Cambridge IGCSE International Mathematics 0607/62 is the May/June 2026 Paper 6 (Investigation and Modelling, Extended), variant 2. It lasts 1 hour 30 minutes, is worth 50 marks and is sat by every Extended candidate, alongside Papers 2 and 4. It had two sections of 25 marks each: an investigation into the game of Nim and a modelling task on regular investments with compound interest, with 45 minutes advised for each. This guide goes through the paper question by question and shows what the official mark scheme gave marks for, including the communication marks. The paper and its mark scheme are free on the Tutopiya Learning Portal.

For an overview of all six 0607 papers from this series, see IGCSE International Maths 0607 May/June 2026 past papers: what came up and how it was marked.

Section A used the same Nim investigation as Paper 5 (Core). The 0607 Paper 52 guide shows the Core version, which has more game tables and is good warm-up practice.

0607/62 at a glance

Section	Question	Topic	Marks
A: Investigation (Nim)	1	Taking 1 or 2 sticks: game tables for 6 and 18 sticks, multiples of 3	6
	2	Sequences and nth terms for up to k sticks, the values of k for 30 sticks	7
	3	You start: why you take 1 stick from 19	1
	4	You start, taking 1, 2 or 3: nth term, moves to win, formula for m in terms of p	7
	5	50 sticks: first move and total moves for k = 2 to 6	4
B: Modelling (regular investment)	6	Compound interest on a single $500 deposit, years to reach $1000	5
	7	Two yearly deposits of $300	3
	8	Five yearly deposits: building and proving the geometric-series model, finding P	8
	9	Applying the general model	2
	10	Monthly deposits: adapting the model, sketching T, a linear model	7
		Total	50

How 0607/62 is marked

The cover says marks are given for full reasons, examples and steps that communicate your mathematics clearly. These are the C marks (communication marks), shown as C1 or C2. There were 12 of them: 4 in Section A and 8 in Section B, and in Section B almost every calculation question had them. A correct final answer with no working loses them.

The rest are answer marks and B marks (marks for a correct answer or statement, given independently). The scheme also uses FT (follow through: credit for correct work based on your own earlier answer), oe (or equivalent), SC (special case: a mark for a specific partial answer named in the scheme), soi (seen or implied), nfww (not from wrong working) and dependent (a mark that can only be given if an earlier mark was earned). There are no M (method) or A (accuracy) labels on this paper; the method credit comes through the C marks.

For the whole modelling section, the scheme allowed 4% to be written as 4/100 and 1.04 as 1 + 4/100 or 104/100, and condoned missing or wrong dollar signs and a missing final bracket.

Section A: the Nim investigation (25 marks)

Two players take turns removing sticks from a pile, and whoever takes the last stick wins. In Questions 1 and 2 your friend moves first; in Questions 3 to 5 you do.

Question 1: taking 1 or 2 sticks (6 marks)

• 6 sticks, two ways to win (2 marks, B1 for each table): friend 1, you 2, friend 1, you 2 (piles 5, 3, 2, 0); or friend 2, you 1, friend 2, you 1 (piles 4, 3, 1, 0).
• Why you win from 9 on your friend’s move (1 mark): 9 is a multiple of 3, or after two moves there will be 6 left. “Divisible by 3” and “3 is a factor of 9” were accepted; “a sequence in threes” was not.
• Sticks to take from 20 on your move (1 mark): 2, leaving 18.
• The 18-stick table (2 marks): your moves are 2, 1, 1, 2, 1, 2, leaving 15, 12, 9, 6, 3 and 0 after each of your turns. B1 for the piles 3, 6 and 15, or for at most 4 incorrect cells.

