Unfortunately, nobody got this week's challenge right! Here is the answer:
For how many positive integers N is the remainder 6 when 111 is divided by N? 111-6=105, so N must divide 105. The factors of 105 are 1, 3, 5, 7, 15, 21, 35, and 105. 1, 3 and 5 are all less than 6, so the remainder would be less than 6 when 111 is divided by any of them. The possible values of N are therefore 7, 15, 21, 35 and 105, so there are 5 positive integers N that satisfy the above.
Week 86: Monday 6th December
Unfortunately, nobody got this week's challenge right! Here is the answer:
How many minutes does it take? Let Derin work at d units per minute, let Saba work at s units per minute and let Rayan work at r units per minute. Let there be w units of work to be done to complete the task. Derin’s time to complete the task is w/d. When all three work together we have eq. 1 w = 5d + 5s + 5r. When Saba and Derin work together we have eq. 2 w = 7d + 7s. Finally, when Rayan and Derin work together we have eq. 3 w = 15d + 15r. Combining eq. 1 and eq. 3 to eliminate d and r gives 2w = 15s so w = 15s/2 . Combining this with eq. 2 gives s/2 = 7d so d = s/14 . Therefore w/d = (15s/2) × (14/s) = 15×7 = 105. Hence Derin would take 105 minutes.
Week 85: Monday 29th November
Unfortunately, nobody got this week's challenge right! Here is the answer:
How many miles is it by train from London to Edinburgh? By train, the distance in miles of the second sign from Edinburgh is 200 - 3.5. This sign is halfway between London and Edinburgh, so the distance in miles between the two cities is 2(200-3.5)=393.
Week 84: Monday 22nd November
Unfortunately, nobody got this week's challenge right! This problem was much tougher than usual and here is the answer:
What is the area of the triangle? First split the inscribed triangle into three isosceles triangles, each with two vertices on the circle and the third vertex at the centre of the circle. Let the base angles in the isosceles triangles be 𝑥°, 𝑦° and 𝑧°, where 𝑥 + 𝑦 = 60, 𝑦 + 𝑧 = 45 and 𝑥 + 𝑧 = 75. Adding the equations gives 2(𝑥 + 𝑦 + 𝑧) = 180 and therefore 𝑥 + 𝑦 + 𝑧 = 90. Subtracting the earlier equations from this, one at a time, gives 𝑧 = 30, 𝑥 = 45, and 𝑦 = 15.
The angles of the isosceles triangles at the centre of the circle are then 150°, 120°, and 90°. Using the formula ‘area = 1/2𝑎𝑏sin𝐶, with 𝑎 = 𝑏 = 2, gives the total area as 1/2 × 2 × 2 × sin 150° + 1/2 × 2 × 2 × sin 120° + 1/2 × 2 × 2. Since sin 150° = sin 30° = 1/2 and sin 120° = sin 60° = sqrt(3)/2 , the total area = (2 × 1/2) + (2 × sqrt(3)/2) + 2 = 1 + sqrt(3) + 2 = 3 + sqrt(3).
Week 83: Monday 15th November
Unfortunately, nobody got this week's challenge right! Here is the answer:
What is the smallest number of rectangles? Consider the areas of the small rectangles and the larger rectangle. Each small, 2 cm by 3 cm, rectangle has area 6 cm^2 . The larger rectangle has sides in the ratio 4 : 5 so has sides of length 4k and 5k, for some positive integer k, giving an area of 20k^2 . Values of 20k^2 are 20, 80, 180, 320, .... The smallest of these which is a multiple of 6 is 180, when k = 3. The sides of the larger rectangle are then 12 cm and 15 cm. Now we can check to see that the 180/6 = 30 small rectangles can be arranged.
Week 82: Monday 8th November
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
How many animals are there in Jacob's flock? x+12=2x x=12 x*3=36
Week 81: Monday 1st November
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
What was Bertie's approximate average speed in km per hour? 5.5 m=0.0055 km 19.6 s = 19.6/3600 hrs 0.0055*3600/19.6=1.01 km/hr
Week 80: Monday 25th October
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
How many children are in Tegwen's family? Since Tegwen has an equal number of brothers and sisters, there is one more girl than boy in the family. This means that each brother has two more sisters than brothers, so 2 sisters = 50% more. This means that the number of brothers they have is 4, and they have 6 sisters. This gives a total of 5 brothers and 6 sisters, so there are 11 children in total.
