Numbers help us organise, compare and describe everyday life. Five common uses are:

1. telling time,
2. writing dates and calendars,
3. counting objects or marks,
4. measuring height, weight, length or distance,
5. dealing with money.

Phone numbers, house numbers, scores, temperatures and vehicle numbers are other examples.

Each child says the number of taller neighbours standing immediately next to them.

• 0: neither neighbour is taller.
• 1: exactly one neighbouring child is taller.
• 2: both neighbouring children are taller.

A child at an end has only one neighbour, so an end child can say only 0 or 1.

No. A child at an end has only one neighbouring child. To say 2, a child must have two taller neighbours, one on each side.

Therefore an end child can never say 2.

The supplied solution gives yes if all children are of the same height: then nobody has a taller neighbour.

So every child would say 0.

Yes. Adjacent children can both have the same count of taller neighbours.

For example, in a suitable height arrangement two adjacent children may each have one taller neighbour and therefore both say 1.

Yes. Arrange them in increasing order of height from left to right.

The first four children each have a taller child immediately to their right, so each says 1. The tallest child is at the right end and has no taller neighbour, so that child says 0.

No. The tallest child has no taller neighbour anywhere. So the tallest child must say 0, whether placed at an end or in the middle.

Hence a sequence made entirely of 1s is impossible.

Yes. Choose the middle child to be shorter than both immediate neighbours, so the middle child says 2. Choose the two end children to be taller than their only neighbours, so they say 0.

The remaining two positions can then each have exactly one taller neighbour, giving the sequence 0, 1, 2, 1, 0.

A child can say 2 only when both immediate neighbours are taller. Such a child is a local low point in the height arrangement.

Two children saying 2 cannot stand next to each other, so with five positions the maximum is 2.

Use an alternating high–low–high–low–high arrangement. The two low children then each have two taller neighbours.

A cell is a supercell when the number in it is greater than all its adjacent cells.

For a middle cell, compare with both left and right neighbours. An end cell has only one adjacent cell, so it is a supercell if it is greater than that single neighbour.

Compare each number only with its immediate neighbour(s):

• 6828 > 670 → supercell.
• 9435 > 670 and 3780 → supercell.
• 8000 > 7308 and 5583 → supercell.

No other entry is larger than all its adjacent entries.

One valid filling from the supplied solution is:

The bold/green-style positions are the intended supercells. Each is greater than its adjacent entry or entries, while every uncoloured cell fails the supercell test.

To maximise supercells, alternate high and low values, beginning and ending with a high value.

Every highlighted entry is greater than its neighbour(s). This gives 5 supercells.

For 9 cells, the maximum is 5.

In general, supercells must be separated by non-supercells. The best arrangement alternates high and low entries.

So for 2,4,6,… cells the maximum is n/2; for 1,3,5,7,… it is (n+1)/2.

No. Look at the largest number in the entire table. Since no other number is larger, every neighbour of that cell is smaller.

Therefore the cell containing the largest number must be a supercell.

The largest number is always a supercell when all entries are distinct, because all its neighbours are smaller.

The smallest number cannot be a supercell in a multi-cell table of distinct numbers, because any neighbour it has is larger.

Put the largest number immediately beside the second-largest number. Then the second-largest entry has a larger neighbour and fails the test.

Here 8 is the second-largest number, but it is adjacent to 9, so 8 is not a supercell.

Yes. One example is:

The number 2 is at an end and is greater than its only neighbour 1, so 2 is a supercell. The number 8 is beside 9, so 8 is not a supercell.

Examples suggested by the supplied solution include:

• Can a 9-cell table have more than 5 supercells?
• Can a 9-cell table have exactly 4 supercells?

The first is impossible because the maximum is ⌈9/2⌉ = 5. The second is possible by breaking one peak in an alternating arrangement.

In a multi-row table, only cells immediately left, right, above and below count as neighbours. Diagonal cells do not count.

A cell is a supercell if its number is greater than every neighbour that exists in those four directions.

One complete arrangement from the supplied solution is:

Every entry uses exactly the digits 1, 0, 6, 3 and 9 in some order.

