Maths Wizard

Here are your challenges for this week:

Here are the answers to last week’s mathematical conundrums:

The sides of a triangle have lengths 4, 7 and k, where k is a whole number. How many possible values are there for k?

Hint: k is between $7-4$ and $7+4$.

Solution: In a triangle, we require the sum of the lengths of the two shorter sides to be greater than the length of the third side. So the possible values of k are 4, 5, 6, 7, 8, 9 and 10; that is 7 possible values.

 

What is the smallest positive integer with three square factors?

Hint: Think about perfect square divisors of a number. Start listing small numbers and count how many square numbers divide each one.

Solution: The first few square numbers are 1, 4, 9, 16, 25, … . The smallest number with three square factors is 16, which is divisible by 1, 4 and 16.

 

Tweedledum was supposed to add 999 to a number; instead he subtracted 999, ending up with 55. What number should he have ended up with?

Hint: Think in reverse: if subtracting 999 gave 55, add 999 to 55 to get the starting number—then add 999 again.

Solution: The starting number must have been 55+999=1054, so Tweedledum’s final number should have been 1054+999=2053.

 

Tweedledee was given a two-digit number, and was supposed to count the number of different primes that divided it; instead he just counted the number of positive integers that divided it, arriving at a total count of 7. What should his count have been?

Hint: Think: which kind of number has exactly 7 divisors? (It’s a power of a single prime)

Solution: There is only one two-digit number with 7 factors: 64 (with factors 1, 2, 4, 8, 16, 32, 64). The only prime dividing 64 is 2, so his count should have been 1.

 

A chocolate bar is split between Somel and Tyrese. Somel received three and a half times as much chocolate as Tyrese. What fraction of the bar did Somel receive?

Hint: Think of the ratio: Somel : Tyrese = 3.5 : 1. Now ask: what fraction of the total is Somel’s part?

Solution: We can think of the bar as being split into 4.5 parts, with Somel receiving 3.5 parts. So the answer is 3.5/4.5=7/9.

 

What is the difference between 4% of 85 and 85% of 4?

Hint: Use this idea: $a\%$ of $b$ = $b\%$ of $a$

Solution: Zero (the two quantities are equal).

 

Find two consecutive square numbers that differ by exactly 99.

Hint: The difference between consecutive squares is always an odd number.

Solution: The difference between 49×49 and 50×50 is 49+50=99, so the consecutive squares we are looking for are 2401 and 2500.

How many other pairs of square numbers differ by exactly 99?

Hint: a2−b2=99 — so just think: how many ways can 99 be written as a product of two odd numbers?

Solution: The other pairs of squares that differ by 99 are (1=1×1 and 100=10×10) and (225=15×15 and 324=18×18); so there are two other pairs.

 

In how many ways can 25 be written as the sum of a positive even number and a positive multiple of 3?

Hint: Try thinking of small even numbers you can add to something to make 25.
Then check: can the other number be a multiple of 3.

Solution: There are 4 ways: 25=3+22=9+16=15+10=21+4.

 

In how many ways can 100 be written as the sum of a positive multiple of 4 and a positive multiple of 5?

Hint: Try just the smallest positive multiple of 4… Does what’s left over look like a multiple of 5?

Solution: There are only 2 ways: 100=20+80=40+60.

 

In a class of 30 pupils, 16 like pizza, 19 like pasta and 27 like panna cotta.

1) What is the largest possible number of pupils who dislike pizza, pasta and panna cotta?

Hint: Think about how many pupils you can make be in all three groups at the same time.

The more overlap you create, the more space you leave for pupils who like none of the foods.

Solution: As only 3 pupils dislike panna cotta, it cannot be the case that 4 or more pupils dislike all three foods. It is possible that all those who like pizza and pasta also like panna cotta, so the largest possible number that dislike all three foods is 3.

 

2) What is the smallest possible number of pupils who like pizza, pasta and panna cotta?

Hint: Add the three numbers: 16 + 19 + 27. Ask yourself: How much bigger is this than 30?