Week 79: Monday 18th October
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
How many positive cubes less than 5000 end in the digit 5? Numbers ending in 5 with cubes <5000 are 5 and 15, so there are only two numbers
Week 78: Monday 11th October
Unfortunately, nobody got this week's challenge right! Here is the answer:
What is the product of these three integers? First note that the triangular numbers less than 20 are 1, 3, 6, 10, 15. As we are looking for a sequence of consecutive integers which are respectively prime, even, and triangular, the required triangular number is odd. So it is 3 or 15. It is not 3 as 1 is not prime, but it is 15 as 13 is prime. So the required product = 13 × 14 × 15 = 2730.
Week 77: Monday 4th October
Unfortunately, nobody got this week's challenge right! Here is the answer:
How many ways are there to arrange the letters: I, N, T, E, G, R, A, L, M, A, T, H, S? Total number of letters: 13, number of unique letters: 11.
Thus the number of unique arrangements is given by: 13!/(2!*2!) = 13!/4 = 1556755200 different ways
Week 76: Monday 27th September
Unfortunately, nobody got this week's challenge right! Here is the answer:
How many different voucher codes like this are there? There are 3 different possibilities for the first character. The second character may be any digit from 0 to 9 inclusive, so it has 10 different possibilities. The third character differs from the second digit, so has 9 different possibilities. Once the second and third characters are determined, the fourth character is also determined since it is the units digit of the sum of the second and third characters. So, the number of different codes is 3 × 10 × 9 = 270
Week 75: Monday 20th September
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
What is the total weight? x+2y+3z=4.8, x+4y+5z=8.4, 2y+2z=3.6, y+z=1.8, x+2y+3z-y-z=4.8-1.8, x+y+2z=3
Week 74: Monday 13th September
Congratulations to Xavier Chanfor winning this week's challenge! His answer was as follows:
What is the value of a x b? 2.5+4.5=7 4.5-2.5=2 2.5*4.5=11.25
Week 73: Monday 6th September
Unfortunately, nobody got this week's challenge right! Here is the answer:
What is the second term? Suppose the first three terms of the sequence are a, b, c. Then c=1/2(a+b) and so a+b=2c. The mean of the first three terms is then 1/3(a+b+c) = 1/3(2c+c) = c, so the fourth term is c. Similarly, the following terms are all equal to c. Since one of these terms is 26 and a = 8 then b = 2c - a = 52 - 8 = 44
Week 72: Monday 30th August
Unfortunately, nobody got this week's challenge right! Here is the answer:
What is the sum of these two integers? Let the two positive integers be a and b. Then ab = 2(a + b) = 6(a − b). So 2a+2b=6a-6b, that is 8b=4a. Therefore a=2b. Substituting for a gives: (2b)b = 2(2b+b) . Solving the quadratic, b=0 or 3. However, b is positive so the only solution is b=3. Therefore a=2x3=6 and b+a = 3+6 = 9.
Week 71: Monday 23rd August
Unfortunately, nobody got this week's challenge right! Here is our answer:
How many sweets did Brooke have originally? Ava's initial number of sweets is a multiple of 3, and can thus be written as 3a. Ava gives 3a/3 = a sweets to Brooke, and after this, Brooke's number of sweets must also be a multiple of 3. Likewise, after Brooke passes on sweets to Chloe, Chloe's total sweets must be a multiple of three. Chloe began with 40 sweets = 3 x 13 + 1. Because of this, we can deduce that Beth must pass on 3b-1 sweets, and thus Brooke had to have 9b - 3 sweets after receiving a sweets from Ava. Her initial number of sweets is thus 9b - 3 - a. After putting these values in a table that indicates the initial value, final value, and values after everytime candies is given, it is evident that Ava ends up with 2a+b+13 sweets, while Brooke ends up with 6b-2 sweets, and Chloe 2b + 26 sweets. Thus by solving 2a+b+13=6b-2=2b+26, we determine a=10 and b=7. Plugging into our initial expression for Brooke's original candies, 9b-3-a, we see that she begins with 9(7) - 3 - 10 =
50 sweets.
Week 70: Monday 16th August
Unfortunately, nobody got this week's challenge right! Here is our answer:
How many ways are there to write 70 as the sum of square numbers? 49+16+4+1=70, 36+25+9=70, so there are 2 ways to write 70 as the sum of square numbers
Week 69: Monday 9th August
Congratulations to Sean Lifor winning this week's challenge! His answer was as follows:
What is the rule? (digits excluding first digit) / (first digit)
Week 68: Monday 2nd August
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the value of the expression? (2+1)^-1 = 1/3, (1/3 + 1)^-1 = 3/4, (3/4 + 1)^-1 = 4/7, 4/7 + 1 = 11/7
Week 67: Monday 26th July
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the greatest number of consecutive integers that have a sum of 45? Summing the numbers from -44 to 44 gives a total of 0, and adding an extra 45 will give a sum of 45. The number of terms in this sequence is 44 + 1 + 44 + 1 = 90.