• Biggest number = 96,310.
• Smallest even number = 10,396.
• Smallest number greater than 50,000 = 60,193.

For evenness, look only at the units digit. For size comparisons, compare place values from left to right.

First locate the thousand interval, then place the number according to its hundreds and tens.

From left to right:

For example, 2180 lies just to the right of 2000, while 9950 lies just to the left of 10,000.

(a) The interval is 5:

Smallest 1990; largest 2035.

(b) The interval is 1:

Smallest 9993; largest 10002.

(c) The interval is 1:

Smallest 15077; largest 15086.

(d) The interval is 1000:

Smallest 83705; largest 92705.

Each new digit length has nine times a power of ten: 9, 90, 900, 9000, 90000.

Add the digits and check that the total is 14. Examples include:

There are many more possibilities.

A one-digit number cannot have digit sum 14. So try a two-digit number.

To make the number as small as possible, make the tens digit as small as possible while the units digit stays at most 9.

Smallest number = 59.

To maximise the number, put as much of the digit sum as possible in the leftmost places.

The first digit can be 9. We then have 14−9=5 left for the next digit, and zeros after that.

Digit sum check: 9+5+0+0+0=14.

There is no largest number if the number of digits is unrestricted.

We can keep inserting zeros without changing the digit sum:

Every extra zero increases the place value of the 9 and makes a larger number while the digit sum remains 14.

The sums rise by 1 as the units digit increases, then drop when a new decade begins.

Example: 49 has sum 13, but 50 has sum 5 because the units digit changes from 9 to 0 while the tens digit increases only by 1.

For the standard increasing examples:

Every sum is a multiple of 3, and each new sum is 3 more than the previous sum.

The pattern in this exact form cannot continue beyond 789 because the next ‘digit’ after 9 would be 10, which is not a single digit.

From 1 to 100: 7 appears 10 times in the units place (7,17,…,97) and 10 times in the tens place (70–79).

From 1 to 1000: in the three-digit block 000–999, each digit appears equally often in each of the hundreds, tens and units positions.

The number 1000 adds no extra 7.

A palindrome reads the same from left to right and from right to left.

Examples from the chapter include 66, 848, 575, 797 and 1111.

For a 3-digit palindrome, the first and last digits must be equal.

A 3-digit palindrome has the form ABA. There are 3 choices for A and 3 choices for B, giving 3×3=9 possibilities:

Start with a 2-digit number, reverse its digits and add. If the sum is not a palindrome, reverse the new number and add again.

Examples:

The chapter states that starting from a 2-digit number this process reaches a palindrome.

Let the units digit be u. Because the number is odd, u must be odd. The tens digit is 2u and the hundreds digit is 4u.

Since 4u must still be a digit, u can only be 1 among positive odd choices.

So u=1, tens=2 and hundreds=4. Palindrome symmetry gives the ten-thousands digit 1 and thousands digit 2.

In words: twelve thousand four hundred twenty-one.

1. Choose a 4-digit number with at least two different digits.
2. Arrange its digits in descending order to make A.
3. Arrange the same digits in ascending order to make B.
4. Subtract: C = A − B.
5. Repeat using the digits of C.

The chapter shows that this process reaches 6174, the Kaprekar constant.

Follow the same arrange-and-subtract rule each round:

Once 6174 is reached, the process repeats: 7641−1467=6174 again.

The supplied solution demonstrates the process beginning with 321:

Now 954−459 gives 495 again. So the 3-digit Kaprekar constant is 495 for the intended non-repdigit cases.

The activity is open-ended. Examples from the supplied solution include:

• Repeated digits: 2:22, 3:33, 4:44.
• Repeated hour/minute pattern: 09:09, 10:10, 11:11, 12:12.
• Palindromic-style displays: 05:50, 10:01, 12:21.

More examples can be found depending on whether a leading zero is included in the hour display.

The supplied solution gives examples such as 20/04/2004 and 20/06/2006.

The aim is to look for a repeated arrangement of digits across day, month and year, not merely identical dates.