Question 2: taking up to k sticks (7 marks)

• Sequences and nth terms (3 marks): 4, 8, 12, 16 (4n); 5, 10, 15, 20 (5n); 6, 12, 18, 24 (6n); and (k + 1)n for taking up to k. The first row, 3, 6, 9, 12, has nth term 3n. B1 for the correct sequences, B1 for the first four nth terms and B1 for the general one. k + 1 × n without brackets was not accepted, and nor was expanding it to something wrong.
• All values of k for 30 sticks (4 marks): 1, 2, 4, 5, 9, 14 and 29. Your friend starts and you win, so 30 must be a multiple of k + 1: k + 1 is a factor of 30. The list earned 2 marks (one value could be missing; B1 for four correct ignoring extras; SC1 for 1 ≤ k ≤ 29 if nothing else scored). Unlike Paper 5, the question didn’t tell you how many values there were.
  • The working earned C2: listing the factors of 30 (1, 2, 3, 5, 6, 10, 15, 30) with no extras, or four correct equations such as 2n = 30 or 5(k + 1) = 30, or a general form such as n = 30 ÷ (k + 1) or k = 30 ÷ n − 1 with two worked examples. C1 was for four correct factors (at most four extras), one correct equation solved, or the general form with one example. 2n = 30 and 2(k + 1) = 30 counted as the same equation, and k + 1 = 30 didn’t count at all.

Question 3: you start, 19 sticks (1 mark)

• Why take 1 stick first (1 mark): it leaves 18, a multiple of 3. “18 is a multiple of 6”, 6 × 3 = 18, and listing the piles after your turns (18, 15, 12, 9, 6, 3, 0) were accepted.

Question 4: you start, taking 1, 2 or 3 (7 marks)

You take 1 stick first, so the possible starting piles are 1, 5, 9, 13, 17 and so on.

• The nth term (2 marks): 4n − 3. B1 for kn − 3 with k ≠ 0 or for 4n + j. Tn = 4n − 3 was accepted, but n = 4n − 3 scored only B1.
• Total moves to win (3 marks): for 1, 5, 9, 13 and 17 sticks the moves are 1, 3, 5, 7 and 9. The answers earned 2 marks (the scheme gives B1 for 5, 7 or 9). C1 was for playing out a game from 9, 13 or 17 in the table with your turns leaving 8, 4, 0 (or 12, 8, 4, 0, or 16, 12, 8, 4, 0).
• A formula for m in terms of p (2 marks): m = (p + 1) ÷ 2 as the final answer. B1 for (p + 1) ÷ 2 seen, for 2m = p + 1, for m = 2n − 1, or for spotting differences of 4 for p and 2 for m. m = 2p − 1 scored nothing.

Question 5: 50 sticks, k from 2 to 6 (4 marks)

You start with 50 sticks and can take up to k, where 1 < k < 7, so k is 2, 3, 4, 5 or 6. To be sure of winning you must leave a multiple of k + 1 after your first move.

k	Sticks you take first	Total moves to win
2	2 (leaves 48, a multiple of 3)	33
3	2 (leaves 48, a multiple of 4)	25
4	X (50 is a multiple of 5, so you can’t be sure)	–
5	2 (leaves 48, a multiple of 6)	17
6	1 (leaves 49, a multiple of 7)	15

• The table (3 marks): B2 for one row correct (other than k = 4) or the whole middle column correct; B1 for one correct number in either column.
• The reason (C1): one row’s link to k, such as “48 is a multiple of 3” for k = 2, or the rule that the total moves are twice the pile after your first move divided by k + 1, plus 1. For k = 2 that is 2 × 48 ÷ 3 + 1 = 33. The link with k had to be clear.

Section B: modelling regular investment (25 marks)

The section gave the compound interest formula A = P × rⁿ with r = 1 + R/100 and a worked example, then built up to a model for saving the same amount every year or month.

Question 6: Rosie’s single deposit of $500 at 4% (5 marks)

• After 5 years (2 marks): 500 × 1.04⁵ = $608.33 (1 mark), with C1 for writing 500 × 1.04⁵. $608.32, 608.3 and 608 were also allowed.
• Complete years to reach $1000 (3 marks): 18 (1 mark). The C2 went to one of three complete methods: the inequality 500 × 1.04ⁿ > 1000 (or 1.04ⁿ > 2) together with n log 1.04 > log 2 or log 2 ÷ log 1.04; a sketch of 500 × 1.04ⁿ crossing a horizontal line; or trials at 17 and 18 years showing $973.95 and $1012.91 (or 1.9479 and 2.0258). C1 was for the inequality or the log step alone, the sketch alone, or two trials that weren’t 17 and 18 with the right answer. A sketch with a decreasing gradient got nothing.

