Puzzle of the Week #209 - Coin vs Dice

Which of the following scenarios should on average* take the most throws:

Repeatedly tossing a coin until you have seen at least 6 heads and at least 6 tails, or

Repeatedly throwing a dice until you have seen each of the numbers 1 to 6 at least once each?

 

*(simple arithmetic mean)

 

Puzzle of the Week #206 - Tumbler

glass tumbler.JPG

A glass tumbler has an outside diameter of 78mm, and inside diameter of 72mm and a solid base that is 30mm deep.

Its centre of gravity is exactly 30mm from the table, ie, at the height of the bottom of the inside of the glass.

What is the overall height of the glass tumbler?

Puzzle of the Week #205 - Formula

I have a sequence. It begins:

1

101

10101

1010101

101010101

etc

As it’s populated entirely with 1s and 0s it is not entirely clear whether these are decimal numbers, or binary numbers, or some other base entirely.

 

Construct a formula to give the nth term, which works no matter what base we are working in.

***EDIT*** You are looking for a simple formula where you plug in the value of n and get the nth term, no recursive formulae or finite sums required.

Puzzle of the Week #203 - Fractions

1/3 = 0.333… and 3/8 = 0.375. No fraction with a single-digit denominator lies between them.

What is the fraction greater than 1/3 and less than 3/8 that has the lowest denominator? Call that fraction ‘x’.

What is the fraction greater than ‘x’ and less than 3/8 that has the lowest denominator? Call that fraction ‘y’.

Finally, what is the fraction greater than ‘x’ and less than ‘y’ that has the lowest denominator?

Puzzle of the Week #202 - Spherical Asteroid

A perfectly spherical asteroid is in orbit somewhere in the solar system. There is a small deposit of Unobtainium buried within it.

Scientists are able to scan the asteroid and determine the position of the deposit of Unobtainium relative to the surface of the sphere, by imposing a perpendicular coordinate system on it:

From the Unobtainium, if you go the 24 metres in the positive x direction you reach the surface

From the Unobtainium, if you go the 60 metres in the negative x direction you reach the surface

From the Unobtainium, if you go the 30 metres in the positive y direction you reach the surface

From the Unobtainium, if you go the 36 metres in the positive z direction you reach the surface

What is the radius of the spherical asteroid?

sphere dimensions.JPG

This may help you: The general equation of a sphere is: (x - a)² + (y - b)² + (z - c)² = r², where (a, b, c) represents the centre of the sphere, r represents the radius, and x, y, and z are the coordinates of the points on the surface of the sphere.

Puzzle of the Week #201 - Two of One and One of Another

I recently investigated which numbers can be represented at the sum of one square number and double a different square number, for instance 9 is 1+2(4), or 38 is 36+2(1).

 

With pencil and paper: 99 is the smallest number that can be represented as a^2 + 2b^2 in three different ways: find them all!

 

With a computer: what is the smallest number that can be represented as a^2 + 2b^2 (with a and b different positive integers) in six different ways?

 

Puzzle of the Week #196 - Out and Back

Bakewell parkrun is a picturesque free, weekly, timed 5km run along the Monsal trail in Derbyshire. The course heads off 2.5km along the trail, then takes a u-turn, before heading back to the start/finish.

Two friends run the parkrun, both starting at the same time (Saturday 9am). They run at different paces, but their paces are consistent. Nicola, the faster of the two, reaches the turnaround point first, then exactly two minutes and twenty seconds later, passes Danny travelling the other way. At the finish, Nicola has exactly twelve minutes to wait before Danny crosses the finish line.

How long do Nicola and Danny take to complete their parkruns?

Puzzle of the Week #195 - Personal Bests

Runners will know, if they want to predict what time they should be aiming for in, say, a 10km race, based on what they can do at 5km, you should do more than merely doubling the time: the longer a race, generally the slower you need to go. One useful rule is known as the 6% rule. It isn’t unfortunately as simple as merely doubling the 5km time and adding 6% - instead it is raising the ratio of distances to a power of 1.06.

So (10/5)^1.06, would give you a factor of around 2.085. Multiply this by your typical 5km time and you should be looking at that for your 10km time.

Of course, this is only a rule of thumb, so will be more true for some people than for others. For one athlete, let’s call him Norm, he finds that his personal best times, at a variety of distances, precisely follow the rule (to the nearest second). His half-marathon best time is exactly 1 hour more than his 10km best time. (A half-marathon in metres is 21097.5).

What are his 10km and half-marathon personal bests?

Puzzle of the Week #194 - Double Birthdays

This conversation took place on a particular day of the (non-leap) year:

 

“Isn’t it curious,” remarked Izzie, “that if you take today’s date, double the date number and double the month number, you get my birthday?”

“Interesting!” replied Leila, “but if you consider today as the ‘n’th day of the year and work out what the ‘2n’th day of the year is, you get MY birthday!”

 

Izzie’s and Leila’s birthdays are in the same calendar month.

 

When did this conversation take place and when are the girls’ birthdays?

 

Puzzle of the Week #191 - One in Five Chance

In a betting game there are five cards placed face down. Each has a number on it.

You win the bet if you end up with the card with the highest number.

You can select any of the cards at random, and having seen the number you can decide whether to accept the card or reject it and choose another, and you may do this as many times as you like, however once you have rejected a card you cannot get it back.

The actual values on the card give you no indication of whether you might have the highest, for instance ‘1’ may be the highest number, or ’50,000,000’ may be the lowest.

Since there are five cards you might think a fair price for the bet would be 4 to 1 against (you place £1 against the house’s £4 and if you win you take all £5). Instead you are offered a measly 3 to 2 against (you place £2 against the house’s £3 and if you win you take all £5).

Is it possible to beat the house? If so how?