Examples provided in the supplied solution include 01/02/2001 and 02/02/2002.

When the slashes are ignored, the digits read the same in reverse.

Yes, some yearly calendars repeat. For a calendar to match, the year must begin on the same weekday and the leap-year structure must also be compatible.

The supplied solution notes repeats after 5 or 6 years in some cases depending on how leap years fall.

(a) Using 7,4,3,1:

(b) Using 7,4,3,3:

Many other choices are possible.

(c)

(d)

The smallest 5-digit palindrome is 10001. The largest is 99999.

The next palindromic time is 11:11.

The following palindromic time is 12:21.

Apply the Kaprekar step repeatedly:

Answer: 8 rounds.

Possible mental decompositions are:

1,000 cannot be made. The only available number below 1000 is 400, and no positive whole-number multiple of 400 equals 1000.

The supplied solution concludes that among exact whole thousands, 1,000 is the only one that cannot be made under this activity’s rules.

One convenient set of solutions is:

Different valid expressions are possible.

• (a) 5-digit + 5-digit gives a 5-digit number: Sometimes. Example 20,000+30,000=50,000, but 20,000+80,000=100,000.
• (b) 4-digit + 2-digit gives a 4-digit number: Sometimes. 1,000+20=1,020, but 9,999+99=10,098.
• (c) 4-digit + 2-digit gives a 6-digit number: Never. Even 9,999+99=10,098.
• (d) 5-digit − 5-digit gives a 5-digit number: Sometimes. 99,999−10,000=89,999, but 12,000−10,000=2,000.
• (e) 5-digit − 2-digit gives a 3-digit number: Never. The smallest possible positive result is 10,000−99=9,901, which is still 4-digit.

Group equal entries and use multiplication.

(a) There are 12 boxes of 40 and 10 boxes of 50:

(b) The 8×8 grid has 20 cells showing 5 dots and 44 cells showing 1 dot:

(c) There are 32 cells of 32 and 16 cells of 64:

(d) Count by rows: six rows total 18 each, and one row totals 15:

(e) By symmetry, 15, 25 and 35 each appear 22 times:

(f) The concentric pattern has sixteen 125s, eight 250s, four 500s and one 1000:

Start with a positive whole number and repeat:

• If the number is even, divide it by 2.
• If the number is odd, multiply it by 3 and add 1.

The famous Collatz conjecture says that every positive starting number eventually reaches 1, but the chapter explicitly notes that this remains unproved.

Starting at 28:

Starting at 19:

If the starting number is a power of 2, it is even at every stage until 1 is reached.

No 3n+1 step is needed. Repeated halving proves the result for every power of 2.

Apply the even/odd rule at each step:

So the sequence starting at 100 does reach 1.

An estimate is a reasonable approximate value used when an exact count is unnecessary, unavailable or too time-consuming.

Good estimation uses a known smaller quantity as a benchmark, then scales it up.

These answers depend on the student and location, so there is no single fixed answer. A useful method is:

1. Estimate your average step length.
2. Estimate the distance.
3. Use steps ≈ distance ÷ step length.

For example, if one step is about 0.6 m and the classroom door is 9 m away, the estimate is 9÷0.6≈15 steps.

First observe yourself for one minute, then scale the result.

If you count b breaths in one minute:

The same multiplication method works for blinks.

This is an open estimation activity. Suitable examples depend on context. The supplied solution suggests items such as four-digit identification/vehicle-style numbers for ‘a few thousand’ and quantities such as salaries or long number labels for ‘more than ten thousand’.

A better classroom response should name countable objects relevant to the student’s surroundings.

The answer depends on quantity and prices. The supplied solution says it may be possible with small quantities and inexpensive fruits, but may be too low for larger servings or costly fruits.

The mathematical lesson is to break the estimate into parts: milk cost + fruit 1 + fruit 2 + fruit 3, then round each cost to an easy nearby amount.

The supplied solution uses an estimate of about 2,500 km.

This is an estimation task: the intention is to locate both cities on a map, judge the scale and choose a sensible order of magnitude rather than produce an exact route distance.