Week 66: Monday 19th July
Unfortunately, nobody got this week's challenge right! Here is our answer: What is the ratio of the area of triangle A to the area of triangle B? Divide triangle A in half, giving two triangles with base 3 and hypotenuse 5. This means the height is 4 (by Pythagoras - 5^2-3^2=16=4^2). Dividing triangle B in half gives two triangles with base 4 and hypotenuse 5, making the height 3 (also using Pythagoras). Since the bases and heights are the same, the areas are as well, so the ratio is 1:1.
Week 65: Monday 12th July
Congratulations to Adam Wangfor winning this week's challenge! His answer was as follows:
What number comes next? What is the rule? The rule is the nth term = n! - 1, so the next number is 6! - 1 = 719.
Week 64: Monday 5th July
Unfortunately, nobody got this week's challenge right! Here is our answer:
How long did the full journey take? In the first two hours, I travelled 55*2=110km. Let the time for the rest of the journey be t. Then the remaining distance is 70t. Average speed is total distance over total time, so 60 = (110+70t)/(2+t). Rearranging gives 120 + 60t = 110 + 70t, so 10t = 10 and t = 1. Therefore the total time is t+2=3 hours.
Week 63: Monday 28th June
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the largest possible number in the set? If the mean of the seven integers is 7, their sum is 7*7=49. Since the integers are distinct, the smallest possible sum of the 6 smallest ones would be 1+2+3+4+5+6=21. Therefore, the largest possible number would be 49-21=28.
Week 62: Monday 21st June
Congratulations to Aaditya Vorafor winning this week's challenge! His answer was as follows:
How many different orders are there for Bob to complete his tasks in? Treat the 2 tasks that must be in order as one, and then find the possible number of orders, which is 3! = 6
Week 61: Monday 14th June
Congratulations to Gabriel Sidhan Löwyfor winning this week's challenge! His answer was as follows:
What is the rule? Square of the sum of digits
Week 60: Monday 7th June
Unfortunately we made a typo in this week's challenge. Will's statement should read: Yasmine cannot say that Xavier is lying. Given this correction, here is our answer:
Who is lying and who is telling the truth? Assume that Xavier (X) is telling the truth. This implies Y is lying, so her statement is false. Therefore, a truth-teller could say that Z is lying, so Z is also lying. That means her statement is also false, which implies that W must also be lying. For this to be the case, his statement must be false, i.e. Y can say that X is lying. Since Y is lying and X is telling the truth, this is the case, so our initial assumption holds and the only truth-teller is X.
Week 59: Monday 31st May
Unfortunately, nobody got this week's challenge right! Here is our answer:
If there are 104 days of summer vacation, what days of the week can the vacation start on to maximise the number of full weekends (both Saturday and Sunday) in the vacation? 104 mod 7 = 6, so if the vacation starts on, say, a Monday, all days will be included in the 6 extra days except Sunday. The only starting days that result in an excluded weekend day are Sunday and Monday, so the other five days (Tuesday, Wednesday, Thursday, Friday, Saturday) all work, as they also don't result in a split weekend at either end of the vacation.
Week 58: Monday 24th May
Congratulations to Sean Lifor winning this week's challenge! His answer was as follows:
How many distinct subsets does S have? Number of subsets = 10C0 + 10C1 + 10C2 + ... + 10C9 + 10C10 The sum of choose operations is the sum of a row in Pascal's triangle, which is 2^n, so the number of subsets = 2^10 = 1024
Another way to think about this problem is the following: each integer from 1 to 10 can either be in a subset or not, so there are 2 states for each of the 10 integers, so there are 2^10=1024 distinct subsets.
We will soon be posting an article proving the link between the sum of choose operations and powers of 2.
Week 57: Monday 17th May
Congratulations to Leo Webbfor winning this week's challenge! His answer was as follows:
What number comes next? What is the rule? The rule is the nth term = n^n, so the next number is 5^5=3125.
Week 56: Monday 10th May
Congratulations to Sean Lifor winning this week's challenge! His answer was as follows:
What was my overall average speed in km/h? Total distance d = (30/60)*20 + (90/6)*5 = 0.5*20 + 1.5*5 = 10+7.5 = 17.5 Average speed = (total distance)/(total time) = 17.5/(120/60) = 17.5/2 = 8.75 km/h
Week 55: Monday 3rd May
Congratulations to Jade Poonfor winning this week's challenge! Her answer was as follows:
How many paths are there from X to Y that are 3 edges long? If each edge is only used once, there are 4 paths that are 3 edges long. If each edge can be used more than once, then the edge that connects X directly to Y can be used, with one of the edges connected to X or Y being used exactly twice. This gives 5 paths, so the total number of paths is 4+5=9.
