[credit provider=”YouTube/In Betweeners” url=”http://www.youtube.com/watch?v=Lo_QKHMCtj8″]
If you’re not good at maths, stop reading right now.Braingle, a brain teaser website, shared its most challenging riddles with us, as voted on by users.
Wake up your brain and give these a try. If you get stumped, don’t worry. The answers are on the slides following each question.
We bet you don’t get more than 3 right.
title=”1. Six Villages”
content=”There are six villages along the coast of the only perfectly round island in the known universe. The villages are evenly distributed along the coastline so that the distance between any two neighbouring coastal villages is always the same. There is an absolutely straight path through the jungle connecting every pair of villages. These paths create thirteen crossings in the interior of the island, one of which is in the middle of the island where paths from every village meet.
The island has a strange courtship custom. Before a father will give permission for his daughter to marry, her suitor must bring the father a fish each day until he has traveled by every route from his village to the father’s village. The young man only travels along routes where he is always getting closer to his destination. The young man may visit other villages along the way.
On April first a father’s three sons come to tell him of their intent to woe a bride, each from a different village. The brides’ villages are the first three villages encountered when travelling clockwise around the island.
If the sons begin their courtship today and the couples are married on the day following each son’s last trip, what are the three wedding dates?
Bonus Question: If the coastline of the island is 10 miles long, how long is the longest route that any of the sons takes to reach their betrothed’s village?“
content=”The wedding dates are April 6th, May 12th, and August 1st.
Bonus question: ~ 6.86 miles.
All villages are crossings. Here are the steps to determine the number of routes to reach any point:
1) First set the starting point to one.
2) Select the unvisited crossing that is adjacent to a visited crossing and is closer to the destination, but which is the farthest such unvisited crossing from the destination.
3) Set this crossing to the sum of the values for all of the adjacent visited crossings. (If you do this in the correct order, all the adjacent visited crossings are farther from the destination.)
4) If the last selected crossing is not the destination then continue with step two.
Which points are closer or farther from the destination can be determined using the right triangles formed by the intersecting paths. The shortest path from a point to a line is perpendicular to the line. Moving toward the right angle formed by the path and the line is moving closer to the point; moving away from this right angle is moving farther from the point.
Note that when moving from a village to the opposite village that all of the crossings to the left of the direct path between the villages are symmetrical with the corresponding crossing to the right.
This is a little hard to follow without a diagram, so you might want to create one. The diagram is a regular hexagon with lines connecting each pair of vertices. There are three intersections created along each of these connecting lines, with the three lines connecting opposite vertices meeting in the middle. Label one vertex A and moving clockwise, label the remaining vertices B-F. Label the middle intersection M. Label the intersection along the path from A to M (path AM) as Ai; label the similar intersections for the remaining vertices as Bi to Fi. Label the intersection of paths AC and BF as ABi. Label the similar intersections BCi, CDi, DEi, EFi and FAi.
Routes to an adjacent village (A to B):
To determine the routes from A to B first set A to 1.
The farthest point from B that is closer than A and adjacent to A is Ai. Ai can only be reached from A, so set Ai to 1.
The next farthest point is ABi. ABi can be reached from A and Ai so set ABi to 1+1=2.
The next farthest point is Bi. Bi can only be reached from ABi, so set Bi to 2.
The next farthest point is the destination B. B can be reached from A, ABi and Bi, so set B to 1+2+2=5.
There are five routes to an adjacent village, so that son will make his last trip on April 5th and the wedding will be the next day on April 6th.
Routes to the second village along the coast (A to C):
A = 1
AFi = A = 1
Fi = AFi = 1
Ai = A + AFi = 2
ABi = Ai + A = 3
M and B are equidistant from C, so in either order:
M = Ai + Fi = 3
B = A + ABi = 4
Bi and Di are equidistant from C, so in either order:
Bi = B + ABi + M = 10
Di = M = 3
BCi and CDi are equidistant from C, so in either order:
BCi = B + Bi = 14
CDi = Di = 3
Ci = BCi + CDi + M = 20
C = B + BCi + CDi + Ci = 41
There are 41 routes from A to C, so that son will make his last trip on May 11th and the wedding will be the next day on May 12th.
Routes to the opposite village (A to D):
Since the intersections to the left and right of the direct path are mirrors, the left-side calculation is assigned to both.
A = 1
F, B = A = 1
FAi, ABi = A + B = 2
Ai = A + ABi + FAi = 5
Fi, Bi = ABi + B = 3
EFi, BCi = B + Bi = 4
M, C and E are equidistant from D, so in any order:
M = Ai + Fi + Bi = 11
E, C = B + BCi = 5
Ei, Ci = BCi + M + C = 20
DEi, CDi = C + Ci = 25
Di = M + CDi + DEi = 61
D = C + E + CDi + DEi + Di = 121
There are 121 routes from A to D, so that son will make his last trip on July 31st and the wedding will be the next day on August 1st.
The longest route is a route to the opposite village. Given the above diagram, the longest routes the son will have to travel from A to D is:
and the mirror route
(ABi-C is the same as ABi-Bi-BCi-C)
If the circumference is 10 miles then the diameter is 10/Pi miles and the radius is 5/Pi miles. Since a regular hexagon is formed from six equilateral triangles, the length of a side of the hexagon is equal to its radius.
The path A-B is then 5/Pi miles. The path B-D bisects the path C-M, so both C-Ci and Di-D are half the radius. A-B + C-Ci + Di-D is then equal to the diameter 10/Pi.