Solution: As 16 like pizza, 19 like pasta and 16+19=35, there must be at least 5 pupils who like both pizza and pasta (the class only contains 30 pupils). Similarly, as 27 like panna cotta and 5+27=32, there must be at least 2 pupils who like all three foods. In fact it is possible that exactly 2 pupils like all three foods – among the other 28 pupils, 3 like pizza and pasta only, 11 like pizza and panna cotta only and 14 like pasta and panna cotta only. So the answer is 2.

 

How many different combinations of coins add to 7 pence?

Hint: Only three coins matter: 1p, 2p, and 5p. First decide how many 5p coins you use (0 or 1).
Then see how many ways the remaining amount can be made using 1p and 2p coins.

Solution: 7=5+2=5+1+1=2+2+2+1=2+2+1+1+1=2+1+1+1+1+1=1+1+1+1+1+1+1. There are six combinations overall.

 

In a shop, the price of a jumper is increased by a third. A week later, the price falls by a third. What fraction of the original price of the jumper is the current price?

Hint: Let the original price be 1 (or £1). First increase it by one third, then decrease the new price by one third. Be careful: the second third is taken from the new price, not the original.

Solution: After the price increase, the jumper’s price is 4/3 of the original price. This is equivalent to 12/9. So after the fall in price, the new price is 8/9 of the original price.

What is the size of each interior angle in a regular 100-sided polygon?

Hint: As a polygon gets more sides, each interior angle gets closer and closer to 180°. A 100-sided polygon has a lot of sides — so each interior angle will be just a little less than 180°.

Solution: The sum of the interior angles of a 100-sided polygon is 180x(100-2)=18000-360=17640 degrees. So each interior angle measures 17640/100=4 degrees.

 

A pizza is shared between three people. One person takes 4/7 of the pizza; the remainder is split in the ratio 5:6. What fraction of the pizza does the person with the smallest portion receive?

Hint: First, subtract 4/7 from 1 whole pizza to see what’s left. Then split that leftover into 11 equal parts (because 5 + 6 = 11). The smallest share is 5 of those 11 parts.

Solution: Once 4/7 of the pizza is taken, 3/7 remains. This is split in the ratio 5:6, so the smallest portion is 5/11 of 3/7 of the whole pizza, and the fraction received is (5/11)x(3/7)=15/77.

 

How many hours and minutes are there in 3/16 of a day?

Hint: A day has 24 hours. Just find 3/16 of 24.

Solution: There are 24 hours in a day, so 3/16 of a day is 24×3/16=72/16=9/2=4.5 hours, or 4 hours 30 minutes.

What is the smallest positive whole number containing four different digits?

Hint: Think about the first number that has four digits, and ask: do any digits repeat?

Solution: 1023.

 

How many square numbers are of the form 222…2 (only containing the digit 2)?

Hint: Look at the last digit of square numbers and compare it with the last digit of numbers like 222…2.

Solution: No square numbers end in a digit 2 (if you multiply any digit by itself, the result never ends in a digit 2). So the answer is zero.

How many square numbers are of the form 666…6 (only containing the digit 6)?

Hint: Check what digits square numbers can end with, and compare that with the last digit of numbers like 666…6.

Solution: The square of an odd number is odd, and the square of an even number is always a multiple of 4; but 666…6 is neither odd nor a multiple of 4. So no number of this form is a square, and the answer is zero.

 

In how many ways can 5 different pieces of fruit be split between two children, if each child receives at least one piece of fruit?

Hint: For each fruit, you only have two choices: give it to child A or child B.
Just make sure at the end that both children get at least one fruit.

Solution: For each piece of fruit, there are two possible children who could receive it, so there are 2x2x2x2x2=32 ways to split the pieces of fruit between them; however, 2 of these 32 ways result in one of the children empty-handed. So the final answer is 32-2=30.

 

In how many ways can 2 identical apples and 3 identical oranges be split between two children, if each child receives at least one piece of fruit in total?

Hint: Ignore who gets which specific fruit. Just decide how many apples and how many oranges one child gets — but not zero of everything.

