Chapter 3: Number Play (Numbers)

• Time: clocks, timetables, train and bus timings.
• Dates: calendars, birthdays, holidays.
• Money: prices, bills, savings, change.
• Measurement: height, weight, temperature, distance, cooking quantities.
• Counting and identification: marks in a test, scores in games, roll numbers, phone numbers, house numbers, vehicle numbers.

Time, dates, money, measurement, and counting or identification (marks, scores, phone numbers).

No. A child at an end has only one neighbour, so they can say at most 1.

No: an end child has only one neighbour, so can say only 0 or 1.

Not if the children have different heights. The shortest child is shorter than each of their neighbours, so they must say 1 or 2. All could say 0 only if they were all of exactly the same height (then nobody has a taller neighbour).

No (for children of different heights): the shortest child always has a taller neighbour. Only children of equal heights would all say 0.

Yes. For example, if 5 children stand in increasing order of height, shortest first, the first four children all say 1 (each has only the next child taller than them). So neighbours can say the same number.

Yes, e.g. children standing from shortest to tallest say 1, 1, 1, 1, 0.

Yes. They should stand in increasing order of height. Each of the first four has exactly one taller neighbour (the next child), so says 1; the tallest child at the end has no taller neighbour, so says 0.

Yes: shortest to tallest gives 1, 1, 1, 1, 0.

No. The tallest child has no taller neighbour, wherever they stand, so the tallest child always says 0.

No: the tallest child always says 0.

Yes. Number the children by height, 1 = shortest and 5 = tallest, and stand them as 5, 3, 1, 2, 4:

• 5 (at an end): neighbour 3 is shorter → says 0.
• 3: neighbours 5 (taller) and 1 (shorter) → says 1.
• 1: both neighbours taller → says 2.
• 2: neighbours 1 (shorter) and 4 (taller) → says 1.
• 4 (at an end): neighbour 2 is shorter → says 0.

Yes, e.g. heights in the order tallest, middle, shortest, second shortest, second tallest (5, 3, 1, 2, 4).

A child says 2 only if both neighbours are taller. The end children cannot say 2, and two neighbours cannot both say 2 (one of them would have to be taller than the other). So with 5 children at most 2 children (in the 2nd and 4th places) can say 2.

Arrangement (by height, 5 = tallest): 5, 1, 4, 2, 3. The children say 0, 2, 0, 2, 0.

At most 2 children can say 2, e.g. heights 5, 1, 4, 2, 3 give 0, 2, 0, 2, 0.

A cell is a supercell if its number is larger than all its neighbours.

• 6828 > 670; 9435 > 670 and 3780; 8000 > 7308 and 5583.
• 7308 is not a supercell because 8000 is bigger; 3780 is smaller than 9435.

Supercells: 6828, 9435 and 8000.

One way:

Check: 6000 > 5346 and 1000; 1258 > 1000 and 1100; 9800 > 9635. No other cell is bigger than both its neighbours (1200 < 1300, 1300 < 9635, 9635 < 9800).

For example: 5346, 6000, 1000, 1258, 1100, 1200, 1300, 9635, 9800.

Put large numbers in the 1st, 3rd, 5th, 7th and 9th cells and small numbers in between:

For example: 950, 150, 850, 250, 750, 350, 650, 450, 550 (supercells in bold).

5 supercells (cells 1, 3, 5, 7 and 9). This is the most possible, because two cells next to each other cannot both be supercells.

5

Pattern: for an even number of cells, the maximum is half the number of cells; for an odd number, it is half of (number of cells + 1).

Method: start with a large number in the first cell and then alternate large and small numbers (large, small, large, small, …). Every "large" cell is then bigger than its neighbours.

Maximum = half the cells (even), or half of one more than the cells (odd). Fill large and small numbers alternately, starting with a large one.

No. The cell with the largest number is bigger than every other number, so it is bigger than its neighbours. There is always at least one supercell.

No: the largest number is always a supercell.

• Largest number: yes, always a supercell, since it is bigger than all other numbers, including its neighbours.
• Smallest number: no (unless the table has just one cell), since it is smaller than its neighbours.

The largest is always a supercell; the smallest can never be one (in a table with more than one cell).

Place the second largest number next to the largest:

800 is the second largest, but its neighbour 900 is larger, so 800 is not a supercell.