Use the chapter’s rough assumptions of 6 school hours per day and about 200 working days per year.

Even over about 8 school years:

So 13,000 hours is considerably too high under these assumptions.

Use the relationship:

A normal walking speed can be approximated for the activity, then used consistently. Nearby destinations may take minutes or hours; a neighbouring state capital would take many hours or days; walking the length of India would take many weeks or months.

The student’s chosen location and route determine the actual estimate.

The key is to control numbers spaced 4 apart:

The first player begins by saying 1. After the opponent adds k (1, 2 or 3), add 4−k. Together the two moves always advance by 4, so the first player keeps landing on the control numbers and eventually says 21.

Now each pair of moves should total 11. The control numbers are:

Because the first player must begin with 1–10, the second player can always respond with enough to make 11. Thereafter, if the opponent adds k, respond with 11−k.

This keeps the second player on every multiple of 11 up to 99.

Change the centre number 62,871 by swapping its digits 6 and 1:

The grid becomes:

The four highlighted cells are each greater than their left/right/up/down neighbours.

For 1980:

1980 takes 6 rounds. Students should repeat the same process using their own birth year.

Using repetition of odd digits:

• Largest = 73,999.
• Smallest = 35,111.
• Closest to 50,000 = 51,111.

The supplied solution also gives non-repeating versions: 73,951; 35,179; and 51,379.

This is school-specific. A good method is to estimate in parts:

1. weekend days,
2. festival/public holidays,
3. summer/winter breaks,
4. other school closures.

Add the estimates, then compare them with the exact school calendar. The point is to measure how close the estimate was.

This is an order-of-magnitude estimation question. Think of familiar reference containers.

A mug is typically measured in fractions of a litre or around 1 litre, a bucket in several litres, and an overhead tank in hundreds or thousands of litres. The exact values depend on the actual objects being observed.

One example from the supplied solution is:

18,000 is a 5-digit number, while 300 and 370 are both 3-digit numbers.

The supplied example chooses 250. One pattern is:

Another is a 5×5 array of 10s:

The idea is to use symmetry or repeated groups rather than an unstructured list.

Every power of 2 is even, so the rule repeatedly halves it:

The same argument works for 2, 4, 8, 16, 32, …

The full sequence is:

Therefore 100 reaches 1.

The first player should begin by adding 2, then aim for numbers 4 apart:

If the opponent adds k, respond by adding 4−k. The two moves together increase the total by 4, so the first player always returns to the winning sequence and finally reaches 22.

Therefore the first player can force a win.

Supercells must alternate with non-supercells.

Answer: 6.

Answer: 6.

Make the thousands digit as small as possible while the remaining three digits can total the rest. Three digits can contribute at most 27. Try thousands digit 1; remaining sum 22 can fit. To minimise further, make hundreds as small as possible while tens+units can total at most 18. Hundreds must be 4.

Smallest = 1499.

Put the largest possible digit at the left.

Largest = 992000.

A 3-digit palindrome has form ABA. A has 9 choices (1–9), B has 10 choices (0–9).

Answer: 90.

A 5-digit palindrome is ABCBA. To be just above 34,000, start with 34. The form must be 34C43. Choose the smallest C=0.

Answer: 34,043.

Result after one round: 3087.

495 is reached.

From 12:21 to 1:00 is 39 minutes, then 1 more minute.

Answer: 40 minutes.

Both addends are 5-digit numbers.

The largest possible sum is:

So no, 20,000 is impossible.

Choose consecutive 5-digit numbers.

Answer: 1.

The next control number is 17.

You should add 3 and say 17.

The second player wants to reach 11.

Add 4, so the new number is 11.

Because 16 is a power of 2, only halving is needed.

7 is odd:

22 is even:

Answers: 22 and 11.

Answer: 1200.

Round 49≈50 and 198≈200.

A good estimate is about 10,000.

About 8,400 hours.

Each interior high position has a lower number on both sides, and each high end position has its only neighbour lower. Therefore every high position is greater than all of its adjacent cells.

Five supercells are obtained, which is the maximum for 9 cells.