Week 54: Monday 26th April
Unfortunately, nobody got this week's challenge right! Here is our answer:
How many ways are there to choose 4 books such that there is at least one fiction book and at least one non-fiction book? There are 4C1=4 ways to choose one fiction book and another 4C1=4 to choose one non-fiction book. This leaves 6 books, from which we have to choose two so we have total of 4 books. 6C2=15, so the total number of ways is 4x4x15 = 240.
Week 53: Monday 19th April
Congratulations to Valerie Ang for winning this week's challenge! Her answer was as follows:
What is the rule? All of the inputs are two-digit numbers. The rule is to take the first digit to the power of the second digit.
Week 52: Monday 12th April
Congratulations to Sean Li for winning this week's challenge! His answer was as follows:
What day of the week were we born on? There are 365 days between April 2020 and April 2021 (no leap day). 365 mod 7 = 1, so there is a difference of one day, so Integral was born on a Thursday.
Week 51: Monday 5th April
Congratulations to Dan Moore for winning this week's challenge! Their answer was as follows:
What is the probability that I am accepted to at least one university? The probability that you are accepted to at least one university is 1 minus the probability of you not being accepted anywhere, which is (0.1+0.8)^4 = 0.656. So the probability is 1 - 0.656 = 0.34.
Week 50: Monday 29th March
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is (a,b)? Working backwards, the reflection of (2,4) in y=x would be (4,2). Rotating this clockwise (the opposite direction) by 90° about (1,3) gives (0,0), so (a,b) is (0,0), which is the origin.
Week 49: Monday 22nd March
Congratulations to Sean Li for winning this week's challenge! His answer was as follows:
What number comes next? What is the rule? The third difference between the numbers is constant at 1. The second differences are 3, 4, 5... and the first differences are 3, 6, 10, 15... so the next number is 36+(15+6)=56.
While the sequence has a constant third difference (meaning it can be represented as a third degree (cubic) polynomial) there is another way of representing the same sequence. What we had in mind was the sequence nC3 (the choose function) for n>=3, which gives the same sequence. It's really interesting to think about why a sequence based on factorials can also be represented by a polynomial, so we'll be writing an article about this soon. Stay tuned!
Week 48: Monday 15th March
Unfortunately, nobody got this week's challenge right! Here is our answer:
Approximate the area of the pi symbol. There is 1 square that is completely filled and 48 squares that are partially filled. Assuming that the more-than-half-filled squares will cancel out with the less-than-half-filled squares, we get the area as 1+(48/2) = 25 square units.
Week 47: Monday 8th March
Congratulations to Aarush Garg for winning this week's challenge! His answer was as follows:
Who is a liar and who is a truth-teller? Let's assume Alex is telling the truth. If he is telling the truth, then Dylan must be telling the truth as well, and Claire must be lying. If Claire is lying, then neither Dylan nor Bob must be telling the truth, which is contradictory, as we already know that Dylan is telling the truth. Hence Claire must be telling the truth. If so, Alex is lying and Dylan is also lying. Furthermore, since one of Dylan and Bob is telling the truth and Dylan is lying, it must be Bob telling the truth, implying that Ellie is a liar. Hence, Bob and Claire are telling the truth and Alex, Dylan and Ellie are lying
Week 46: Monday 1st March
Congratulations to Aarush Garg for winning this week's challenge! His answer was as follows:
How many ways are there to get from S to E by only moving right or up? To move to the horizontal squares and vertical squares from the row of S, there is only 1 way, as you can only go right and up, and cannot go left or down. Then you can compute there are 2 ways to reach the center square. From there, there are 3 ways to reach each of the adjacent squares to S, and hence there are 2*3 = 6 ways to reach the square E.
Week 45: Monday 22nd February
Congratulations to Xavier Chan for winning this week's challenge! His answer was as follows:
What number comes next? What is the rule? The next term is the product of the previous two terms, so the next number is 91*1183 = 107653.
Week 44: Monday 15th February
Congratulations to Justin Kim for winning this week's challenge! His answer was as follows:
What is the units digit of N? P(19) is 9 and S(19) is 10, S(19)+P(19) is 19, so the units digit is 9.
While using an example solves the problem, we had an algebraic solution in mind when setting the problem, which we've described here:
What is the units digit of N? Let N = 10a + b where a and b are positive integers less than 10. Then P(N) = ab and S(N) = a + b. From the problem statement, we can say that 10a + b = a + b + ab. Rearranging gives 9a = ab. Since a cannot equal 0 (this would mean N is not a two-digit number), b (the units digit) must equal 9.
