Find a set of four different positive whole numbers whose sum is a factorial number and whose product is also a factorial number.
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?
I have a sequence. It begins:
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.
I’ve taken the name of a particular sub-national region of a large country and split it apart. It is possible to use each of the following as a filling in the ‘sandwich’ to create three words:
What is the name of the region?
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?
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?
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.
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?
I have a number.
I multiply each of the digits together to get another number.
I multiply each of the digits of that number together, and I find that I get the number 36.
If you disallow the use of the digit 1 (for obvious reasons) there are only two numbers I could have started with, what are they?
Which is greater:
the 789th number in the 123 times table to begin with the digits 456,
or the 123rd number in the 789 times table to begin with the digits 456?
An expedition is being planned to climb a truncated conical volcano that is 40km diameter at the base, and 24km diameter at the crater edge. The slope length is 12km.
The path starts at A at the base and ends at point B on the crater edge, exactly halfway round the volcano from A.
What is the length of the shortest possible route around the conical face of the volcano (to the nearest km)?
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?
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?
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?
Can you add two more bars to this digital display to make it a multiple of 13 that is NOT 78?
Place each of the numbers 1 to 20 in a row, in such an order that the sum of any pair of adjacent numbers is a number that appears in the Fibonacci Sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, etc.
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?
There are seven circles of various sizes, each touching its neighbours.
Connecting the centres of the circles creates a heptagon. The edges of the heptagon are given, except for one.
The only other information we have is that the two bold circles are the same radius.
What is the missing length?
Consider a deck of n numbered cards. You play the following game: you turn over the first card and retain it. Then you turn the next one: if it ranks less than the first one, discard it; if it ranks greater then keep it and make it the new benchmark. Repeat this step for each subsequent card. At the end of the deck, count the number of cards you have retained. For a given ‘n’, what is the expected number of cards? If there were 11 cards? If there were 12,367 cards?
We’ll start with a few small values of n: trivially when n=0, the number of cards retained, x, will also be 0. For n=1, x will also be 1. For n=2, the two cards will either be in descending or ascending order, so x could be either 1 or 2, with equal probability, so X (big x, the average value of x) will be 1.5.
Since no cards will be retained after the maximum card has been seen, it is useful to consider where in the deck the maximum card might be and what the consequences are:
Consider the case with three cards. If the 3 card comes up first, which it will do in 1/3 of cases, x will be 1. If the 3 card comes up second, x will be 2. If the 3 card comes up last, x may be 2 or 3, depending on the order of the 1 and 2 cards.
In other words, when the maximum 3 card is first, there are 0 cards before it so the average number of cards retained before reaching the maximum card is 0, so the count will be 1 more than that, 1.
If the maximum card is second, there is 1 card before it so the average number of cards retained before reaching the maximum card is 1, so the count will be 1 more than that, 2.
If the maximum card is third, there are 2 cards before it so the average number of cards retained before reaching the maximum card is 1.5, so the count will be 1 more than that, 2.5.
Those three cases are equally likely, so we can express X3 in terms of previous X values:
X3 = ((X0) + 1 + (X1) +1 + (X2) +1)/3
We can generalise this so that any Xn can be expressed as an equation involving previous X values by extending the idea of considering where the maximum card is and how many cards are retained prior to reaching it:
Xn = ((X0) + (X1) + … + (Xn-1) + n)/n
Now a very interesting thing happens if you compare the equation for Xn with that of Xn-1
Xn-1 = ((X0) + (X1) + … + (Xn-2) + n-1)/(n-1)
In the first equation you multiply both sides by n, and in the second equation by (n-1) to get the following two equations:
n(Xn) = (X0) + (X1) + … + (Xn-1) + n
(n-1)(Xn-1) = (X0) + (X1) + … + (Xn-2) + n-1
Now if you take the difference between those two equations almost all the terms cancel out:
n(Xn) - (n-1)(Xn-1) = (Xn-1) + 1
Which then simplifies to:
(Xn) – (Xn-1) = 1/n
This is great as it means that since X0 = 0, the sequence of Xn is merely the partial sum of the harmonic series:
1 + 1/2 + 1/3 + 1/4 +1/5 …
… a series about which we know a great deal.
We know for instance that the sequence does not converge but tends to infinity. This rings true when we consider the numbered cards, as we would expect Xn to keep creeping up as the number of cards increased and it would seem very odd for it to converge to an upper bound.
What is surprising I feel, when considering it in terms of the cards, is just how slowly Xn does increase.
It seems quite reasonable that Xn first exceeds 3 when n is 11, that is when there are 11 numbered cards, the average number of cards retained is around 3.
However, when we calculate when the expected number of retained cards first exceeds 10, the number of cards would be 12367 (!).
I recently stumbled upon an intriguing theorem that stated that every positive whole number can be expressed as the sum of three palindromic numbers, although it is often far from straightforward to find them.
Can you express the number 6878592 as the sum of three palindromes?
It is the sum of three numbers of 7, 6 and 5 digits respectively, and no digit is repeated except where being a palindrome dictates it:
ABCDCBA + EFGGFE + HJKJH = 6878592