This is an open-ended prompt. Any five real situations involving counting, measuring, ordering, labelling or calculating are valid.
Five situations are: reading time, using a calendar, counting objects or marks, measuring height and weight, and using money.
From the activity, each child says a number based on the heights of the children next to them in the line.
The numbers tell how many neighbouring children are taller than the child saying the number.
A child at an end has only one neighbouring child, so the end child cannot have two taller neighbours and therefore cannot say 2.
No.
If no child has a taller neighbour, every child says 0. This can happen when all children have equal height.
Yes, if all the children are of the same height.
Two adjacent children can have the same count of taller neighbours depending on the arrangement of heights.
Yes.
In an increasing order from one end, each child except the tallest has one taller neighbour beside them. The tallest child has no taller neighbour.
Yes. They can stand in increasing or decreasing order of height so that four children have exactly one taller neighbour and the tallest child says 0.
The tallest child cannot say 1 because no neighbouring child can be taller than the tallest child.
No.
Arrange the children so that the middle child is shorter than both neighbours, the two next to the ends each have one taller neighbour, and the two end children have no taller neighbour.
Yes, it is possible.
Only middle positions have two neighbours. To say 2, a child must have both neighbours taller, so the maximum possible number is limited by such interior low positions.
At most two children can say 2. Arrange the heights so that two non-end children are each shorter than both their neighbours.
A supercell is greater than its adjacent cells. Checking the row $6828,670,9435,3780,3708,7308,8000,5583,52$, the listed numbers are greater than their neighbours.
The supercells are 6828, 9435, 7308 and 8000.
The chosen 4-digit numbers make exactly the required coloured cells larger than their adjacent cells. Other valid fillings are possible.
One valid filling is: 5346, 5347, 1000, 1258, 1100, 1200, 1300, 9635, 9636.
To maximise supercells in a row, place larger numbers in alternate cells so that each is greater than its neighbours.
One way is to arrange high and low numbers alternately, starting with a high number, for example: 900, 100, 800, 200, 700, 300, 600, 400, 500.
For 9 cells, the maximum number of alternate high positions is $(9+1)\div2=5$.
There can be 5 supercells.
Supercells cannot be adjacent because each would need to be larger than the other. Therefore the best strategy is to use alternate cells.
For $n$ cells, the maximum number of supercells is $n\div2$ if $n$ is even, and $(n+1)\div2$ if $n$ is odd. Start with a supercell and then place supercells in alternate positions.
The cell containing the largest number will always be greater than its adjacent cell or cells, so it must be a supercell.
No.
The largest number is greater than all adjacent numbers. The smallest number is less than any adjacent number, so it cannot be greater than its neighbours.
The largest number will always be a supercell. The smallest number cannot be a supercell.
Here the second largest number, 8, is adjacent to 9, so it is not greater than all its neighbours and is not a supercell.
One valid row is $1,2,3,4,5,6,7,9,8$.
The second smallest number 2 is at the end and is greater than its only neighbour 1, so it is a supercell. The second largest number 8 is beside 9, so it is not a supercell.
Yes. One valid row is $2,1,3,4,5,6,7,9,8$.
A good variation changes one condition while keeping the same supercell rule.
Examples: Fill 9 cells so that there are exactly 4 supercells. Fill 9 cells so that the largest number is at one end. Fill 9 cells so that the second smallest number is a supercell.
All entries use the digits 1, 0, 6, 3 and 9 exactly once. The arrangement follows the required coloured-cell condition.
One valid completion is: 96,310; 96,301; 36,109; 39,160; 96,103; 13,609; 60,319; 19,306; 13,906; 10,396; 60,193; 60,931; 10,369; 10,963; 10,936; 69,031.
Compare the completed table entries. 96,310 is greater than the other listed 5-digit numbers.
The biggest number in the table is 96,310.
Among the even entries in the completed table, 10,396 is the least.
The smallest even number in the table is 10,396.
Compare the table entries greater than 50,000; 60,193 is the smallest among them.
The smallest number greater than 50,000 in the table is 60,193.
Place the numbers in increasing order along the number line from 1000 to 10000.
From left to right, the numbers are 1050, 1500, 2180, 2754, 3050, 3600, 5030, 5300, 8400, 9590 and 9950.
Read the equal intervals on each number line and continue by the step size shown: 5, 1, 1 and 1000 respectively.
a. 1990, 1995, 2000, 2005, 2010, 2015, 2020, 2025, 2030, 2035. b. 9993, 9994, 9995, 9996, 9997, 9998, 9999, 10000, 10001, 10002. c. 15077, 15078, 15079, 15080, 15081, 15082, 15083, 15084, 15085, 15086. d. 83705, 84705, 85705, 86705, 87705, 88705, 89705, 90705, 91705, 92705.