Put the second largest number beside the largest, e.g. …, 700, 900, 800.

Yes. Put the second smallest number at an end with only the smallest number next to it, and put the second largest beside the largest:

200 (second smallest) is a supercell because its only neighbour is 100; 800 (second largest) is not, because 900 is next to it.

Yes, e.g. 200, 100, 300, 400, 500, 600, 700, 900, 800.

Some ideas:

• Fill 9 cells so that there are exactly 3 supercells.
• Fill 9 cells so that the third smallest number is a supercell.
• Fill a table using only the digits 1, 2, 3, 4 (each number uses all four digits) so that the supercells are in the 2nd, 5th and 8th cells.
• Can 8 cells have 4 supercells? (Yes, e.g. 9, 1, 8, 2, 7, 3, 6, 4.)

For example: exactly 3 supercells in 9 cells; the third smallest number as a supercell; 4 supercells in 8 cells.

One way (supercells in bold; the given numbers are 96,301, 36,109, 13,609, 60,319, 19,306, 60,193 and 10,963):

• 96,310 is the only arrangement of the digits larger than 96,301, so the top-left cell must be 96,310.
• The three neighbours of 10,963 must be smaller than it: 10,369, 10,396 and 10,936.
• The biggest number in the table is 96,310.
• The smallest even number (last digit 0 or 6) is 10,396.
• The smallest number greater than 50,000 is 60,193.

Biggest: 96,310; smallest even number: 10,396; smallest number above 50,000: 60,193 (for the table above).

Each part of the line between two thousands is 1000; find the two thousands a number lies between and how far it is from the smaller one (2180 is 180 past 2000, a little less than a fifth of the way to 3000).

See the number lines: e.g. 1050 just after 1000, 1500 halfway between 1000 and 2000, 9950 just before 10,000.

First find the gap between two neighbouring marks from the given numbers, then count forwards and backwards:

• a. 2010 and 2020 are two marks apart, so each step is 5.
• b. 9996 and 9997 are next to each other: steps of 1.
• c. 15,077 to 15,083 is 6 steps: steps of 1.
• d. 86,705 and 87,705 are next to each other: steps of 1000.

The smallest number (to be circled) is the first one on each line and the largest (to be boxed) is the last one: a. 1990 and 2035; b. 9993 and 10,002; c. 15,077 and 15,086; d. 83,705 and 92,705.

a. 1990, 1995, …, 2035 b. 9993, …, 10,002 c. 15,077, …, 15,086 d. 83,705, 84,705, …, 92,705. The smallest is at the left end and the largest at the right end of each line.

(For example, 2-digit numbers: 99 − 10 + 1 = 90.) Each count is 10 times the previous one.

9, 90, 900, 9000 and 90,000.

a. For example: 77, 86, 95, 149, 239, 455, 707, 1166, 2048, 30,029.

b. 59. A 1-digit number has digit sum at most 9, so we need 2 digits; to make the number small, the tens digit should be as small as possible: 5 and then 9 (5 + 9 = 14).

c. 95,000. Put the biggest possible digit first: 9, then 5 (9 + 5 = 14), and zeros after.

d. There is no biggest such number. Adding zeros keeps the digit sum the same: 950, 9500, 95,000, 9,50,000, … (or 9005, 90,005, …). We can always make an even bigger number by writing one more 0.

a. e.g. 77, 149, 455, 2048 b. 59 c. 95,000 d. There is no largest: put more zeros, e.g. 95,00,000; we can always make a bigger one.

Observations:

• Within each ten, the digit sum goes up by 1 each time.
• When we move to the next ten (49 → 50), the digit sum drops by 8 (13 → 5).
• Each row starts one higher than the previous row (4, 5, 6, 7).
• Many numbers share the same digit sum: 49, 58 and 67 all have digit sum 13; 45, 54 and 63 have 9.
• The smallest digit sum is 4 (for 40) and the largest is 15 (for 69).

Sums go up by 1 within each ten and drop by 8 at the next ten; numbers like 49, 58, 67 share a digit sum; the range is from 4 (40) to 15 (69).

Pattern: the digit sums go up by 3 each time and are all multiples of 3; each sum is 3 times the middle digit (for 345, 3 × 4 = 12).