The paths B-ABi, ABi-BCi, and BCi-C are all sides of equilateral triangles with a height of one half of the radius. The paths Ci-CDi and CDi-Di are each half of the side of one of these same triangles. The side of one of these equilateral triangles is
So the total distance traveled is
10/Pi + 4*(5/(sqrt(3)*Pi)) = 10/Pi + 20/(sqrt(3)*Pi) ~ 6.86 miles”
title=”2. Half Again As Big”
content=”What is the smallest integer such that if you rotate the number to the left you get a number that is exactly one and a half times the original number?
(To rotate the number left, take the first digit off the front and append it to the end of the number. 2591 rotated to the left is 5912.)”
content=”1,176,470,588,235,294 x 1.5 = 1,764,705,882,352,941
Here’s how you find the number:
You needed to find a repeating fraction such that the number of repeating digits is one less than the number. This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e., every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).
Now you need to find any occurrences where the nth digit is followed by the 1.5nth digit. ‘n’ will obviously be even.
The repeating digits in 1/7 and their corresponding ‘n’ where the digits are the first digit after the decimal point are:
142857 – repeating digits
132645 – n
The ordered pairs of adjacent values of n are (1,3), (3,2), (2,6), (6,4), (4,5) and (5,1). None of these has the second number equal to one and a half times the first number.
The next prime number p with a repeating fraction for 1/p containing p-1 repeating digits is 17. The repeating digits for 1/17 are:
The easiest way to assign ‘n’ for each digit is to list all of the digit pairs in order:
XX – n (digit)
05 – 1 (1)
11 – 2 (11)
17 – 3 (12)
23 – 4 (5)
29 – 5 (8)
35 – 6 (6)
41 – 7 (10)
47 – 8 (15)
52 – 9 (7)
58 – 10 (2)
64 – 11 (14)
70 – 12 (16)
76 – 13 (13)
82 – 14 (4)
88 – 15 (3)
94 – 16 (9)
Checking the even-numbered values of n in the above table reveals that ALL even values of n less than 2/3 of 17 are followed by the 1.5n digit. The values of n are and their digit places are:
So the five smallest numbers with this property are:
Notice that 2352941176470588 can be rotated TWICE, yielding 1.5 times the number each time since the 4->6->9 n digits are in order.
title=”3. The Palindrome”
content=”A palindrome is a word that reads the same when spelled backwards (eg rotavator).
How could the following word be considered a palindrome?
content=”Footstool is a palindrome when written in Morse code.
It looks like:
..-. — — – … – — — .-..”
content=”If you use a certain formula on 13, you end up with 7.
Under the same formula, 2352 becomes 16, 246 becomes 14, 700 turns into 16, and 1030 becomes 14.
What would 9304 become?”
Convert the number to binary, add one for every zero, and add two for every 1.
9304 becomes 10010001011000, which has 9 zeros and 5 ones.”
title=”5. Executive Travel Woes”
content=”Ms. Langston and four other executives are experiencing the downturn in the economy. While travelling on Wednesday of last week, they had to fly on commercial aircraft instead of enjoying the comfort of private jets. Each executive is an expert in a particular area of business. Use the following clues to determine the first and last name of each executive along with her position, company and her special expertise. No two executives share the same name, position, company or forte.
First Names: Ann, Arlice, Meg, Patricia, Rene
Last Names: Barnes, Langston, Mulcahy, Russo, Whitman
Positions: Chief Executive Officer (CEO), Chief Financial Officer (CFO), Chairperson, Commissioner, President
Companies: ieTrade, Lenamar Tech, Licent Co., Sabancci Ltd., The Surat Group
Fortes: Invention, Leadership, Political Connections, Profit Margin, Stock Options
1. Neither the CEO (who is not the leadership guru) nor the one with the excellent political connections (who is not a CFO or with Licent Co.) is surnamed Russo (who is neither Patricia nor Rene.)
2. Both Ann and Ms. Langston boarded in San Francisco and deplaned in Chicago; the leadership specialist and the chairperson flew from Chicago to Washington, D.C.; the flight of the Sabancci Ltd. executive (who is not surnamed Mulcahy) was from D.C. to New York.
3. The Lenamar Tech. executive (who has no special political connections) and the inventor (who is neither CEO nor CFO) are either the commissioner and chairperson in some order, or are surnamed Russo and Barnes, in some order; only one of these alternatives is true, the other is false.
4. Arlice is neither the ieTrade CFO nor Ms. Barnes; the profit margin specialist is not amongst the three executives.
5. Neither Meg (who is neither the stock options specialist nor the inventor) nor Ms. Mulcahy (who is not Arlice, who is not the chairperson) is affiliated with either Licent Co. or The Surat Group (who is not Rene).
6. Of the CEO (who is not Ms. Russo) and the profit margin specialist (who is not Rene, who is not the chairperson), one is Ms. Whitman and the other is associated with Lenamar Tech (who is not Ann).
7. The stock option specialist is either Ms. Barnes or the executive that works with Sabancci Ltd. (who is neither Patricia, who is not the commissioner, nor Rene.)
8. The president (who is not with Sabancci Ltd.) is not the inventor (who is neither surnamed Whitman nor Barnes.) Ms. Barnes is not the leadership guru.
9. The Surat Group executive is neither surnamed Langston nor Whitman.”
content=”First, Last, Position, Company, Forte:
Rene, Langston, Commissioner, Licent Co., Invention
Arlice, Whitman, CEO, Sabancci Ltd., Stock Options
Meg, Russo, CFO, ieTrade, Leadership
Ann, Barnes, President, Surat Group, Political Connections
Patricia, Mulcahy, Chairperson, Lenamar Tech., Profit Margin”
title=”6. Word Grid”
content=”Arrange 25 five-letter words in a 5-by-5 grid. The five words in each row share a common letter. The five words in each column share a common letter. The five words in each diagonal share a common letter. There are 12 different common letters.