Solution: The first child can receive 0, 1 or 2 apples; that is three choices. Similarly there are four choices for the number of oranges they receive: 0, 1, 2 or 3. So there are 3×4=12 ways to distribute fruit to the first child. And whatever the first child doesn’t receive goes to the second child; so there are 12 ways to split the apples and oranges between them. However, two of these ways leave one child empty-handed. So the final answer is 12-2=10.

 

Over a period of four days, Ellie eats several strawberries. Each day after the first day, she eats twice as many strawberries as she did the day before. In total she eats 90 strawberries. How many strawberries did she eat on the last day?

Hint: If the last day is some number, then the day before is half of that, the day before that is half again, and the first day is half again. All four days together add up to 90.

Solution: Ellie ate 6, then 12, then 24, then 48 strawberries across the four days. So she ate 48 on the last day.

Over a period of four days, Taran eats lots of peas. Each day after the first day, he eats two thirds as many peas as he did the previous day. In total he eats 325 peas. How many peas did he eat on the first day?

Hint: If the last day is some number, then the day before is 3/2 times that, the day before is 3/2 times again, and the first day is 3/2 times once more. All four days together make 325.

Solution: Taran ate 135, then 90, then 60, then 40 peas across the four days. So he ate 135 on the first day.

A rectangle with perimeter 40 can be split into three identical squares. What is the area of the rectangle?

Hint: Imagine placing three equal squares side-by-side in a straight line.
Now think about how the perimeter of that shape adds up to 40.

Solution: The rectangle must be constructed from three 5×5 squares; the sides of the rectangle are 5 and 15. So the area of the rectangle is 5×15=75.

 

A rectangle with perimeter 40 can be split into three squares, not all identical. What is the area of the rectangle?

Hint: Try drawing the rectangle and splitting it into three squares of different sizes.
Notice that the small sides must add up to the same length as the big side somewhere in the picture.

Solution: The rectangle must be constructed from one 8×8 square and two 4×4 squares; the sides of the rectangle are 8 and 12. So the area of the rectangle is 8×12=96.

 

A Christmas pudding costs £1.50, but there is an offer on: ‘Buy puddings for the price of two!’. How many puddings can be bought with a £20 note?

Hint: Focus on what you actually pay for, not how many items you take. Read the offer very literally.

Solution: With the offer, £3 buys 3 packs, so £18 buys 18 packs. Then one more pack can be bought for a further £1.50. So the answer is 19.

 

The price of a gift box was reduced by a third in a sale. Then the price was reduced again, this time by a quarter. The final price was £7.50. What was the original price?

Hint: Work backwards from £7.50 by undoing each reduction one at a time.

Solution: Before the final reduction, the price was £10, so the original price was £15.

 

The time is now 11am on 14th Dec. How many hours are there until Christmas Day?

Hint: Count how many days there are until Christmas Day, then multiply by 24, and finally adjust for the 11am start time.

Solution: There are 13 hours left on 14th Dec, then 10 full 24-hour days until Christmas Day, so the answer is 13+240=253.

 

Ares eats several mince pies, starting on 1st December and finishing on Christmas Day, but some days he eats only a fraction of a mince pie. On average he eats exactly three mince pies a week. What fraction of the final mince pie he starts eating does Ares consume?

Hint: Work out how many weeks there are from 1st December to Christmas Day, then multiply by 3.

Solution: An average of 3 mince pies a week is an average of 3/7 of a mince pie each day. So in 25 days, exactly 25×3/7=75/7 mince pies are eaten. That is 10 whole mince pies and 5/7 of the last pie.

 

What is the smallest positive number divisible by 4, 5 and 6?

Hint: Think of the smallest number that appears in the multiplication tables of 4, 5, and 6.

Solution: Since 5 and 6 have no common factors (except for 1), any number divisible by both 5 and 6 must be divisible by 5×6=30. However, 4 is not a factor of 30. The next multiple of 30 works: 60 is divisible by 4, 5 and 6. So the answer is 60.