Will it continue? No: after 789 there is no 3-digit number with three increasing consecutive digits (8, 9, 10 is not possible). (If we also allow 012 or decreasing digits such as 543, the same rule "3 × middle digit" still holds.)

6, 9, 12, …, 24: they increase by 3 (3 × the middle digit). The pattern stops at 789.

1–100: as a units digit in 7, 17, 27, …, 97 (10 times) and as a tens digit in 70, 71, …, 79 (10 times). Total = 20 (77 is counted twice, once for each 7).

1–1000: in every place (units, tens, hundreds), each digit appears equally often among 000–999: 7 is in the units place 100 times, in the tens place 100 times and in the hundreds place 100 times. Total = 300.

20 times in 1–100, and 300 times in 1–1000.

A 3-digit palindrome has the same first and last digit:

111, 121, 131, 212, 222, 232, 313, 323, 333 (9 palindromes).

111, 121, 131, 212, 222, 232, 313, 323, 333

Examples:

• 25 + 52 = 77 (1 step)
• 57 + 75 = 132, 132 + 231 = 363 (2 steps)
• 68 + 86 = 154, 154 + 451 = 605, 605 + 506 = 1111 (3 steps)
• 89 and 98 are the hardest: they need 24 steps to reach the palindrome 8,813,200,023,188.

Yes, every 2-digit number eventually gives a palindrome (as the textbook notes). For 3-digit numbers nobody knows: starting with 196, no palindrome has ever been found!

Yes, for every 2-digit number (89 and 98 take 24 steps). For some 3-digit numbers, such as 196, it is still unknown.

• Odd number, so the units digit u is odd: 1, 3, 5, 7 or 9.
• t = 2u and h = 2t = 4u must be single digits, so 4u ≤ 9: only u = 1 works. Then t = 2 and h = 4.
• Palindrome: the tth digit equals u = 1, and the th digit equals t = 2.

The number is 12,421: twelve thousand four hundred twenty-one.

12,421 (twelve thousand four hundred twenty-one)

Example with 418: 841 − 148 = 693 → 963 − 369 = 594 → 954 − 459 = 495 → 954 − 459 = 495 …

Example with 527: 752 − 257 = 495 straight away.

Starting with any 3-digit number whose digits are not all the same, we always reach 495, which then keeps repeating. (If a result such as 99 appears, treat it as 099: 990 − 099 = 891, and so on.)

495 (the 3-digit Kaprekar number).

• All digits the same (like 4:44): 1:11, 2:22, 3:33, 4:44, 5:55 and 11:11 (6 times).
• Minutes repeat the hour (like 10:10): 1:01, 2:02, 3:03, …, 9:09, 10:10, 11:11, 12:12 (12 times).
• Palindromes (like 12:21), reading the same both ways: for each hour h from 1 to 9, the times h:0h, h:1h, h:2h, h:3h, h:4h and h:5h (for example 1:01, 1:11, 1:21, 1:31, 1:41, 1:51), which is 54 times, plus 10:01, 11:11 and 12:21, making 57 in all.

Same digits: 1:11, 2:22, 3:33, 4:44, 5:55, 11:11. Hour repeated: 1:01, …, 12:12. Palindromes: h:0h, h:1h, …, h:5h for h = 1 to 9 (54 times) plus 10:01, 11:11 and 12:21 (57 in all).

Dates like 20/12/2012 (day and month together repeat the year): 20/01/2001, 20/02/2002, …, 20/09/2009, 20/10/2010, 20/11/2011; also 19/01/1901, …, 19/12/1912 and 18/01/1801, …, 18/12/1812.

Palindromic dates (DDMMYYYY reads the same backwards) since 2000: the year must be 20xy and the month 02:

(Earlier ones include 10/01/1001, 01/11/1110, 11/11/1111 and 21/11/1112, but no year from 1200 to 1999 gives a valid date.)

Like 20/12/2012: 20/01/2001 … 20/11/2011, 19/12/1912, etc. Palindromic dates since 2000: 10/02/2001, 20/02/2002, 01/02/2010, 11/02/2011, 21/02/2012, 02/02/2020, 12/02/2021, 22/02/2022.

Yes. A calendar repeats when the year starts on the same day of the week and both years have the same number of days (both ordinary years or both leap years).