The words are: align, blind, chase, choir, climb, close, covet, death, depot, dowel, drake, drive, flame, giant, infer, knave, mauve, murky, prank, remit, renal, score, token, vault, waist.
The grid cells are identified by row (1 to 5, top to bottom) and column (1 to 5, left to right). Example: Cell 3.2 identifies Row 3 Column 2.
One letter of each word is provided in the grid.
Row 1: u h w d p
Row 2: f m b i y
Row 3: i o v e r
Row 4: c s h o t
Row 5: a n l g e”
content=”vault chase waist death prank
flame mauve climb remit murky
blind dowel drive depot drake
close score choir covet token
renal infer align giant knave
Rows 1 to 5: a, m, d, o, n
Columns 1 to 5: l, e, i, t, k
Diagonals 1 and 2: v, r
1) Cell 2.5 contains ‘y’. There is only one word that has a ‘y’. Cell 2.5 must be ‘murky’. Cell 1.5 contains ‘p’ and must share a letter with ‘murky’. Cell 1.5 must be ‘prank’. Cell 1.3 contains ‘w’ and must share a letter with ‘prank’. Cell 1.3 must be ‘waist’. The words in Row 1 must share ‘a’.
2) Cell 2.1 contains ‘f’. There are only two words that have a ‘f’. If it is ‘infer’, then the word in Row 2 would have to share ‘r’. However, Cell 2.3 contains ‘b’. There is no word that has both ‘r’ and ‘b’. Cell 2.1 must be ‘flame’. The words in Row 2 must share ‘m’. Cell 2.3 must be ‘climb’. Cell 2.4 contains ‘i’ and must be ‘remit’. Cell 2.2 must be ‘mauve’.
3) The words in Column 3 must share ‘i’. Cell 3.3 contains ‘v’, so it is ‘drive’. Cell 3.4 contains ‘h’, so it is ‘choir’.
4) Cell 1.1 contains ‘u’, so it is ‘vault’. The words in Diagonal 1 share ‘v’. Cell 4.4 contains ‘o’, so it is ‘covet’. Cell 5.5 must be ‘knave’.
5) The words in Diagonal 2 must share ‘r’. Cell 4.2 contains ‘s’, so it is ‘score’.
6) The words in Column 5 must share ‘k’. Cell 3.5 contains ‘r’, so it is ‘drake’. Cell 4.5 contains ‘t’, so it is ‘token’.
7) The words in Column 1 must share either ‘a’ or ‘l’. Since ‘a’ is already used, then the words in Column 1 must share ‘l’. The words in Column 2 must share ‘e’. The words in Column 4 must share ‘t’.
8) The words in Row 3 must share ‘d’, ‘e’, or ‘r’. Since ‘e’ and ‘r’ are already used, then the words in Row 3 must share ‘d’. The words in Row 4 must share ‘o’. The words in Row 5 must share ‘n’.
9) Cell 3.1 is ‘blind’. Cell 4.1 is ‘close’. Cell 5.1 is ‘renal’.
10) Cell 5.2 is ‘infer’. Cell 5.3 is ‘align’. Cell 5.4 is ‘giant’.
11) Cell 1.4 is ‘death’. Cell 3.4 is ‘depot’.
12) Cell 1.2 is ‘chase’. Cell 3.2 is ‘dowel’.”
title=”7. Home For Christmas”
content=”It was good to have the house full of people again; it had been far too quiet since the last of his seven daughters had left for college–seven daughters with seven different majors at seven different schools. About the only thing they had in common was that they were each dating an athlete, although none of them played the same sport–one was even on his school’s fencing team.
He was surprised all the boyfriends had come for Christmas dinner, but he supposed that since he was picking up the tab for the airfare, that a free trip to Hawaii in the winter had something to do with it. Each of his first three daughters had gone to Ivy League schools–one even to his alma mater at Harvard; his next two ended up at the University of California, at Los Angeles and Berkeley; and his two youngest got accepted to schools abroad–the University of Tokyo had been a bit of a surprise. Next year, after seven straight years–and for the first time this millennium–he wouldn’t be attending a high-school graduation.
‘Honey, we’re ready to eat. Can you get Hank, Ian, James and Kyle? They’re in the backyard playing football. The other boys wanted to test themselves against the star quarterback,’ his wife called from the dining room.
The maths major said, ‘I’ll get them Dad!’
As everyone filed into the dining room, the oldest sister was rehashing the whose-school-and-major-is-better argument with her younger sister about whether Yale or Princeton was a better school and whether history or philosophy was a more useful major. The argument was practically a Christmas tradition by now.
The table was arranged with the parents at the two ends and each of the couples seated across from each other with boys seated next to girls along each side. As usual, Angelina and Danica were sitting next to their parents. Curiously, all the athletes that used a ball in their sport were sitting on the same side of the table. Christie, Electra and Edward took their seats first.
Once everyone was seated the youngest coyly observed, ‘I see the biologist is sitting between the two fighters but I’m sitting between the two best passers at the table.’
‘You’d better not make any passes yourself!’ Farrah icily replied. She hadn’t forgiven her sister for stealing her boyfriend when they were in high school and was not happy to see her seated next to her boyfriend now.
‘Lighten up, Farrah’ the biology major joked. ‘He wasn’t even that cute.’
‘He was adorable!’ the youngest interjected.
‘Everyone’s entitled to their opinion,’ said Brooke. ‘Let’s eat!’
The daughter in the middle seat stood up and said, ‘We have an announcement to make!’ After a dramatic pause she screamed ‘we’re getting married!’ and flashed the new diamond ring on her finger. The youngest sister squealed with delight at the news. Her fiance joked to his future mother-in-law, ‘I think she said yes just so she can wear the monogrammed sweater you gave me for Christmas.’ The baseball player next to her asked her fiance, ‘Wow, what’d that rock set you back?’