 

How many pairs of positive numbers (a,b), where a is less than b, have a lowest common multiple of 20?

Hint: Start by listing all the divisors of 20. Any pair (a,b) whose LCM is 20 must come from that list. Then check which pairs actually give LCM = 20.

Solution: The pairs are (1,20), (2,20), (4,20), (5,20), (10,20), (4,5) and (4,10). So the answer is 7.

In a quadrilateral, one angle is twice as big as each of the other three angles. What is the size of the largest angle?

Hint: Think of the quadrilateral as having three equal angles and one angle that is twice as big.
All four angles together must add up to 360°. Use that fact to find the size of one small angle.

Solution: Split 360 into 5 equal parts; then the largest angle takes up 2 parts – 144 degrees.

 

A cake is split into 7 small slices and 3 large slices. Each small slice is equal in size, each large slice is equal in size, and each large slice is twice as large as each small slice. What fraction of the whole cake is one large slice?

Hint: Count the small slices as 1 and the large slices as 2. Add them all up. Now ask: What part of that total is a “2”?

Solution: Each large slice corresponds to 2 small slices, so the split is equivalent to 13 small slices, and each large slice takes up 2/13 of the entire cake.

 

One apple and two pears cost £1.36. Two apples and one pear cost £1.31. How much do nine apples and nine pears cost?

Hint: Try setting up two equations with two unknowns. Once you know the price of one apple plus one pear, it becomes very easy to scale up to nine of each.

Solution: Three apples and three pairs cost £1.36+£1.31=£2.67, so nine apples and nine pears cost 3x£2.67=£8.01.

 

One lemon and one lime cost 71p. One lemon and three limes cost £1.47. How much do three lemons and one lime cost?

Hint: Subtract the cheaper combination from the more expensive one — this will tell you the price of two limes. Then work backward to find the rest.

Solution: Two limes cost £1.47-£0.71=£0.76, so one lime costs £0.76/2=£0.38. Then one lemon costs £0.71-£0.38=£0.33, and so three lemons and one lime cost 3x£0.33+£0.38=£1.37.

What is the smallest multiple of 10 whose digits sum to 10?

Hint: The number must end with 0 (since it’s a multiple of 10).
Now find the smallest number ending in 0 whose other digits add up to 10.

Solution: A number is divisible by 10 if and only if it ends with a 0. The answer is 190.

 

What is the smallest multiple of 60 whose digits sum to 60?

Hint: The number must end with 0 (because it’s a multiple of 60).
So the other digits must add up to 60 — try using as many 9s as possible to reach 60 quickly.

Solution: A number is divisible by 60 if and only if it is divisible by 3, 4 and 5. This requires the digits to have a sum divisible by 3, the number formed by the last two digits to be divisible by 4, and the last digit to be 0 or 5. The answer is 79999980.

 

In a hexagon, what is the greatest possible number of interior right-angles?

Hint: A hexagon’s interior angles must add up to 720 degrees. Try experimenting by adding right angles, but keep in mind that having all six right angles won’t work. How many right angles can you fit while keeping the total sum correct?

Solution: Six is impossible – the interior angle sum would then be 540 degrees instead of the required 720 degrees. The answer is five (make the hexagon look like a letter L).

 

Arthur thinks of a whole number. He squares it, then adds his original number, ending up with a prime. What number or numbers could he have been thinking of?

Hint: Start by trying small whole numbers. Square the number, then add the same number again. Check if the result is a prime number (a number that can only be divided by 1 and itself). Keep trying until you find the answer!

Solution: Whether Arthur begins with an odd number or an even number, the result will be even. But the only even prime is 2. It quickly becomes clear that large numbers (positive and negative) produce results that are also large; an analysis of numbers close to zero yields 1 and -2.

Divide 111,111,111 by 1,001,001 without a calculator.

Hint: Here’s a cool trick! 1,001,001 can actually fit into 111,111,111 exactly 111 times. So: 111,111,111=1,001,001×111

To divide 111,111,111 by 1,001,001, you just reverse the multiplication: 111,111,111÷1,001,001=111

So, when you divide 111,111,111 by 1,001,001,

Solution: the answer is 111!

