On this page we will be adding some quiz questions that we post in our monthly client updates - every time we post a new teaser we will update the previous one with a solution, so look out for these each month.
Here is an ordinary cross. Drawing only 2 straight lines what is the greatest number of pieces you can divide the cross into?
A. 3 B.4 C. 5 D. 6 E. 8
What is the next number in the sequence?
1, 11, 21, 1211, 111221, …
A. 1122221 B. 1212221 C. 312211 D. 1331212 E. 22
Each number is a description of the digits which make up the previous number.
So with a starting point of 1:
1 has one 1, so we get 11
11 has two 1s, so we get 21
21 has one 2 followed by one 1, so we get 1211
1211 has one 1 followed by one 2 followed by two 1s, so we get 111221
111221 has three 1s followed by two 2s followed by one 1, so we get 312211
The first six volumes of Encyclopaedia of Mathematics are arranged in order on my shelf from left to right. The six volumes contain mathematicians with surnames beginning; A-Ba, Be-Ca, Ce-Ei, Ek-Fe, Fee-Fi and, Fo-Fum respectively. If one ignores the covers, which of the following encyclopaedia entries could be on the page ‘next to’ the page with the entry for Einstein as they sit on my shelf?
A. Abel B. Bernoulli C. Cantor D. Euler E. Fibonacci
Ignoring mathematicians who have been omitted form this problem and skating over the fact that calling Einstein a mathematician has offended some readers' sensibilities, Einstein could be last entry in volume 3 (surnames Ce-Ei). As the book sits on the shelf this would make his page the left most page.
This would make him 'next to' the first page of volume 2 (surnames Be-CA) as this is the right-most as it sits on the shelf. Of the options given only Bernoulli could be the first entry in this volume meaning he could be 'next to' Einstein.
Create the shortest possible formula that can be copied from cell A1 downwards to replicate the game "FizzBuzz".
The game works like this:
- The players count up from 1.
- Each multiple of 3 is replaced with the word "Fizz".
- Each multiple of 5 is replaced with the word “Buzz”.
- If a number is both a multiple of 3 and a multiple of 5, then "FizzBuzz" should be used.
- No VBA allowed.
- Your formula has to be written into the cell - no hiding it in a named range or similar.
- That means no row numbers in a separate column, no inputs in fixed cells, no range names, no variables, no formulae in conditional formatting or text formatting, etc.
- Any Excel version allowed, but no add-in or other non-standard functions.
- Array formulae are fine, but the curly braces add 2 to your character count.
There are two equally valid solutions for this teaser depending on the version of Excel available to you:
Excel 2019 Solution - 58 characters
Pre 2019 Excel Solution - 61 characters
A tennis player has a 60% chance of winning any given point in her service game. Assuming that all points are independent of each other, what is the probability that she will win the service game?
(The scoring system in tennis is such that a player must score a minimum of 4 points and be at least 2 clear points ahead of their opponent in order to win a game.)
P(win) = P(win without reaching deuce) + P(reach deuce and then win)
= P(win without reaching deuce) + P(reach deuce)*P(win from deuce)
P(win without reaching deuce) = P(win at least 4 out of first 6 points)
= P(win 4 out of 6) + P(win 5 out of 6) + P(win 6 out of 6)
= (6*5/2!)*(0.4^2)*(0.6^4) + 6*0.4*(0.6^5) + 0.6^6
= 0.31104 + 0.186624 + 0.046656
P(reach deuce) = P(win exactly 3 out of first 6 points)
To calculate P(win from deuce), define the function F as follows:
F(0) = win from deuce
F(1) = win from advantage
F(-1) = win from opponent’s advantage
Then F(0) = 0.6*F(1) + 0.4*F(-1)
= 0.6*(0.6 + 0.4*F(0)) + 0.4*(0.6*F(0) + 0)
= 0.36 + 0.48*F(0)
Therefore F(0) = 0.36/(1 – 0.48) = 9/13
Hence, P(win) = 0.54432 + 0.27648*9/13
One day a jailer brought three prisoners out and had them stand in a line, one behind the other. He then showed them five hats (three red and two white) which he hid in a bag. From these five hats, he selected three and put one on each of the prisoners’ heads. None of the men could see what colour hat he himself wore, but could see the colour of any hats in front of him.
Starting from the back of the line, the jailer asks the first man what colour his hat is, offering him his freedom if he is correct but doubling his sentence if he gets it wrong; he declines to answer. The jailer then asks the second man the colour of his hat, making the same offer; he also declines to answer.
What colour is the third man’s hat?
The answer is red.
Assuming all prisoners are logical and only answer if they are certain.
Starting at the back the last prisoner can see the other two hats. If they are both white then he will know he is wearing red. As he does not answer at least one of the front two prisoners must be wearing a red hat.
The second prisoner now knows that at least one of himself or the prisoner in front is wearing a red hat. If he can see a white hat on the front prisoner then he will deduce his is red and answer.
Given the second prisoner does not answer the third prisoner knows his hat is not white and so must be red.