‘This calls for a toast!’ roared Dad.
‘Oh Honey, that’s wonderful news!’ mum gushed. ‘I just happen to have some champagne right here. Lars, I know you don’t drink alcohol, but would you mind pouring some champagne for Gabrielle next to you?’ the mother asked, handing the bottle to Lars on her left. ‘Gabrielle, please pass the bottle around,’ she asked and then quickly added, ‘But no alcohol for my three babies, or anyone under 21 for that matter. Honey, there’s a store downtown that’s having a sale on wedding dresses. Will you have time to go look at some wedding dresses before you leave?’
‘Sure mum,’ her daughter answered. ‘We have a late flight out, so I can do that on the day we leave.’
‘Michael and I are celebrating New Years Eve at the Sheraton, Waikiki,’ one of the girls announced.
‘What a coincidence,’ said the boy sitting next to her. ‘We’re celebrating New Years at the Sheraton, Los Angeles. It’s within walking distance of our apartment so we won’t have to worry if we drink too much.’
‘That sounds like a better plan than we have. Somehow, just because my car holds a lot of people, I got roped into being the designated driver for the team’ Danica said, looking pointedly at her boyfriend, who gestured to the boy from Princeton saying, ‘he gave me the idea!’
‘Girls, your grandmother will be here on Sunday,’ the father announced. ‘She’d really like to see you all.’
‘Sorry Dad,’ the business major replied, ‘we have to leave tomorrow. There’s a huge storm predicted to hit the Eastern United States this weekend and we need to get home before they close the airports. I noticed that each couple is leaving on a different day for six days after we leave.’
‘Sir, we’re leaving before Sunday also, so we won’t be able to make it either. I’m just glad we don’t have to deal with that storm’ said the wrestler.
The swimmer and water polo player announced their travel plans. One of them remarked that even though his girlfriend and he were departing for Oxford University two days earlier, both couples would get home on the same day.
The physics major said ‘I’ll be here to see grandma, but unfortunately we’re heading back to California before you go look at the wedding gowns.’
‘Well I’m glad you’ll at least be here to see your grandma, but I wish you could stay longer’ the dad said. ‘You know you’re my favourite since you were my first to major in a physical science like me,’ he joked. ‘I have a tee time at the club on Monday, it’s too bad you won’t be here,’ the father remarked to the golfer. ‘Kyle’s got a game next week but you don’t,’ he said to the history major’s boyfriend. ‘How about you join me on the links?’
‘Sure!’ the boyfriend replied enthusiastically.
‘Since we’re leaving the day after him, if you’ve got room for two more Dad, we just happened to bring our clubs’ the second oldest said, winking at her boyfriend.
‘It’s a foursome then!’ her dad said happily.
Ian grumbled, ‘Hank you lucky dog, I’ve wanted to play that course for years!’
The boyfriend of the political science major laughed, ‘I’d like to play there also, but Ian, at least I won’t have to travel as far as you to get another opportunity.’
Identify the boyfriend and his sport, the school, and major for each daughter. Name the order that the daughters were born in and the day of the week that they leave to go home. What is the seating arrangement at the dinner table?
Bonus question: what year is it?”
content=”From oldest to youngest the sisters are:
Who, Boyfriend, Sport, Major, School, Departure day
Danica, Hank, Water Polo, History, Yale, Tue
Electra, Edward, Fencing, Philosophy, Princeton, Wed
Angelina, Lars, Golf, Business, Harvard, Fri
Gabrielle, James, Wrestling, Poly Sci., UCLA, Sat
Farrah, Kyle, Football, Physics, Berkeley, Mon.
Brooke, Michael, Baseball, Biology, Tokyo, Thu
Christie, Ian, Swimming, maths, Oxford, Sun
The seating arrangement is:
———- Dad ———-
———- mum ———-
The year is 2003.
This is the only year from 2000 to 2007 where Christmas is on a Thursday.”
title=”8. Pick a Pizza”
content=”There is a new pizza place in town called Pick-A-Pizza. They have five wheels that, for just a few dollars, you can spin to decide the pizza chef, crust, cheese, meat topping, and veggie topping. When the wheels stop spinning, you have your pizza order! There have only been five customers so far, and each time, a different chef, a different crust (crisp, Greek, Sicilian, thin, or whole wheat), a different cheese (Gorgonzola, mozzarella, Parmesan, provolone, or Romano), a different meat topping (bacon, chicken, meatball, pepperoni, or sausage), and a different veggie topping (broccoli, mushrooms, olives, onions, or peppers) have been chosen. From the information provided, can you determine the crust, cheese, meat and veggie toppings on the pizza made by each chef?
1. The pizza topped with onions (which wasn’t made by Nunzio) didn’t have a Greek crust. Alfredo’s pizza didn’t have a whole wheat crust and wasn’t topped with pepperoni. No pizza had both olives and mozzarella.
2. Mario didn’t make the bacon pizza (whose crust was neither crisp nor thin). The sausage-topped pizza (which didn’t have a whole wheat crust) didn’t include peppers. Luigi’s pizza (which didn’t have a crisp crust) wasn’t topped with broccoli.
3. Alfredo’s pizza (which didn’t include chicken) didn’t have a Sicilian crust. Neither the pizza with a whole wheat crust (which wasn’t topped with onions) nor Nunzio’s pizza was topped with provolone cheese. Neither the pie topped with broccoli nor the pepperoni pizza was the one that had a crisp crust.
4. Neither Giuseppe’s pizza nor the sausage pizza was the one topped with onions. The Sicilian pizza (which wasn’t Luigi’s) wasn’t topped with peppers. Alfredo’s pizza (which didn’t include olives) wasn’t topped with provolone cheese.