Insert two pairs of brackets into the expression 1-2-3-4 to make the result as large as possible.

Hint: Think about how brackets change the order and remember that doing the maths inside the brackets first might change your answer.

Solution: The largest result is 6, so the best way to add brackets is: 1−((2−3)−4).

 

The letters A, A, N, N are written on four tiles (one letter per tile) and placed in a bag. They are then drawn out of the bag at random, one at a time, and placed in a line from left to right. What is the probability that the tiles spell ANNA?

Hint: Start by thinking about the probability that the first tile you draw is an A. Then think about the probability of drawing two N’s in a row and what then is probability that the last letter will be A?

Solution: The probability that the first letter is A is 1/2. With an A removed, the probability the next two letters are both N’s is 2/3 x 1/2 = 1/3. With ANN removed, the last letter is always an A. So the probability of spelling ANNA is 1/2 x 1/3 = 1/6.

 

What is the highest common factor of 117, 171 and 711?

Hint: Start by looking for a simple pattern, like the sum of the digits. Can each be divided by that number?

Solution: 117, 171 and 711 all have a digit sum of 9, so they are all multiples of 9. Dividing each by 9 leaves 13, 19 and 79, all of which are prime so have no common factors (other than 1). So the highest common factor is 9.

 

How many square factors does 300 have?

Hint: Start by finding the factors of 300 (the numbers we can multiply together to get 300), then find the square numbers from these.

Solution: The square factors of 300 are 1, 4, 25 and 100, so there are four factors.

 

How many square numbers less than 1000 are also cube numbers?

Hint: A square number is when you multiply a number by itself. A cube number is when you multiply a number by itself twice. For example, 2×2×2=8. Some numbers can be both a square and a cube. These numbers are special and are called sixth powers. This is because to be both, a number has to be like multiplying the same number six times. For example: 2×2×2×2×2×2=64, so 64 is both a square and a cube.

Solution: There are only three such numbers less than 1000: 1, 64 and 729.

 

If x is 2/3 of y, and z is 4/5 of x, what fraction of z is y?

Hint: Work backwards from z to y using the given relationships.

You’re given:

x=(2/3)y

z=(4/5)x

So… just ask: “What do I multiply z by to get y?”

Solution: z=(4/5)x=(4/5)(2/3)y=(8/15)y. So y=(15/8)z. That is, y is 15/8 of z.

 

What is the largest three-digit multiple of 6 that is still a multiple of 6 when its digits are reversed?

Hint: Start with 996 and go down by 6 each time.
For each number, reverse the digits and check if the reversed number is also a multiple of 6.

Solution: A number is divisible by 6 precisely when it is even and its digits add to a multiple of 3. So we require the first and last digits to be even. The largest number satisfying the given conditions is 894.

 

How many factors of 333 contain at least one digit 3?

Hint: First, list all the numbers that divide 333 exactly (no remainder).
Then, look at those numbers and see which ones have the digit 3 in them.

Solution: The factors of 333 containing a digit 3 are 3, 37 and 333 itself; that is three factors.

 

Which positive integer(s) below 100 has/have the most factors containing at least one digit 3?

Hint: Try numbers under 100. For each, look at its factors. Count how many of those factors have a 3 in them. The number with the most like that is your answer.

Solution: There are three positive integers below 100 that have the most factors containing a digit 3: 39, 78 and 93 each have exactly three factors containing a digit 3.

 

Arthur walks one mile north, then half a mile south, then half of half a mile north, then half of half of half a mile south. How far north or south is he from where he started?

Hint: Just add and subtract the distances he walks, in order:

➕ first step
➖ second step
➕ third step
➖ fourth step

Add them up — that’s how far north he ends up.

Solution: Arthur is 1-1/2+1/4-1/8=8/8-4/8+2/8-1/8=5/8 of a mile north of where he started.

Which integer is closest to the cube root of 15?