Week 43: Monday 8th February
Congratulations to Riya Maiya for winning this week's challenge! Her answer was as follows:
What is the value of f(1)? A possible function f(n) is f(n) = log x because log(ab) = log a + log b (logarithm product rule). f(1) = log 1 = 0 because any number to the power of 0 is 1.
We had a slightly different solution in mind when we set the problem, which we've described here:
What is the value of f(1)? Let a = 1. Then f(a x b) = f(1 x b) = f(b). So if f(a x b) = f(a) + f(b), then f(b) = f(1) + f(b), so f(1) must equal 0.
Week 42: Monday 1st February
Congratulations to Aaditya Vora for winning this week's challenge! His answer was as follows:
What is the value of one of the five integers? Arithmetic sequence can be an increase of x each time, so if a = 8-2x , b = 8-x, c = 8, d = 8+x, and e = 8 + 2x, the sum of terms is 40, hence 8 is guaranteed to be a term
Week 41: Monday 25th January
Congratulations to Rocco Jiang for winning this week's challenge! His answer was as follows:
What is the rule? The number of vowels in each number spelt out in English, e.g. 14 = fourteen = 4
Week 40: Monday 18th January
Congratulations to Justin Kim for winning this week's challenge! His answer was as follows:
What proportion of the water did I drink? Me: 1/2 + 1/8 + 1/32.... You: 1/4+1/16+1/64..... Me: ratio = 1/4 --> use the sum of geometric series formula, insert in a1/(1-r) = 1/2/(1-1/4) --> 2/3
Week 39: Monday 11th January
Unfortunately, nobody got this week's challenge right! Here is our answer:
Who is lying and who is telling the truth? Assume that Alex is telling the truth. Then Claire is lying, so Bob must also be lying. However, if Alex is telling the truth, then Bob's statement that either he or Alex is telling the truth is true, so we have a contradiction. Therefore, Alex must be lying. Following this through means that Bob and Claire are both telling the truth, which is contradiction-free. So Alex is lying and Bob and Claire are telling the truth.
Week 38: Monday 4th January
Congratulations to Leo Webb for winning this week's challenge! His answer was as follows:
What was the most recent year (excluding 2021) in which January 4th was a Monday? Week shifts by two days forward on leap years (2020, 2016), and one day forward on other years: Monday in 2021, Saturday in 2020, Friday in 2019, Thursday in 2018, Wednesday in 2017, Monday in 2016.
Week 37: Monday 28th December
Congratulations to Carlos Rosales for winning this week's challenge! His answer was as follows:
What number comes next? What is the rule? The nth term of the sequence is given by the formula n^2 - 1 (because second derivative is 2 and we are given some values), so the next number is 35.
Week 36: Monday 21st December
Unfortunately, nobody got this week's challenge right! Here is our answer:
How many ordered pairs (x, y, z) are there? x, y and z are different numbers. If the smallest number was 2, then the smallest the next two would be 3 and 4. This would give a sum of 9, which is bigger than 8, so the smallest number must be 1. The options for the next two numbers are then 2,5 and 3,4 to give a sum of 8. Therefore, the number of ordered pairs is the number of ways to order (1, 2, 5) and (1, 3, 4), giving a total of 12 ordered pairs.
Week 35: Monday 14th December
Congratulations to Monique Gaviraghi for winning this week's challenge! Her answer was as follows:
What is the rule? Add the last digit amount to the total number, e.g. 31+1=32
Week 34: Monday 7th December
Congratulations to Sean Li for winning this week's challenge! His answer was as follows:
What is angle EDA? BAD is 90° since the angle at the circumference in a semicircle is always 90°. Since ABDE is a cyclic quadrilateral, opposite angles add up to 180°. Therefore, EDA = 180 - 31 - 90 - 18 = 41°.
Week 33: Monday 30th November
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the rule? We notice that all the prime numbers give an output of 1, so this is likely something to do with the factors of the input numbers. Pursuing this leads us to realise that the output is the sum of the factors of the input, excluding the number itself.