• An ordinary year has 365 days = 52 weeks + 1 day, so the next year starts one day later in the week; after a leap year (366 days) it starts two days later.
• An ordinary year's calendar repeats after 6 or 11 years. For example, the calendar of 2026 will be repeated in 2037, and that of 2025 in 2031.
• A leap year's calendar repeats after 28 years (2024's calendar will come again in 2052).

Yes. An ordinary year repeats after 6 or 11 years (2026 → 2037); a leap year repeats after 28 years.

a. Digits 9, 5, 2, 1: 9521 − 1259 = 8262 (> 5085). (Spread-out digits give a big difference.)

b. Digits 5, 4, 4, 3: 5443 − 3445 = 1998 (< 5085). (Close digits give a small difference.)

c. Digits 9, 8, 7, 6: 9876 + 6789 = 16,665 (> 9779). (Large digits give a large sum.)

d. Digits 1, 2, 3, 4: 4321 + 1234 = 5555 (< 9779). (Small digits give a small sum.)

For example: a. 9521 − 1259 = 8262 b. 5443 − 3445 = 1998 c. 9876 + 6789 = 16,665 d. 4321 + 1234 = 5555

Smallest 5-digit palindrome = 10,001; largest = 99,999.

$$10,001+99,999=1,10,00099,999-10,001=89,998$$

Sum = 1,10,000; difference = 89,998

The palindromic times after 10:01 are 11:11 and then 12:21.

• 10:01 → 11:11 is 1 hour 10 minutes = 70 minutes.
• 11:11 → 12:21 is another 70 minutes, so 12:21 comes 140 minutes after 10:01.

70 minutes (11:11); the one after that, 12:21, is 140 minutes from 10:01.

It takes 7 rounds. (Here the smallest number made from 5, 0, 8, 5 is taken as 0558, as in Kaprekar's process. If instead you use 5058 as the smallest 4-digit number, you reach 6174 in 8 rounds, which is the answer given in the NCERT key.)

7 rounds (8 rounds if a 0 is not allowed at the start of the smallest number, as in the NCERT key).

See the table, e.g. 28,000 = 25,000 + 1,500 + 1,500 and 61,600 = 60,000 + 400 × 4.

• 1,000: not possible. The only middle number smaller than 1,000 is 400, and 400 + 400 = 800 while 400 + 400 + 400 = 1,200; we jump over 1,000.
• 14,000 = 13,000 + ? needs 1,000, which is impossible, so use 1,500 × 4 + 400 × 20 = 6,000 + 8,000, or 1,500 × 8 + 400 × 5 = 12,000 + 2,000.
• 15,000 = 13,000 + 400 × 5 (= 13,000 + 2,000).
• 16,000 = 13,000 + 1,500 + 1,500.

Which thousands cannot be made? Since 400 × 5 = 2,000 and 1,500 × 2 = 3,000, by adding 2,000s and 3,000s we can make every thousand from 2,000 upwards (4,000 = 2,000 + 2,000, 5,000 = 2,000 + 3,000, and so on). So only 1,000 cannot be made.

1,000 cannot be made (400s jump from 800 to 1,200). 14,000 = 1,500 × 8 + 400 × 5, 15,000 = 13,000 + 400 × 5, 16,000 = 13,000 + 1,500 × 2. Every other thousand can be made.

45,000 = 40,000 + 7,000 − 1,500 − 800 + 300; 5,900 = 7,000 − 800 − 300; 17,500 = 12,000 + 7,000 − 1,500; 21,400 = 12,000 + 7,000 + 1,500 + 300 × 3.

Other questions to try: 5-digit + 5-digit = 7-digit sum (never possible, since the largest sum is 1,99,998); 4-digit + 4-digit = 2,900 (possible: 1,000 + 1,900). Make your own and check them like this.

Three cases are impossible: 4-digit + 4-digit can't reach 6 digits (max 19,998); two 5-digit numbers can't add to 18,500 (min 20,000); two 5-digit numbers can't differ by 91,500 (max 89,999). Examples for the others are in the table.

a. Sometimes b. Sometimes c. Never d. Sometimes e. Never

Use groups, rows and multiplication instead of adding one by one:

The quicker ways use repeated addition as multiplication, and spotting equal groups (like the rings in f, each worth 2000).

a. 980 b. 144 c. 2048 d. 123 e. 1650 f. 7000, found by grouping equal numbers and multiplying.