Hint: What number cubed is just less than 15? What number cubed is just more than 15?
The cube root of 15 is closest to the integer in between those two.

Solution: The cube root of 8 is 2 and the cube root of 27 is 3. So the cube root of 15 is between 2 and 3. To decide which is closer, we can cube 2.5. We have 25x25x25=625×25=15625, so 2.5×2.5×2.5=15.625. Hence the cube root of 15.625 is 2.5. This means that the cube root of 15 is between 2 and 2.5. Therefore the closest integer to the cube root of 15 is 2.

 

Two fifths of a number is 5. Find two thirds of the number.

Hint: If 2/5 of a number is 5, ask yourself: What is the whole number? Then, find 2/3 of that number. Finally, two thirds of 12.5 is 2×12.5/3=25/3 or 8 1/3.

Solution: One fifth of the number is 2.5, so the number is 5×2.5=12.5.

 

A two-pence piece is glued to a table, with Heads facing up. Another two-pence piece is placed next to it, so that the two coins are touching. Then the second coin rolls around the first (fixed) coin without slipping, until it returns to its starting position. Through what angle does the second coin turn?

Hint: Imagine the edge of the rolling coin touching and turning not just around the first coin — but also because it rolls, it spins extra. Think: It doesn’t just go around — it also rolls as it moves.

Solution: The second coin makes two complete revolutions (try it for yourself!), so it turns through 2×360=720 degrees.

 

Amber thinks of a number. She doubles it, then subtracts 7, then halves the result, ending up with -2. What number did she start with?

Hint: Try working backwards from the final number, which is -2.

Think:

What number, when you halve it, gives -2?

Then, what number, when you subtract 7 from it, gives that result?

And finally, what number, when you double it, gives that?

Take it one step at a time, in reverse.

Solution: Amber starts with 1.5. She doubles it to get 3, subtracts 7 to get -4, then halves this to end up with -2.

Samuel thinks of a number. He cubes it, then adds 30, then multiplies the result by -1, ending up with the number he first thought of. What is this number?

Hint: Start by trying 0. Follow the steps in the problem exactly.

Then try 1, then -1, then 2, then -2…

Just test each number one by one until the final result equals the starting number. You’re looking for the number where: after cubing it, adding 30, and multiplying by -1, you get the same number you started with.

Solution: Samuel starts with -3. He cubes it to get -27, adds 30 to get 3, then multiplies this by -1 to get back to -3.

 

How many of the first 100 square numbers are divisible by 4?

Hint: If a number is even, its square is divisible by 4. Now ask yourself:
How many even numbers are there from 1 to 100?

Solution: If you square an even number (a multiple of 2), the result is a multiple of 2×2=4. If you square an odd number, the result is still odd, so not divisible by 4. Hence, exactly half of the first 100 square numbers are divisible by 4, that is 50 in total.

A fair six-sided die, with faces numbered 1, 2, 3, 4, 5, 6, is rolled twice. What is the probability that the product of the two numbers rolled is greater than the sum?

Hint: Try rolling small numbers like 1 and 2, or 2 and 3. Ask yourself each time:
Is multiply bigger than add?

Solution: There are 6×6=36 possible pairs of numbers that could be rolled. Of these, most have a product greater than their sum. The exceptions are when one or both numbers are equal to 1 (this occurs in 11 pairs), and when both numbers are 2 (this occurs in 1 pair). So there are 36-11-1=24 pairs for which the product is greater than the sum, and the required probability is 24/36=2/3.

 

What is the product of all the even numbers between -49 and 49?

Hint: One of the numbers is 0. What happens when you multiply anything by 0?

Solution: One of the even numbers between -49 and 49 is 0, so the product of all the numbers is 0.

 

Alfie draws two regular polygons P and Q. Each interior angle of P is 100 degrees more than each interior angle of Q. How many sides do P and Q have in total?

Hint: Try plugging in small numbers of sides (like 3, 4, 5…) and find the interior angles using this formula:

Interior angle = (n−2)×180 / n

Then look for two of them where the angles differ by 100°.