Week 32: Monday 23rd November
Congratulations to Xavier Chan for winning this week's challenge! His answer was as follows:
How many odd digits appear in the list? 1,3,5,7,9,10,11,12,13,14,15,16,17,18,19,21,23 All contain odd digits. 1,3,5,7,9 and 21,23 contain only one odd integer, so they are counted once. (7) 10,12,14,16,18 also only contain one odd digit (5) 11,13,15,17,19 contain two odd integers so they are each counted twice (5*2=10) 7+5+10=22 so there are 22 odd digits in the list
Week 31: Monday 16th November
Congratulations to Leo Gierhake for winning this week's challenge! His answer was as follows:
What is the ratio of the area of the square to the area of the circle? Let x be the side length of the square. Then the square's diagonal, and the circle's diameter, is sqrt(2) x. The radius is that divided by two, so the area of the circle is pi (sqrt(2) x / 2)^2 = pi x^2 / 2 The area of the square is x^2 So the ratio is 2/pi
Week 30: Monday 9th November
Congratulations to Charlotte Leong for winning this week's challenge! Her answer was as follows:
Who is guilty? It won’t be anyone who claims that someone else is innocent, so Abby and David are out. If Brian is guilty, that would make Chloe innocent, meaning David is guilty, which doesn't work. So through process of elimination, Chloe is guilty.
Week 29: Monday 2nd November
Congratulations to Sean Li for winning this week's challenge! His answer was as follows:
What is the ratio of the area of the shaded rectangle to the area of the octagon? Divide the octagon into 8 isosceles triangles with their tips at the centre of the octagon and their bases at the sides of the octagon. These triangles have equal area, so each triangle is 1/8th the area of the octagon. Two of these triangles are completely within the shaded rectangle, leaving behind two more triangular regions in the rectangle. These triangles are of equal area to the other two triangles, as they have the same base and height (just swapped around). So the area of the rectangle is equivalent to the area of 4 of the isosceles triangles, which we said were each 1/8th the area of the octagon. Therefore, the area of the rectangle is 4*(1/8) = 1/2 the area of the octagon.
Week 28: Monday 26th October
We made a mistake with the phrasing of this problem! In its published form, the problem is too difficult to answer in the space here. We will be writing an article about its solution soon though, so stay tuned!
Week 27: Monday 19th October
Unfortunately, nobody got this week's challenge right! Here is our answer:
How many ways can the letters in "WEEK TWENTY SEVEN" be reordered? The phrase contains 5 Es, 1 K, 2 Ns, 1 S, 2Ts, 1 V, 2 Ws, and 1 Y, making a total of 15 letters. If all of the letters were distinct, there would be 15! ways to reorder them, but this overcounts the Es, Ns, Ts and Ws. To account for them, we have to divide by (5! 2! 2! 2!), as this allows the identical letters to swap places within themselves, which has no effect on the ordering of the letters. Our answer is therefore 15! / (5! 2! 2! 2!) = 1,362,160,800.
Week 26: Monday 12th October
Congratulations to Brigit Lee for winning this week's challenge! Her answer was as follows:
What number comes next? What is the rule? Each number corresponds to a letter (according to alphabet order) and the letters spell out integral (9=i, 14=n, 20=t, 5=e, 7=g, 18=r.....), so the next number is 18.
Week 25: Monday 5th October
Congratulations to Vikram Singh for winning this week's challenge! His answer was as follows:
What is the smallest value of N such that all of N's digits are the same? N must be divisible by 3 and 5, so it must end in 5 (cannot end in 0 since all digits are the same). For it to be divisible by 3, its number of digits must be a multiple of 3. The smallest one is 3, so N = 555 (15*37).
Week 24: Monday 28th September
Congratulations to Xavier Chan for winning this week's challenge! His answer was as follows:
What is the sum of all of the different 4-digits numbers that can be made by using the digits of 2020 exactly once? 2002+2020+2200=6222, since all must start with 2
Week 23: Monday 21st September
Congratulations to Jade Poon for winning this week's challenge! Her answer was as follows:
What colour shirt is Alice wearing? If only one of them is lying, it would contradict the person who is telling the truth. Since the questions says that at least one of them is lying, it implies both of them are lying. So, the person who says their name is Beatrice is Alice, and the person who says their name is Alice is Beatrice. This means that Alice is wearing a red shirt.
Week 22: Monday 14th September
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the rule? We notice that inputs 1, 4 and 9 all produce an output of 0. One thing these numbers have in common is that they are all square numbers. We therefore try square rooting all of the inputs - we get 1.00, 1.41, 1.73, 2.00, 2.24, 2.45, 2.65, 2.83, 3.00, and 3.16, all to two decimal places. We notice that the first decimal place of all of these numbers corresponds to the listed output, so the rule is: the first decimal place of the square root of the input.