• 7: 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
• 9: 9, 28, 14, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1
• 15: 15, 46, 23, 70, 35, 106, 53, 160, 80, 40, 20, 10, 5, 16, 8, 4, 2, 1

Yes, every sequence we try reaches 1 (and then repeats 4, 2, 1). Computers have checked this for an enormous range of starting numbers and it has always worked, so most people believe it is true. But nobody has been able to prove that it must happen for every whole number, which is why it is still an unsolved mystery: one exception, somewhere very far out, could still exist.

Every number tried reaches 1 (e.g. 7 takes 16 steps). It has been checked for huge numbers, so it is believed to be true, but it has never been proved for all numbers.

A child's step is about half a metre, so 1 metre ≈ 2 steps (100 m ≈ 200 steps; 1 km ≈ 2000 steps).

a. About 10–20 steps (5–10 m). b. About 150–300 steps (a ground 75–150 m long). c. About 50–200 steps, depending on the school. d. For a home 1 km away, about 2000 steps; 2 km, about 4000 steps.

Use about 2 steps per metre: a. 10–20 b. 150–300 c. 50–200 d. about 2000 steps per km.

Count your own blinks or breaths for one minute and multiply.

About 15–20 a minute, about 1000 an hour, and roughly 16,000 blinks or 20,000–25,000 breaths a day.

a. A few thousand: pages of all the books in your school bag together, bricks in a classroom wall, leaves on a small plant or bush, seats in a cinema hall, students in a big school.

b. More than ten thousand: hairs on your head (about 1 lakh), grains of rice in a 1 kg packet (about 50,000), grains of sand in a handful, leaves on a big tree, people in a stadium.

A few thousand: pages in your school books, bricks in a wall. More than ten thousand: hairs on your head, rice grains in 1 kg, leaves on a big tree.

a. More than 5000. A page has about 200–300 words and the book has about 300 pages, so it has tens of thousands of words.

b. This depends on your school. Estimate: if about one-third of a school of 900 students come by bus, that is 300, so more than 200; in a small school with 300 students it would be less than 200.

a. More than 5000 (tens of thousands). b. Depends on the school; estimate from its size.

Probably not. For 5 people he needs about 1 litre of milk (about ₹60–70) and, say, 2–3 bananas, 2 apples and some grapes (together about ₹150–200), plus sugar and custard powder. The total is roughly ₹250–300, so ₹100 is too low (it would be enough only for very small servings with cheap fruits).

No: milk alone costs about ₹60–70 and the fruits ₹150 or more, so the total is about ₹250–300.

On the map of India, Gandhinagar is in the west and Kohima in the far north-east, almost on the same latitude; the gap is about two-thirds of India's east–west width (about 3000 km). So the straight-line distance is about 2200 km; by road it is about 3000 km because roads curve around (through the Siliguri corridor).

About 2200 km in a straight line (about 3000 km by road).

No. A school day has about 5–6 hours and there are about 200 school days a year, so one year ≈ 1000–1200 hours. By Grade 6 she has been in school about 6 years (or 8–9 years if pre-school is counted), which is about 6000–10,000 hours. To reach 13,000 hours she would have to be in school for about 11–13 years.

No: at about 1000–1200 hours a year for about 6–8 years, she has spent roughly 6000–10,000 hours, not 13,000.

a. A place 2 km away: about 25 minutes (2 ÷ 5 h).

b. For example, Delhi to Jaipur is about 280 km: 280 ÷ 5 = 56 hours of walking, i.e. about 7 days walking 8 hours a day.

c. Kanyakumari to the northern tip of Ladakh is about 3200 km in a straight line and about 3700 km by road: 3700 ÷ 5 ≈ 740 hours, about 3 months walking 8 hours a day.

At 5 km/h: a. about 25 minutes for 2 km b. about a week (e.g. Delhi–Jaipur, 280 km) c. about 3 months (about 3700 km).

• How many students are there in your school?
• How many litres of water does your family use in a day?
• How many rotis does your family eat in a year?
• How many hours do you sleep in a year?
• How many steps do you climb in a week?

For example: litres of water your family uses in a day, rotis eaten in a year, hours of sleep in a year.