5. Mario didn’t make the pepperoni pizza. The sausage pizza (which wasn’t made by Nunzio) wasn’t topped with mushrooms. The broccoli pizza was neither the one with a thin crust nor the one topped with mozzarella.
6. The meatball pizza (which wasn’t topped with onions) didn’t have a crisp crust. The olive pizza wasn’t topped with Romano. Luigi’s pizza (which didn’t have a thin crust) wasn’t topped with provolone.
7. The pizza with a Greek crust wasn’t topped with broccoli. The bacon pizza had neither Parmesan nor Romano cheeses. There were no peppers on the thin crust pizza.”
content=”Alfredo, Thin, Romano, Meatball, Mushrooms
Giuseppe, Whole Wheat, Mozzarella, Pepperoni, Peppers
Luigi, Greek, Parmesan, Sausage, Olives
Mario, Crisp, Provolone, Chicken, Onions
Nunzio, Sicilian, Gorgonzola, Bacon, Broccoli
The hints are not necessary to solve the puzzle. Here is an explanation of the first hint:
Luigi’s crust is not crisp (clue 2), Sicilian (clue 4), or thin (clue 6), so it must be Greek or whole wheat. However, onions are not on the pizza with a Greek crust (clue 1) or a whole wheat crust (clue 3), so Luigi cannot have onions on his pizza.
That leaves only Mario or Alfredo who can have onions. Assuming Alfredo has onions leads to a contradiction: Alfredo did not use chicken (clue 3) or pepperoni (clue 1), and onions are not with meatballs (clue 6) or sausage (clue 4), so if Alfredo has onions, he must also have bacon. Alfredo does not have a Sicilian crust (clue 3) or a whole wheat crust (clue 1), and onions are not on the pizza with a Greek crust (clue 1), so he must have a crisp crust or a thin crust. However, bacon is not on the pizza with a crisp crust or a thin crust (clue 2), so Alfredo cannot have bacon. This is a contradiction, so the assumption that Alfredo has onions is false. Thus Mario has onions.”
title=”9. Olympic Swim Team”
content=”Five swimmers (Adam, Brad, Carl, Doug, and Eric) have been preparing for the Olympics. It is now time for the swimming time trials. The five swimmers each compete in the four different strokes (backstroke, breaststroke, butterfly, and freestyle). The top three finishers in each event will qualify for the Olympic swim team in that stroke. Using the following clues, determine the order of finish in each of the four strokes.
1) Only one contestant qualified in all four strokes.
2) No contestant finished last in more than one event.
3) Adam finished better in the backstroke than he did in the butterfly.
4) Brad finished better than Doug in the butterfly.
5) Adam finished just behind Brad and just ahead of Eric in the breaststroke.
6) Doug finished just ahead of Carl in the freestyle.
7) Neither Brad nor Eric finished third in any event.
8) Eric’s finish in the backstroke was the same as Doug’s in the butterfly.
9) Doug only finished in the same position in the backstroke and the freestyle.
10) Carl finished in a different position in each event.
11) Brad finished only two events in the same position.
12) The contestant who finished second in the butterfly beat Doug in the freestyle.
13) The contestant who finished first in the freestyle did not qualify in the backstroke.
14) The contestant who finished fifth in the backstroke did not finish third in the butterfly.
15) No contestant finished in the same position in both the breaststroke and the butterfly.”
content=”Backstroke: Adam, Carl, Doug, Brad, Eric
Breaststroke: Doug, Brad, Adam, Eric, Carl
Butterfly: Eric, Adam, Carl, Brad, Doug
Freestyle: Eric, Adam, Doug, Carl, Brad“
title=”10. Rex and Ralph: Mystery Number”
content=”Two mathematicians, Rex and Ralph, have an ongoing competition to stump each other. Ralph was impressed by the ingenuity of Rex’s last attempt using clues involving prime numbers, but he thinks he’s got an even better one for Rex. He tells Rex he’s thinking of a 6-digit number.
‘All of the digits are different. The digital sum matches the number formed by the last two digits in the number. The sum of the first two digits is the same as the sum of the last two digits.’
‘Take the sum of the number, the number rotated one to the left, the number rotated one to the right, the number with the first three and last three digits swapped, the number with the digit pairs rotated to the left, and the number with the digit pairs rotated to the right. The first and last digits of this sum match the last two digits of the number, in some order.’
Ralph then asks, ‘If each of the three numbers formed by the digit pairs in the number is prime, then what is the number?’
Rex looks confused, and for a moment Ralph thinks he’s finally gotten him. Then Rex smiles, scribbles a few things down on a pad of paper and then says, ‘Very nice, Ralph!’
Rex then tells Ralph his number.
What did Rex say?”
Here’s how Rex determined Ralph’s number:
The insight Rex needed to solve this involves the number produced by the sum of the six numbers created from configurations of the digits in the number. Assign ABCDEF to the digits in Ralph’s number. The six numbers Ralph had Rex add together were:
Notice that the six digits in each column in this summation are the six digits in the number. This means that the sum of each column will be the digital sum of the number. If that digital sum is represented by A+B+C+D+E+F = XY, then the problem is equivalent to adding:
The first and last digits will be X and Y only if 10 > X + Y. If X + Y is greater than 9, then the first digit will be X+1. Since both the digital sum of the number (A+B+C+D+E+F) and the first and last digit of the sum of the numbers match the digits XY, you know that 10 > X + Y. When Rex realised this relation he smiled because he knew that he had enough information now.