Solution: Suppose Q had 4 or more sides. Then each interior angle of Q would be at least 90 degrees, so each interior angle of P would be at least 190 degrees, which is impossible. Therefore Q must have 3 sides (with interior angles each 60 degrees), and P must have interior angles each 160 degrees. Then each exterior angle of P is 180-160=20 degrees, and P has 360/20=18 sides. Hence, in total, P and Q have 3+18=21 sides.

 

How many positive numbers are factors of both 45 and 60?

Hint: Find the GCD (greatest common divisor) of 45 and 60, then count how many positive divisors that number has.

Solution: The common factors of 45 and 60 are precisely the factors of 15, namely 1, 3, 5 and 15. So the answer is 4.

 

How many positive numbers are factors of 100 but not 50?

Hint: First, list all the positive factors of 100 and positive factors of 50. Then, find which ones are in 100 but not in 50. You’re looking for the difference between the two sets!

Solution: We require those factors of 100 that are divisible by 4, namely 4, 20 and 100. So the answer is 3.

 

Two numbers have a sum of 2 and a product of -960. What is their difference?

Hint: Think of two numbers that add up to 2 and multiply to -960. Try setting up a quadratic equation based on those facts and solving it. Once you find the two numbers, subtract one from the other to get the difference.

Solution: The two numbers are 32 and -30, so their difference is 62.

 

Two numbers have a sum of 1 and a product of -5/16. What is their difference?

Hint: Try using the formula for the difference of two numbers based on their sum and product.

If you know the sum is 1 and the product is −5/16, you can plug those into a known identity to find the difference. Think about squaring the difference first.

Solution: The two numbers are 5/4 and -1/4, so their difference is 3/2 (or 5).

 

On 27/09/21 the sum of the digits of the day (2+7) was equal to the number of the month (9). How many more times will this happen in 2021?

Hint: Look for other dates where (sum of the two digits of the day) = (month number). Check all months and possible days in 2021 where this can happen.

Solution: This will happen again on 19/10, 28/10 and 29/11; that is three more times.

 

Does there exist a number that is larger than its cube?

Hint: Think about what happens when you cube numbers greater than 1, between 0 and 1, and less than 0. Try plugging in a few examples to see when the number is bigger than its cube!

Solution: There exist lots of numbers larger than their cube: any number between 0 and 1, and any number less than -1; so the answer is yes.

 

The reciprocal of a number is the result of dividing 1 by the number. For example, the reciprocal of 4 is 1/4. What is the reciprocal of 1/2 + 1/3?

Hint: Think about how you add fractions and what a reciprocal means. What do you need to do before finding the reciprocal?

Solution: 1/2 + 1/3 = 3/6 + 2/6 = 5/6, and the reciprocal of 5/6 is 6/5.

 

A positive integer (whole number) n is divisible by 14 and 15. How many other positive integers are guaranteed to be factors of n?

Hint: Think about the numbers that divide both 14 and 15. What must a number divisible by both 14 and 15 also be divisible by? Consider the factors of 14 and 15, and how they combine.

Solution: Since n is divisible by 14 and 15, it must be divisible by the lowest common multiple of 14 and 15, which is 210. It must also be divisible by all the factors of 210: 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105, 210. So there are 16 factors of 210, of which we have already counted the factors 14 and 15; hence there are 14 other positive integers guaranteed to be factors of n.

 

In how many ways can 5 apples be distributed between 3 boys, George, Kieran and Rohan, so that each child receives at least 1 apple?

Hint: First, give 1 apple to each child, because they all need at least 1. Now, how many apples are left? Think about how you can share the leftover apples with the children.

Solution: The possible distributions to George, Kieran and Rohan respectively are (1,1,3), (1,3,1), (3,1,1), (1,2,2), (2,1,2) and (2,2,1). So there are 6 ways to distribute the apples.

 

In how many ways can the letters in the word NOON be arranged?

Hint: Count the total arrangements of 4 letters, then divide by the factorial of the repeated letters’ counts to avoid overcounting identical arrangements.