Week 21: Monday 7th September
Unfortunately, nobody got this week's challenge right! Here is our answer:
What is the probability that the toddler finishes the puzzle? If the toddler picks the middle piece (call it M) first, then she is guaranteed to finish the puzzle. The probability that this happens is 1/5. If she picks a different piece (4/5 chance of this happening), there are 3/4 outcomes that allow her to continue with the puzzle. Of these 3 outcomes, all of them guarantee that she can finish the puzzle, since all remaining pieces share at least one edge with one of the two already-chosen pieces. So the probability that the toddler finishes the puzzle is 1/5 + (4/5)(3/4) = 1/5 + 3/5 = 4/5.
Week 20: Monday 31st August
Congratulations to Muhammad Sinan C K for winning this week's challenge! His answer was as follows:
How many times from 12:00am to 12:00pm will be palindromes? 101...151 202...252 ....... 909...959 Total 6*9=54 3 more: 1001, 1111 and 1221, so the answer is 54+3=57
Week 19: Monday 24th August
Congratulations to Muhammad Sinan C K for winning this week's challenge! His answer was as follows:
How many different routes are there from X to Y? The possible routes are: Xady, Xacdy, Xbady, Xbacdy, Xbdy and Xby, so there are 6 different routes.
Week 18: Monday 17th August
Congratulations to Arya Killa for winning this week's challenge! Her answer was as follows:
What is the largest possible value of S? The sum is greatest when the three largest numbers (13, 14 and 15) are in the corners. The sides can be balanced by putting the smallest number (10) between the biggest sum (14 and 15), giving a sum of 39. Putting 11 between 13 and 15, and then 12 between 13 and 14, gives a sum of 39 for all three sides, so the largest possible value of S is 39.
Week 17: Monday 10th August
Congratulations to Frank Lee for winning this week's challenge! His answer was as follows:
What is the sum of the six numbers on the cube? 11+16=27, 12+15=27, 13+14=27, 27*3=81
Week 16: Monday 3rd August
Congratulations to Ishika Tulsian for winning this week's challenge! Her answer was as follows:
How many speeder bike owners do not own a repulsorlift speeder? 351-(331+45) = 25 so that’s the intersection of the two subsets Hence, only speeder bike = 351-25=306
Week 15: Monday 27th July
Congratulations to Daniel Ciesla for winning this week's challenge! His answer was as follows:
What letter come next? A,C,F,J,O are the 1,3,6,10,15 th letters of the alphabet, which are the triangle numbers. The next triangle number is 21, so the next letter is U (21st letter).
Week 14: Monday 20th July
Congratulations to Leo Webb for winning this week's challenge! His answer was as follows:
What is the minimum number of tests to find an infected person in a group of 8 people? Firstly, I will show that you can always find the infected in 8 people. Say the eight people are named 1, 2, 3, ... 8. First, you test 1, 2, 3, 4. If positive, 1, 2, 3, 4 have the infected among them. Otherwise, 5, 6, 7, 8 have the infected among them. The 4 who have the infected are renamed 1x, 2x, 3x, 4x. Secondly, you test 1x, 2x. Through similar logic as last time, either 1x, 2x have the infected among them, or 3x, 4x do, depending on the test result. The 2 who have the infected among them are renamed 1&, 2&. Thirdly, you test 1&. If 1& is positive, he's infected. If not, 2& is infected. Since the naming system is here is arbitrary, this system will work no matter who's the infected case among the 8. So the infected can be found in three tests.
That's not enough to claim to have answered the question, though. Secondly, I will prove that you can never find the infected among 8 with only 2 tests. For any given person, there are 4 possibilities:
Not tested at all
Tested in test 1
Tested in test 2
Tested in both tests
and you can therefore organise the people into 4 sets, which correspond to which tests are positive or negative (e.g. if test 1 is positive and test 2 is negative, the infected must be in the set of people who were tested in test 1 only) By the pigeon-hole principle, there will always be at least one of those sets with more than 1 person, and so there will always be a "worst-case scenario" in every possible testing method using 2 tests that doesn't clarify which of a (>1) number of people has the virus. Thus, you can't guarantee you'll find the positive person in 2 tests.
Week 13: Monday 13th July
This week's challenge was won by an anonymous reader! Their answer was as follows:
For what fraction of a day do the clocks show the same time? Clocks are same for x:10 to x:50 every hour, which is 40/60 mins. 2/3 of the time
Week 12: Monday 6th July
Congratulations to Vikram Singh for winning this challenge! His answer was as follows:
How many spars did Ahsoka win? Total spars = sum of spars lost = 8. 8-4-3 = 1
Week 11: Monday 29th June
Congratulations to Arya Killa for winning this challenge! Her answer was as follows:
What is N? If all the digits are even and there can't be the same digit twice, then the choices are 0,2,4,6 and 8. If you have all of them it doesn't work because 27 is divisible by 3, but the sum of the digits isn't. The smallest digit you can lose for it to be divisible by 3 is 2, since then it equals 18. So you try 8640, which is the largest arrangement of those digits and it is divisible by 27. So N equals 8640.