Two-digit numbers run from 10 to 99, giving $99-10+1=90$. Similarly, 100 to 999 gives 900, 1000 to 9999 gives 9000, and 10000 to 99999 gives 90000.
There are 90 two-digit numbers, 900 three-digit numbers, 9000 four-digit numbers and 90000 five-digit numbers.
From 1-100, 7 appears 10 times in the ones place and 10 times in the tens place. From 000 to 999, each digit appears 100 times in each of three places, so 7 appears $3\times100=300$ times; 1000 adds no 7.
From 1-100, the digit 7 occurs 20 times. From 1-1000, it occurs 300 times.
To make the smallest number with digit sum 14, use the fewest place value while keeping the first digit nonzero: 59. To make the largest 5-digit number, put as much of the sum as possible in the leftmost digit: 95000.
a. Examples are 248, 653, 356, 815, 833 and 12335. b. The smallest number is 59. c. The largest 5-digit number is 95000. d. There is no largest possible number; examples include 95, 9005, 900005 and 90000005, and more zeros can be inserted to make still larger numbers.
Within each decade, the ones digit increases by 1, so the digit sum also increases by 1. At the next decade, the tens digit increases and the ones digit resets.
The digit sums are 4,5,6,7,8,9,10,11,12,13 for 40-49; 5,6,7,8,9,10,11,12,13,14 for 50-59; 6,7,8,9,10,11,12,13,14,15 for 60-69; and 7 for 70.
For consecutive digits $n,n+1,n+2$, the sum is $3n+3=3(n+1)$, a multiple of 3.
For 123, 234, 345, 456, 567, 678 and 789, the digit sums are 6, 9, 12, 15, 18, 21 and 24. They are multiples of 3 and increase by 3 each time. This pattern continues for three consecutive increasing digits only up to 789.
A 3-digit palindrome has the form ABA. Using digits 1, 2 and 3, choose the first/last digit in 3 ways and the middle digit in 3 ways, giving 9 palindromes.
The possible palindromes are 111, 121, 131, 212, 222, 232, 313, 323 and 333.
Try several 2-digit numbers by reversing and adding. The official solution states that the process always gives a palindrome for 2-digit starts.
For 2-digit numbers, this process gives a palindrome. For example, $12+21=33$ and $47+74=121$.
In the place-value wording, units digit $u=1$, tens digit $t=2u=2$, and hundreds digit $h=2t=4$. A 5-digit palindrome then has digits 1, 2, 4, 2, 1.
The number is 12421.
Example: $321-123=198$, $981-189=792$, $972-279=693$, $963-369=594$, $954-459=495$, and then $954-459=495$ repeats.
The number 495 starts repeating.
Look for digit patterns in the hour and minute display of a 12-hour clock.
Examples include repeated-digit times such as 2:22, 3:33, 4:44, 10:10, 11:11 and 12:12, and palindromic times such as 12:21, 05:50 and 10:01.
Form the largest and smallest possible numbers from the chosen digits, then subtract or add and compare with the given values.
a. Digits 7, 4, 3, 1: $7431-1347=6084$. b. Digits 7, 4, 3, 3: $7433-3347=4086$. c. Digits 7, 4, 3, 3: $7433+3347=10780$. d. Digits 7, 4, 3, 1: $7431+1347=8778$.
The smallest 5-digit palindrome is 10001 and the largest is 99999. Sum: $10001+99999=110000$. Difference: $99999-10001=89998$.
The sum is 110000 and the difference is 89998.
From 10:01 to 11:11 is 1 hour 10 minutes, or 70 minutes. From 10:01 to 12:21 is 2 hours 20 minutes, or 140 minutes.
The next palindromic time is 11:11, after 70 minutes. The one after that is 12:21, after 140 minutes from 10:01.
$8653-3568=5085$; $8550-0558=7992$ is not the appendix route, so using the official digit rearrangement chain: $8653-3568=5085$, $8550-5058=3492$, $9432-2349=7083$, $8730-3078=5652$, $6552-2556=3996$, $9963-3699=6264$, $6642-2466=4176$, $7641-1467=6174$. Thus it takes 8 rounds.
It takes 8 rounds to reach 6174.
The available smaller step 400 cannot combine to exactly 1000 with the other given numbers, while 14000, 15000 and 16000 can be formed as shown.