The digital sum of the number must be between 0+1+2+3+4+5 = 15 and 9+8+7+6+5+4 = 39. Since each of the digit pairs in the number form a prime number, the digital sum must be a prime number in this range. There are only six prime numbers in this range: 17, 19, 23, 29, 31, and 37. 19, 29 and 37 are eliminated since 1+9 > 9, 2+9 > 9 and 3+7 > 9.
The sum of the first two digits must match the sum of the last two digits (X+Y). For 31, 3+1=4, but the only other way to make four without repeating any digits is 0+4. Zero and four can’t form a prime number, so the digital sum can’t be 31. This leaves 17 and 23.
For the digital sum to be 17, the digits must be either 0+1+2+3+4+7 = 17 or 0+1+2+3+5+6 = 17. Only 0, 1, 2, 3, 4, 7 has both a 1 and 7 to make the last two digits be 17. However, there is no pair of remaining digits with a sum of 1+7 = 8, so the digital sum can’t be 17 and therefore must be 23.
If the last two digits are 23, then the first two digits must total 2+3 = 5. The possibilities are 0+5 = 5 and 1+4 = 5. There is one prime number possible with each of these pairs: 05 and 41. The first two digits of the number can’t be 05 because then the number would be a 5-digit number and Ralph’s number has six digits. So the first two digits are 41 and the last two digits are 23.
Since the sum of all the digits is 23, then the sum of the middle two digits must be 23 – (2+3) – (4+1) = 13. There are two pairs of the remaining digits that total 13: 5+8 = 13, and 6+7 = 13. 60-seven is the only prime number that can be formed from these pairs, so the middle two digits are 67 and Ralph’s number is 416723.”
title=”11. Triathlon Competition”
content=”A triathlon consists of three events: swimming, cycling, and running. The contestants are: Adam, Bert, Carl, Doug, and Eric. From the following clues, you must determine: the order of finish in swimming, the order of finish in cycling, the order of finish in running, and the overall rankings.
Points are awarded for each event: 1st place earns 4 points, 2nd earns 3, 3rd earns 2, 4th earns 1, and 5th earns 0. Each contestant’s three event scores are added together to obtain that person’s total score. The total score determines the order of the overall rankings.
1) Adam did not finish 5th in any individual event.
2) Bert finished 2nd in only one individual event.
3) Carl finished one position better in swimming than in cycling.
4) Doug finished 4th in only one individual event.
5) Eric did not finish 3rd in any individual event.
6) No contestant who won an individual event had the best overall ranking.
7) The contestant who finished 3rd in cycling finished better in swimming than the contestant who finished 3rd in running.
8) The contestant who finished 4th in running finished better in cycling than the contestant who finished 4th in swimming.
9) The contestant who finished 5th in swimming finished better in running than the contestant who finished 5th in cycling.
10) Only one contestant finished each event in a different position, and that contestant finished 4th overall.”
content=”Swimming: Eric, Carl, Bert, Adam, Doug
Cycling: Doug, Bert, Carl, Adam, Eric
Running: Adam, Carl, Bert, Doug, Eric
Overall: Carl, Bert, Adam, Doug, Eric
The scoring system provides 10 points per event, or a total of 30 overall points. Since the overall winner did not win any individual event (clue 6) and Bert finished 2nd in one event (clue 2), that means that the best total score could only be 8 points (two 2nd place and one 3rd place). The only way to achieve 30 points with five unique scores, 8 being best, is 8 + 7 + 6 + 5 + 4 = 30.
Only the contestant who finished 4th overall had three different finishes, (clue 10), so the only way to achieve the proper overall point totals is as follows:
1st place (8 points) – 2nd, 2nd, 3rd (no 1st, 4th, or 5th individual finishes)
2nd place (7 points) – 2nd, 3rd, 3rd (no 1st, 4th, or 5th individual finishes)
3rd place (6 points) – 1st, 4th, 4th (no 2nd, 3rd, or 5th individual finishes)
4th place (5 points) – 1st, 4th, 5th (no 2nd or 3rd individual finishes)
5th place (4 points) – 1st, 5th, 5th (no 2nd, 3rd, or 4th individual finishes)
Adam did not finish 4th or 5th overall (clue 1). Bert finished 2nd overall (clue 2). Doug finished 4th overall (clue 4). Eric did not finish 1st or 2nd overall (clue 5). The contestant who finished 3rd in cycling is not the same as the one who finished 3rd in running (clue 7), so the contestant who finished 2nd overall had to finish 3rd in swimming. Carl finished one position better in swimming than in cycling, and the only combination that works is 2nd in swimming and 3rd in cycling.”
title=”12. Mystery Number”
content=”There is a 10-digit mystery number (no leading 0), represented by ABCDEFGHIJ, where each numeral, 0 through 9, is used once. Given the following clues, what is the number?
1) A + B + C + D + E is a multiple of 6.
2) F + G + H + I + J is a multiple of 5.
3) A + C + E + G + I is a multiple of 9.
4) B + D + F + H + J is a multiple of 2.
5) AB is a multiple of 3.
6) CD is a multiple of 4.
7) EF is a multiple of 7.
8) GH is a multiple of 8.
9) IJ is a multiple of 10.
10) FE, HC, and JA are all prime numbers.”
A = 5, B = 7, C = 3, D = 6, E = 9, F = 1, G = 2, H = 4, I = 8, J = 0
1) 5 + 7 + 3 + 6 + 9 = 30, which is a multiple of 6.
2) 1 + 2 + 4 + 8 + 0 = 15, which is a multiple of 5.
3) 5 + 3 + 9 + 2 + 8 = 27, which is a multiple of 9.
4) 7 + 6 + 1 + 4 + 0 = 18, which is a multiple of 2.
5) 57 is a multiple of 3.
6) 36 is a multiple of 4.
7) 91 is a multiple of 7.
8) 24 is a multiple of 8.
