand was clamped to the handrail in a death grip. I had to hold on even tighter as the sweat of fear made my hands slippery. In such situations I often feel a sense of vertigo or loss of balance. It was then that I thought the handrail was moving faster than the steps so that I was being pulled forward. I couldn’t tell if it was the vertigo or an actual movement. In any case, I periodically let go and repositioned my death grip. After an eternity, it was over, and I staggered out into the street. Needless to say, on our return I sought out the elevator. Fortunately, it was working—not always the case in the Washington Metro.

Once my brain was functioning a bit, I pondered the question of the relative speeds of the handrail and steps. How could they be synchronized? But after a while I left it as an interesting curiosity.

See Escalator Terror

Here is a simple problem from an old *Futility Closet* posting.

“My wife and I walk up an ascending escalator. I climb 20 steps and reach the top in 60 seconds. My wife climbs 16 steps and reaches the top in 72 seconds. If the escalator broke tomorrow, how many steps would we have to climb?”

See Moving Up for solutions.