1000 cannot be made. Examples: $14000=1500\times8+400\times5$, $15000=13000+400\times5$, and $16000=1500\times8+400\times10$. Only 1000 cannot be made.
Check each place-value condition using the smallest and largest possible numbers of the required digit lengths.
Examples: 5-digit + 5-digit gives a 5-digit sum more than 90,250: $45000+45400=90400$. 5-digit - 5-digit gives a difference less than 56,503: $80000-50000=30000$. 4-digit + 4-digit gives a 6-digit sum: not possible. 5-digit - 4-digit gives a 4-digit difference: $12000-2500=9500$. 5-digit + 3-digit gives a 6-digit sum: $99999+999=100998$. 5-digit - 3-digit gives a 4-digit difference: $10000-999=9001$. 5-digit + 5-digit gives a 6-digit sum: $60000+40000=100000$. 5-digit - 5-digit gives a 3-digit difference: $50999-50000=999$. 5-digit + 5-digit gives 18500: not possible. 5-digit - 5-digit gives 91500: not possible.
a. $20000+30000=50000$, but $20000+80000=100000$. b. $1000+10=1010$, but $9999+99=10098$. c. The greatest possible sum is $9999+99=10098$, only 5-digit. d. $90000-10000=80000$, but $12000-10000=2000$. e. Even $10000-99=9901$, which is 4-digit, so a 3-digit result is impossible.
a. Only sometimes true. b. Only sometimes true. c. Never true. d. Only sometimes true. e. Never true.
Use the rule: if the number is even, halve it; if it is odd, multiply by 3 and add 1. These examples reach 1, but the general conjecture is still unproved.
Example sequences: starting at 28 gives 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1. Starting at 19 gives 19, 58, 29, 88, 44, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1.
The answer should give realistic examples of quantities in the stated ranges.
A few thousand: books in a school library, houses in a locality, or vehicle numbers in a small town. More than ten thousand: mobile phone numbers, population of a town, or salaries in rupees.
This is an estimation question. The reasonableness of ₹ 100 depends on local prices and quantities.
It may be possible if he buys small quantities and low-cost fruits, but it may not be enough if the fruits are costly or the serving size is large.
The prompt asks for an estimate, not an exact road distance. The official appendix gives 2500 km.
A reasonable estimate is about 2500 km.
If a student spends about 6 hours per school day and has about 200 school days per year, that is $6\times200=1200$ hours per year. $13000\div1200\approx10.8$ years, which is more than the likely school years completed by a Grade 6 student.
No, 13000 hours is too high for a Grade 6 student.
After changing 62,871 to 12,876, the grid has four cells that are greater than their neighbours.
Swap the digits 1 and 6 in 62,871 to make 12,876.
$9810-1089=8721$, $8721-1278=7443$, $7443-3447=3996$, $9963-3699=6264$, $6642-2466=4176$, $7641-1467=6174$.
For example, the year 1980 takes 6 rounds to reach 6174.
Use only odd digits and keep the number between 35,000 and 75,000. For the closest value above 50,000, the first digit must be 5 and the remaining digits should make the number as small as possible.
If repetition of digits is allowed: largest is 73,999, smallest is 35,111 and closest to 50,000 is 51,111. If digits are not repeated: largest is 73,951, smallest is 35,179 and closest to 50,000 is 51,379.
There are about 104 weekend days in a year. Adding festivals and vacations can bring the total near 120 or more, depending on the school calendar.
A sample estimate is about 120 holidays in a year.
These are everyday estimates; actual capacities vary by object size.
A mug may hold about 1 litre, a bucket about 15 litres, and an overhead tank about 500 to 1000 litres.
Here 18000 is a 5-digit number and 300 and 370 are 3-digit numbers. Their sum is $18000+300+370=18670$.
$18000+300+370=18670$.
Choose repeated addends or a symmetric pattern whose total is the selected number.
For 250, one pattern is five 50s: $50+50+50+50+50=250$. Another is twenty-five 10s.
A power of 2 has the form $2^n$. Since it is even, the Collatz rule halves it to $2^{n-1}$, then $2^{n-2}$, and so on until it reaches $2$, then $1$.
Every power of 2 eventually reaches 1 by repeated halving.
$100,50,25,76,38,19,58,29,88,44,22,11,34,17,52,26,13,40,20,10,5,16,8,4,2,1$.
Yes. Starting from 100, the sequence reaches 1.
Since each player may add 1, 2 or 3, a player can respond so that each pair of moves adds 4. The target numbers are therefore 22, 18, 14, 10, 6 and 2.
The winning strategy is to play first and make the running totals 2, 6, 10, 14, 18 and 22.