Work backwards from 21. If you say 17, the other player can only reach 18, 19 or 20, and you then reach 21. In the same way, saying 13 lets you reach 17, saying 9 lets you reach 13, and so on. The winning numbers go down in steps of 4:

$$21,17,13,9,5,1$$

The first player can always win: start by saying 1, then whatever the other player adds (1, 2 or 3), add enough to make 4 (3, 2 or 1) and say 5, 9, 13, 17 and 21.

The first player wins by saying 1, 5, 9, 13, 17, 21 (always making the two turns add up to 4).

Now the "key" numbers go down in steps of 11 (since 1 + 10 = 11): 99, 88, 77, 66, 55, 44, 33, 22, 11. The first player must start with a number from 1 to 10, so they cannot say 11. The second player can always win: whatever the first player says, add enough to reach 11, then always make the two turns add up to 11: 22, 33, 44, …, 99.

The second player wins by saying 11, 22, 33, …, 88, 99.

At present 62,871 (in the centre) is bigger than all four of its neighbours. For 4 supercells, the four middle-edge cells (39,344, 23,609, 45,306 and 50,319) must become supercells, so the centre must become smaller than 23,609. Swapping two digits of 62,871 so that it starts with 1: swap the 6 and the 1 to get 12,876.

Now 39,344, 23,609, 45,306 and 50,319 are each bigger than all their neighbours, and the corners are not.

Swap the 6 and the 1 in 62,871 to make 12,876; then 39,344, 23,609, 45,306 and 50,319 are supercells.

Example for 2014:

So 2014 takes 7 rounds. (Every 4-digit number with at least two different digits reaches 6174 in at most 7 rounds.) Try it for your own year of birth.

For example, 2014 takes 7 rounds. No year needs more than 7 rounds.

The odd digits are 1, 3, 5, 7 and 9.

• Largest: the first digit can be at most 7, and then the second digit must keep the number below 75,000, so it is 3 (the largest odd digit below 5): 73,999.
• Smallest: the number must be more than 35,000: 35,111.
• Closest to 50,000: the smallest member above 50,000 is 51,111 (1,111 away); the largest below 50,000 is 39,999 (10,001 away). So 51,111 is the closest.

Largest: 73,999; smallest: 35,111; closest to 50,000: 51,111.

Estimate: Sundays 52 + about 15 festival holidays + summer vacation about 45 days + winter/Diwali breaks about 15 days (some of these days are already Sundays). Total ≈ 110–130 days (more if your school has second Saturdays or all Saturdays off). Then count exactly using your school calendar.

Roughly 110–130 days (Sundays, festivals and vacations); check with your school calendar.

• Mug: about 0.5 to 1 litre.
• Bucket: about 15 to 20 litres.
• Overhead tank (home): about 500 to 2000 litres.

Mug about 1 litre, bucket about 15–20 litres, overhead tank about 500–2000 litres.

$$18,070+300+300=18,670$$

Other examples: 17,870 + 400 + 400; 18,000 + 335 + 335; 16,670 + 999 + 1,001 is not allowed (1,001 has 4 digits).

For example, 18,070 + 300 + 300 = 18,670.

Example: 300. A 3 × 4 grid of 25s: 12 × 25 = 300.

Another: 240 as a ring: one 40 in the centre and eight 25s around it (40 + 8 × 25 = 240).

For example, a 3 × 4 grid of 25s makes 300; 40 in the centre with eight 25s around it makes 240.

Every power of 2 (except 1) is even, so the rule says "halve it", and half of a power of 2 is the previous power of 2: 64 → 32 → 16 → 8 → 4 → 2 → 1. We never get an odd number until we reach 1, so the sequence always comes down to 1.

Halving a power of 2 gives the previous power of 2, so the numbers keep halving and reach 1 without ever being odd.

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, it reaches 1 (in 25 steps).

Yes: 100 reaches 1 after 25 steps.

Working back from 22 in steps of 4, the winning numbers are 22, 18, 14, 10, 6, 2. The first player should start by adding 2 (saying 2). Then, whatever the other player adds (1, 2 or 3), add 3, 2 or 1 so that each pair of turns adds 4: say 6, 10, 14, 18 and finally 22.

The first player wins: say 2, then 6, 10, 14, 18, 22 (making each pair of turns add up to 4).