9) 80 is a multiple of 10.
10) 19, 43, and 05 are prime numbers.
A) The sum of digits 0 through 9 is 45. To satisfy Conditions 1 and 2, a pair of multiples must equal 45. The only pair that satisfies those conditions are 30 (multiple of 6) and 15 (multiple of 5). Likewise, to satisfy Conditions 3 and 4, the pair of multiples must be 27 (multiple of 9) and 18 (multiple of 2).
B) Make lists of the multiples of 3, 4, 7, 8, and 10 (for Conditions 5 through 9). Delete multiples whose digits are the same (33, 44, 66, 77, 88, 99).
C) To satisfy Condition 9, J must be 0. Delete multiples that require 0 in a different location. For JA to be prime (Condition 10), A must be: 2, 3, 5, or 7. Delete multiples of 3 that have another digit for A.
D) For FE and HC to be prime (Condition 10), C and E can not be even or 5. Delete multiples of 4 and 7 that have another digit for C and E.
E) Per Condition 4, B + D + F + H + J = 18. D, H, and J must be even (required to form multiples of even numbers). To attain the sum of 18 (an even number), B and F must either be both odd or both even. If both are even, then the sum of those five digits would be 20. B and F must be odd. Delete multiples of 3 and 7 that have an even digit for B and F. The only remaining multiple of 7 is 91. E is 9 and F is 1. Delete multiples that require 1 or 9 in a different location.
F) The only remaining options for AB contain 7. Delete multiples that require 7 in a different location. The only remaining options for CD contain 3. Delete multiples that require 3 in a different location.
G) To satisfy Condition 1, CD must be 36 and AB must be either 57 or 75. C is 3 and D is 6. Delete multiples that require 5, 6, or 7 in a different location.
H) The only remaining options for GH contain 4. Delete multiples that require 4 in a different location.
I) Per Condition 3, A + C + E + G + I = 27. Since C =3 and I = 9, the other three must total 15. There is only one combination of remaining multiples (57, 24, and 80) that can satisfy this condition. A is 5, G is 2, and I is 8. B is 7 and H is 4.”
title=”13. Prisoners and 2 Switches”
content=”The warden meets with 23 new prisoners when they arrive. He tells them, ‘You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
‘In the prison there is a switch room which contains two light switches labelled A and B, each of which can be in either the ‘on’ or the ‘off’ position. BOTH SWITCHES ARE IN THEIR OFF POSITIONS NOW.* The switches are not connected to anything.
‘After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can’t move both but he can’t move none either. Then he’ll be led back to his cell.’
‘No one else will enter the switch room until I lead the next prisoner there, and he’ll be instructed to do the same thing. I’m going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back.’
‘But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, ‘We have all visited the switch room.’
‘If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators.'”
content=”1. Appoint a ‘scorekeeper’ and the other 22 prisoners as ‘transmitters’.
2. When a transmitter enter the switch room for the first time, he will flick switch B ON if he sees it in the OFF position. Alternatively, he will flick switch A if he sees switch B is in the ON position and waits for his next visit. The transmitter’s mission is to find ONE opportunity to flick switch B to the ON position. After he completes his mission, he will flick switch A for all subsequent visits. In short, each transmitter’s role is to send a signal to the scorekeeper that he has been to the switch room.
3. The scorekeeper’s role is to flick off switch B whenever he sees it in the ON position. Each time he flicks off switch B he adds 1 to his score. When he sees switch B is already OFF, he flicks switch A instead and does not add to the score. As soon as he accumulates 22, he reports to the Warden.”
title=”14. Glue Factory Stakes”
content=”Five horses have just competed in the annual ‘Glue Factory Stakes’. With the given clues, determine for each horse: its colour, its number, its jockey, and where it finished.
Finish: Win, Place, Show, 4th, 5th
Horses: Ghoulish Pleasure, Infirm, Roughage, Seattle Slow, Secret Harriet
colours: Black, Red, Yellow, Blue, White
Numbers: #18, #29, #37, #46, #50
Jockeys: Allen, Jerry, Luke, Nathan, Tim
1) The 5 horses are: Infirm, the horse ridden by Tim, the horse that wore black, #50, and the horse that Placed.
2) The 5 jockeys are: Allen, the one that rode Seattle Slow, the one that rode the horse that wore red, the one that rode #18, and the one that rode the horse that Showed.
3) Secret Harriet finished ahead of the horse that wore blue, which finished ahead of the horse ridden by Luke.
4) #50 finished ahead of the horse that wore white, but finished behind Roughage.
5) Jerry finished next to #37.
6) Allen finished next to #29.
7) The horse that wore blue finished two positions away from #50.
8) The horse that wore yellow finished two positions away from #46.
9) Luke finished at least two positions away from #46.
10) Nathan finished at least two positions away from #50.
11) Luke rode a horse that wore an even number.
12) Only one of the horses that wore a primary colour also wore an even number.”
content=”Win: Secret Harriet, Black, #46, Allen
Place: Seattle Slow, Blue, #29, Nathan
Show: Roughage, Yellow, #37, Tim
4th: Ghoulish Pleasure, Red, #50, Jerry
5th: Infirm, White, #18, Luke
Much of the grid can be filled using the normal process of elimination. The following explanation resolves one of the more difficult sections.
If #46 finished 5th, then the horse that wore yellow would have had to Show (clue 8), and Luke would have had to have ridden the horse that Showed (clues 3 and 9). This would mean that Luke rode the horse that wore yellow, which Showed. The horse that Showed could not wear #18 (clue 2) or #50 (clues 3, 4, and 7); Luke could not wear #27 or #39 (clue 11) or #46 (clue 9); therefore, Luke could not have ridden the horse that Showed. Thus, #46 could not have finished 5th, which means that #18 finished 5th.”
title=”15. Rex and Ralph: Polygonal House”
content=”Two mathematicians, Rex and Ralph, have an ongoing competition to stump each other. Ralph has just finished building a custom house and invites Rex to dinner.