Week 10: Monday 22nd June
Congratulations to Leo Webb for winning this challenge! His answer was as follows:
What does 402 become, and what rule does → represent? → represents summing the squares of the digits of the input (arrived at this through intuition and guessing), so 402 becomes 4^2 + 0^2 + 2^2 = 20
Week 9: Monday 15th June
This week's challenge was won by an anonymous reader! Their answer was as follows:
What letter comes next? Each letter corresponds to a number (A=1, B=2 etc). the numbers keep on going while the letters repeat, so A would be 27 in the next round. The letters A, D I, P, Y and J all correspond to squared numbers. A=1, D=4, I=9, P=16, Y=25 and J=36. the next square number is 49 and it corresponds to the letter W.
Week 8: Monday 8th June
Congratulations to Malhar Rajpal for winning this challenge! His answer was as follows:
What is the sum of Alice's three integers? Out of all of the possibilities for 3 numbers that multiply to 36, only two triples have the same sum (6, 6, 1 and 9, 2, 2). This sum is 13, so that is the sum of Alice's integers.
Week 7: Monday 1st June
Congratulations to Nitya Nigam for winning this challenge! Her answer was as follows:
Is it possible to determine what colour will be in cell X? Yes What colour will be in cell X? The 3rd cell in the 1st row can be either red or green, but in both cases, working through the grid methodically results in the cell on the bottom right (X) being coloured red
Week 6: Monday 25th May
Congratulations to Edgar Cheng for winning this challenge! His answer was as follows:
What is the second term in the 13th row? The expression for the middle term of row n is n^2 - n + 1, and the expression for the number of terms in row n is 2n - 1. The middle term of row 13 is therefore 157, the number of terms in row 13 is 25, so if we calculate 157 - 11, we get 146.
Week 5: Monday 18th May
Congratulations to Leo Webb for winning this challenge! His answer was as follows:
How many distinct factors does N^5 have? N = abc, where a, b, c are in the set of primes. Therefore, N^5 = a^5b^5c^5 In each factor, the powers of a, b, or c can be any of 0, 1, 2, 3, 4, 5; 6 possibilities - the power to which a, b, or c is raised is independent of the other two, and so there are 6^3 = 216 possibilities for numbers in the form a^i b^j c^k where i,j,k are in the set {0,1,2,3,4,5}; we know we are not double counting because a, b, c are prime (given), so there are 216 factors.
Week 4: Monday 11th May
Congratulations to Aaditya Vora for winning this challenge! His answer was as follows:
What digits are L, I, V and E? L must be 1, as any other multiplication would result in five digits. Hence, E must be 9 since 9*9=81 is the only digit that, when multiplied by 9, ends in a 1. We now know I is a 0, since it cannot be 1 because E already is a 1, and anything larger would lead to EVIL having 5 digits due to a carryover. The V has to be an 8, since 9*V must end in a 2, to accommodate the 8 carried forward from the 9*9 in the tens place, and 8*9 = 72. 1089 * 9 = 9801, which can be checked using a calculator as well.
Week 3: Monday 4th May
Congratulations to Ken Kuwabara for winning this challenge! His answer was as follows:
What is angle x? Ratio of FG to DG is 1:2 as DG is same length as FH and FG is a half of FH. As the ratio of FG to DG is 1:2 you can make a formula of cosx = 1/2 where x is the angle of <DGF. It is obvious that cos 60 is 1/2 so angle of <DGF is 60 degrees. As this is an mirror image with the BG as the axis, angle <EGH will also be 60 degrees. Therefore you will have angle <DGE be 60 degrees as well. You can imagine to split the kite into half and have isosceles triangle DGB and BGE where angle of <DGB is 30 degrees. s the triangle is an isoscles, the angles of DBG and BDG are 75 degrees. As I have said, this is a mirror image so <DBG = <EBG. Converting the equation, you will have 2 * <DBG = <DBE 2* 75 = 150 THEREFORE x=150
Week 2: Monday 27th April
Congratulations to Aaditya Vora for winning this challenge! His answer was as follows:
What is 17 # 4? 12 Describe the operation that is represented by the # symbol. Subtract the second number from the first and write that down, then add the 2 numbers up and write that sum adjacent to the difference
Week 1: Monday 20th April
Congratulations to Malhar Rajpal for winning this challenge! His answer was as follows:
What number goes in the red circle? 12 What is the rule? Add the digits of the numbers above