Solution: The possible arrangements are NNOO, NONO, NOON, ONNO, ONON, OONN, so the answer is 6.

 

In how many ways can the letters in the word SCHOOL be arranged?

Hint: First, count the total letters, then identify any repeated letters.

Use the formula: total letters! / factorials of each repeated letter’s count

to find the number of distinct arrangements.

Solution: Suppose all six letters were different. Then there would be 6 ways to choose the first letter, then 5 ways to choose the second letter, then 4 ways to choose the third letter, and so on. This yields 6x5x4x3x2x1=720 arrangements. But in the word SCHOOL, two of the letters are identical, so 720 has counted every possible arrangement twice, and the answer is 360.

 

In a triangle, the interior angles are in the ratio 3:4:5. What is the size of the largest angle?

Hint: The sum of all interior angles in a triangle is 180°. If the angles are in ratio 3:4:5, add the parts (3 + 4 + 5), then find the value of one part by dividing 180 by that total. Multiply by 5 to get the largest angle.

Solution: There are 12 parts, and 180 degrees in total, so each part is worth 15 degrees. The largest angle corresponds to 5 parts, so is equal to 75 degrees.

Bananas come in bunches of 3 (small bunch) or 5 (large bunch). Tim wants to purchase exactly 32 bananas; how many different combinations of bunches could he purchase?

Hint: Try using different numbers of small and large bunches to make exactly 32 bananas!

Solution: The possible bunch combinations are 4 small and 4 large, or 9 small and 1 large, so there are just two combinations with a total of 32 bananas.

 

How many three-digit numbers contain the digit 3?

Hint: Think about how many total three-digit numbers there are. Then, figure out how many don’t have the digit 3 in any place.

Solution: There are 9x10x10=900 three-digit numbers, of which 8x9x9=648 do not contain the digit 3, so the required total is 900-648=252.

 

If you really enjoy solving mathematical problems or if you’d just like a bit more practice, here are some web sites you can look at. I’m certain you will find plenty to challenge you!

Click on any of the links to take you straight there.

Discover Hampton podcast – The Magic of Maths

In this episode, we meet Hampton teacher Mr Griller whose love of Maths is infectious. Today’s lesson is all about combinatorics – a kind of clever counting! Join us in the classroom as Mr Griller shows his pupils how to apply strategies and short cuts to help them count without actually needing to do all the hard work – why not have a go at the questions yourself? Click on the link below to listen.

Junior Maths Challenge

Here you can try online challenges from previous Junior Maths Challenges, as well as download past papers and solutions.

UKMT Junior Maths Challenge

Nrich Website / Twitter

The NRICH Project aims to enrich the mathematical experiences of all learners. They have lots of activities, questions and games to develop your mathematical skills, whatever level you are. They definitely make maths fun!

NRICH Primary Maths

They also post questions on twitter for all ages of students.

NRICH Twitter Account

BBC Bitesize

Lots of fun problem solving questions for KS2.

BBC Bitesize Maths

Times Tables Rockstars

Times Tables Rock Stars is a carefully sequenced programme of fun daily times tables practice, which concentrates on a different times table each week. It has a small subscription of £7.20 a year for a family.

Times Tables Rockstars

As well as these web sites, I can also recommend these two brilliant maths books, written by one of Hampton’s own Maths Teachers, Mr Griller.

A Ring of Cats and Dogs 

Mr Griller’s brand new book! With a cast including a frog, a ladybird, a liar among truthtellers and a semicircular pool table, this intriguing and wide-ranging original collection will keep you absorbed, whether you are a beginner or an experienced problem solver, and completing it will take something very special indeed.

Elastic Numbers

This is not an ordinary puzzle book. Beautifully crafted and immensely enjoyable, the problems require minimal technical knowledge, being accessible to young secondary school pupils. However, there is an astonishing range in difficulty; while some of the problems are fairly straightforward, others are significantly tougher, with a great deal of ingenuity and clarity of thought needed to make progress.

Why not have a look at our Bookworms section, with book recommendations, words of the week, and lots more useful resources.

Bookworms