He tells Rex, ‘The lot my house sits on is a regular polygon. My house is a matching regular polygon sitting on a circular foundation in the middle of the yard. The closest part of the foundation to the edge of my property is exactly the same as the diameter of the foundation. The house is two stories, built around a central circular atrium with a diameter that is exactly one tenth the longest measurement of my property.’
‘Sounds like quite the house!’ remarked Rex.
‘Yes, I’ve been building it for almost two years. When I was excavating the foundation I found a section of an old irrigation pipe. The pipe exactly bisected the area of the house, with each end of the pipe at an edge of the house. After the house was completed while I was preparing to put up a fence around the edge of the property, I discovered that the length of the pipe I found was exactly the length of one side of the property. I used the pipe as a gate across the side of the property with the driveway and built the fence around the other sides. I built a total of 400 feet of fence.’
‘So what’s the square footage of my house?’ Ralph asked.
‘Are you counting the space occupied by the walls in the square footage?’ asked Rex.
‘Yes of course,’ Ralph replied. ‘It also includes the stairways, hallways, and basically all of the space inside of the exterior walls.’
Rex smiles and says, ‘Nice try, Ralph! Assuming the information you gave me is correct, it’ll take just a minute or two to calculate it.’
How did Rex know the square footage, and what was his answer?”
content=”2,730 sq. feet
Rex realised that he would need to know the number of sides on Ralph’s house in order to make use of the only measurement Ralph gave him, which was the perimeter minus the length of one side for the property. The length of a side of the property needed to be the length of the pipe that bisected the area of the house. The length of the pipe is the key to figuring out how many sides the house has.
The house is a regular polygon inscribed in the circle that is the foundation. Since the closest edge of the property is equal to the diameter of the foundation, then the property must be a similar polygon circumscribed around a circle with a diameter three times the size of the foundation.
The minimum and maximum length line segments that can bisect the area of a regular polygon are calculated differently for even- and odd-sided polygons.
For an even-sided polygon inscribed in a circle of radius r, the maximum line segment has its ends at two opposite vertices and the length is 2r. The minimum line segment goes from the middle of one edge to the middle of the opposite edge. The distance from the edge to the centre of a polygon is the apothem, so the minimum line segment is two times the apothem a.
For an odd-sided polygon the maximum bisector goes from a vertex to the middle of the opposite side, so the maximum is r + a. As with an even-sided polygon, the minimum bisector for an odd-sided polygon extends through the centre of the polygon and at an angle halfway between two adjacent maximum bisectors.
2Pi radians = 360 degrees. The central angle between any two adjacent vertices on a polygon with n sides is then 2Pi/n and the angle between a vertex radius and the apothem is half that or Pi/n. The apothem and the radius form two sides of a right triangle, with the third side formed by one half of the side bisected by the apothem. For a regular polygon with n sides of length s this gives the following relationships:
a = r*Cos(Pi/n)
r = a/Cos(Pi/n)
s = 2*a*Tan(Pi/n)
Rex assigns one as the radius of the foundation. This makes the apothem of the property three and the side of the property is:
2*3*Tan(Pi/n) = 6*Tan(Pi/n)
So where Min(n) and Max(n) are the minimum and maximum length of a bisector for an n-sided polygon inscribed in a unit circle, Rex needs to find an integer value for n that satisfies the inequality:
Max(n) >= 6*Tan(Pi/n) >= Min(n)
Since Rex is pretty sure there’s only one solution, he figures first he’ll just use the minimum and maximum functions for even-sided polygons. The minimum and maximum for odd-sided polygons will be within this range, so he might find an odd-sided polygon that doesn’t satisfy the equation, but he won’t miss any that do. This gives:
2 >= 6*Tan(Pi/n) >= 2*Cos(Pi/n)
Taking just the first part of the inequality gives:
2 >= 6*Tan(Pi/n)
1/3 >= Tan(Pi/n)
atan(1/3) >= Pi/n
atan(1/3) / Pi >= 1/n
n >= Pi/atan(1/3)
n >= 9.8
He tries 10 and finds that:
Max(10) = 2 >
6*Tan(Pi/10) = 1.95 >
Min(10) = Cos(Pi/10) = 1.90
Now Rex knows that Ralph’s house has 10 sides and is a regular decagon. That makes the length of the pipe and of a side of the larger decagon 400/9 = 44 4/9 = 44.444.. feet. From the equation above:
44.444 = 2*a*Tan(Pi/10)
a = 44.444 / Tan(Pi/10) ~ 68.393
So the radius of the larger circle is 68.393 feet and the radius of the foundation is 68.393/3 = 22.798, which is also the radius of the house decagon. The apothem and length of a side for the house are:
a = 22.798 * Cos(Pi/10) = 21.682
s = 2 * 21.682 * Tan(Pi/10) = 14.090
The side and apothem are the base and height of one of the 10 triangles formed by two vertex radii and a side. The area of one of these triangles is then
14.090 * 21.682 / 2 = 152.75
and the area of the decagon is 10 * 152.75 = 1527.5. Now Rex just needs to know the size of the atrium to calculate the square footage of the house. The radius of the atrium is one-tenth the radius of the large decagon, which is
r = 68.393 / Cos(Pi/10) = 71.913
so the area of the atrium is
(71.913 / 10)^2 * Pi = 162.47
and the area of the two floors in Ralph’s house is
(1527.5 – 162.47) * 2 = 2730.0 square feet.”
title=”For more ways to test your smarts, check out:”
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