Question 7: Abdul’s two deposits of $300 (3 marks)

• Total after 2 years: the first $300 earns interest for 2 years and the second for 1 year: 300 × 1.04² + 300 × 1.04 = 324.48 + 312 = $636.48 (1 mark; 636.5 or 636 accepted). C2 for that full calculation, or for 300 × 1.04 = 312 then 612 × 1.04. C1 for just one of 300 × 1.04, 300 × 1.04², 300 × 1.04 + 300 = 612 or 612 × 1.04.

Question 8: Joe’s five yearly deposits (8 marks)

• Completing the two lines (2 marks, B1 each): the missing first term of T is P × 1.04⁵, and the missing terms of 1.04T are P × 1.04⁶ and P × 1.04⁵.
• Show that T = 1.04P(1.04⁵ − 1) ÷ (1.04 − 1) (3 marks): subtracting the lines cancels the middle terms, so 1.04T − T = P × 1.04⁶ − P × 1.04 (1 mark, FT from your 8(a)). Then factorise: T(1.04 − 1) on the left (1 mark, dependent on the first) and P × 1.04(1.04⁵ − 1) on the right, with both factorisations leading to the given result (1 mark, dependent on the first). Writing 0.04T and then replacing 0.04 with 1.04 − 1 was allowed. Working backwards from the given answer did not earn the first mark, and if you remove brackets the negative signs must be visible.
• Yearly deposit for $700 (3 marks): set 700 equal to the model and rearrange: P = 700 × 0.04 ÷ (1.04 × (1.04⁵ − 1)) = 28 ÷ 0.2253… = $124.27. The answer mark accepted 125 or 124.27 (and 124.3). C2 for the equation and the rearrangement, a sketch with a horizontal line at 700, or trials such as P = 124 giving $698.49 and P = 125 giving $704.12. C1 for the equation alone, the sketch alone, or a table with at least two correct values above P = 100.

Question 9: Maya’s general model (2 marks)

• $250 a year for 9 years at 5%: T = 250 × 1.05 × (1.05⁹ − 1) ÷ (1.05 − 1) = $2894.47 (1 mark; 2890 or 2894.5 accepted), with C1 for the substitution. 0.05 on the denominator was fine.

Question 10: Esa’s monthly deposits (7 marks)

Esa saves $50 a month for 15 months at R% a month.

• Show that T = (5000 ÷ R + 50)((1 + R/100)¹⁵ − 1) (3 marks): substitute P = 50, r = 1 + R/100 and n = 15 into the Question 9 model, with r − 1 = R/100 on the denominator (1 mark). Multiply by 100 ÷ R (1 mark, nfww), so 50(1 + R/100) ÷ (R/100) becomes 5000 ÷ R + 50. Reach the given form with no errors for the last mark. Missing brackets in the working counted as an error.
• Sketch for R between 0 and 0.8 (2 marks): an approximately straight line through (0, 750) and about (0.8, 800). B1 for a straight line with positive gradient through only one of those points, or for a visibly curved line through both. At R close to 0 there is almost no interest, so T is close to 15 × 50 = 750.
• A linear model (2 marks): T = mR + 750 with 60.3 ≤ m ≤ 64.4 (125/2 and the integers 61 to 64 were allowed). B1 for the right form with only one of the gradient or the intercept in range, or for the right numbers with other letters (y = mx + c). x = my + c was not accepted, and if you use T and R they must be the right way round.

Where the marks are on 0607/62

Section A (25 marks):

• Sequences and nth terms: 5 marks (2(a) and 4(a)).
• Values of k for 30 sticks: 4 marks (2(b)), half of them for the working.
• Counting moves: 5 marks (4(b) and the formula in 4(c)).
• The 50-stick table: 4 marks (Question 5).
• Game tables: 4 marks (1(a) and 1(c)).
• Short reasons and next moves: 3 marks (1(b)(i), 1(b)(ii) and Question 3).

Section B (25 marks):

• Compound interest on one deposit: 5 marks (Question 6), including the logarithm or trial method for 18 years.
• Regular deposits by hand: 3 marks (Question 7).
• Building and proving the series model: 5 marks (8(a) and 8(b)).
• Using the model: 5 marks (8(c) and Question 9).
• Adapting the model to months: 3 marks (10(a)).
• Sketching and a linear model: 4 marks (10(b)).

The longest single answers were 2(b) and Question 5 (4 marks each) and the two 3-mark “show that” proofs, 8(b) and 10(a). 12 of the 50 marks were C marks.

How to self-mark 0607/62 with the mark scheme

1. Separate answer marks from C marks. In 6(b), 18 years with no inequality, sketch or trials scores 1 out of 3. Check your working against the C-mark methods listed.
2. For trials, check you tested both sides of the answer. The full C2 in 6(b) needed 17 and 18 years; in 8(c), 124 and 125.
3. Mark the “show that” questions line by line. In 8(b) and 10(a), each step is a mark and later marks depend on earlier ones. If your last line only matches because you worked backwards, or a bracket went missing, don’t give yourself the mark.
4. Use follow through properly. The first mark in 8(b) follows through from your own 8(a), so a slip there doesn’t have to cost you twice.
5. Check the form of formulas. n = 4n − 3 and m = 2p − 1 lose marks; T = mR + 750 needs a gradient between 60.3 and 64.4.
6. Log each lost mark as missing working, a wrong rearrangement, a list error or a sketch slip, and practise in that order.

Sit 0607/62 as a timed mock

The paper allows 90 minutes for 50 marks and advises no more than 45 minutes on each section. A workable plan:

Block	Marks	Time
Question 1	6	10 minutes
Question 2	7	12 minutes
Question 3	1	3 minutes
Question 4	7	10 minutes
Question 5	4	8 minutes
Check Section A	–	2 minutes
Question 6	5	8 minutes
Question 7	3	5 minutes
Question 8	8	13 minutes
Question 9	2	4 minutes
Question 10	7	12 minutes
Check Section B	–	3 minutes

Use your graphic display calculator for the sketches in 6(b), 8(c) and 10(b), and for the trial values.

Other May/June 2026 0607 papers

• 0607 Paper 12 (Core Non-calculator)
• 0607 Paper 22 (Extended Non-calculator)
• 0607 Paper 32 (Core Calculator)
• 0607 Paper 42 (Extended Calculator)
• 0607 Paper 52 (Core Investigation)

Practise and download

• Download 0607/62 and its mark scheme on the Learning Portal →
• All 0607 May/June 2026 papers: overview and marking patterns
• Try a free IGCSE International Mathematics quiz

Need help with a 0607 topic? Book a free trial lesson with an IGCSE International Mathematics tutor →

Frequently asked questions

What was on 0607 Paper 62 May/June 2026? +

Two sections worth 25 marks each. Section A was an investigation into the game of Nim: winning piles, nth terms, the values of k for 30 sticks, a formula for the number of moves and a table for 50 sticks. Section B was a modelling task on investing money regularly in an account paying compound interest.

How many marks is 0607/62 and how long is it? +

It is out of 50 marks and lasts 1 hour 30 minutes. The paper advises spending no more than 45 minutes on each section.

What were the answers to the compound interest questions on 0607/62? +

$500 at 4% grows to $608.33 after 5 years and first reaches $1000 after 18 complete years. Abdul's two deposits of $300 grow to $636.48. Joe needs to invest $124.27 a year (125 was also accepted) to reach $700, and Maya's model gives $2894.47.

What was the linear model in 0607/62 Question 10? +

T = mR + 750, with a gradient m between 60.3 and 64.4. The graph of the monthly investment model is very close to a straight line from (0, 750) to about (0.8, 800) for interest rates between 0 and 0.8% a month.

What are the C marks on 0607 Paper 6? +

Communication marks for showing your method: listing the factors you used, writing the inequality you solved, setting up the equation with the model, or giving trials either side of the answer. There were 12 on 0607/62, 4 in Section A and 8 in Section B.

Where can I download 0607/62 May/June 2026 and its mark scheme? +

Both are free to download and print on the Tutopiya Learning Portal, on the Cambridge IGCSE International Mathematics 0607 past papers page.

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