Notes

M1: Finds the gcd of 36 and 21 or in some way considers the hcf of these two numbers.

A1: States that the gcd is 3 and does not divide into 11 therefore no solutions

Notes

M1: Identifies \(36y\) as \(kay\) and 192 as \(kb\) (may be implied by cancelling).

M1: Finds the gcd of \(k\) and 40. Then divides through by the gcd of \(k\) and 40, Modulus must be reduced to 10.

dM1: Completes the solution by appropriate means. There will be many variations on how they can do this. Accept just \(y = 2\) as an attempt to find the solution. Note the multiplicative inverse of 9 modulo 10 is 9. If a multiplicative inverse is attempted, it must be from a correct method.

A1: Deduces the correct solution

Alternative I

M1: Reduces RHS modulo 40 to give \(36y \equiv 32 \pmod{40}\)

M1: Finds the gcd of 36 and 40 and divides through by 4. This must include the modulus being reduced to 10.

dM1: Depends on previous M. Finds a value for \(y\) by any appropriate means. There will be many variations on how they can do this. Accept just \(y = 2\) as an attempt to find the solution.

A1: Deduces the correct solution

Note: Candidates to start by finding the hcf of 36 and 40 and divide through to get \(9y \equiv 48 \pmod{10}\) will score M2 at the start of the solution.

(ii) Alt II

M1: Writes the congruence as a linear equation with \(40k\)

M1: Divides the equation through by 4

dM1: Completes the solution by appropriate means. One example shown in scheme.

A1: Deduces the correct solution.

Notes

M1: Any two out of 26 or \(10 \times 9 \times 8\) or 4 seen multiplied together

A1: Correct answer

Notes

B1ft: For the correct answer or follow through their answer to (a) multiplied by 20. Note they may start again and use \(26 \times {}^{10}C_3 \times 4 \times 5!\)

Notes

B1: Recalls Fermat’s Little Theorem correctly. May be implicit in the working rather than stated separately. Accept if stated in terms of \(a\) and \(p\).

M1: A complete attempt to write \(4^{50}\) as powers of \(4^{12}\) or \(4^{13}\) and completes the process to find the residue.

A1: Deduces the correct residue.

Notes

B1: Deduces that there are 5 different 1-digit odd numbers that do not contain the digit 2. Variation: if they consider 000 to 999 allow B1 for correct option of 9 for first digit. (corrected from the printed mark scheme: the printed note says 3 different 1-digit odd numbers, but there are 5 (1, 3, 5, 7, 9), as in the scheme)

M1: Finds the number of different 2-digit and 3-digit odd numbers that do not contain the digit 2. (In the variation they would do \(9 \times 9 \times 5\) possibilities without 2, giving 405 directly.)

dM1: For 500 – the sum of the number of 1, 2, 3 digits that don’t contain the digit 2. Must have been full method to find all the digits without 2 first.

A1: Correct answer

Alternative 1

B1: Deduces that there are only two positions for 2, first and second. (May be implied)

M1: Considers first digit is 2 and the possible digits for second and third position and considers second digit is 2 and the possible digits for first and third position. Allow if they have double counted some digits for this mark.

dM1: Full method considering the possibilities for 2 in various positions and adds the results. They must subtract any repeats for this mark.

A1: Correct answer

Alternative 2

B1: Deduces that there are no single digit numbers

M1: Considers the possible 2 and 3 digit numbers. Allow if they have double counted some digits for this mark or omit some cases as long as at least some cases are considered.

dM1: Adds up all the possible digits following consideration of all possible combinations. They must subtract any repeats for this mark. For 3 digit numbers they may have \(1 \times 10 \times 5\) starting 2, \(9 \times 1 \times 5\) with 2 in middle, but need to note 5 of these are repeated (22x), and so get \(5 + 50 + 45 - 5 = \ldots\)

A1: Correct answer.

Other Alternatives are possible. E.g.

B1: Considers 3 digit numbers from 000 to 999. (May be implicit)

M1: Starting with 2 : 50 possibilities (must be odd), middle 2 but not starting 2 : \(9 \times 1 \times 5\) possibilities.

dM1: So total number of possibilities is 50 + 45 = …

A1: Correct answer.

Notes

M1: Applies the divisibility test by summing the digits.

A1: Correct solution and explanation.

Notes

M1: Makes a start with the table, achieving at least 7 correct entries.

A1: At least 18 correct entries.

A1: Completely correct table.

Notes

B1: Cao

Notes

B1: Cao

Notes

B1: any of the three correct subgroups.

Notes

M1: Starts the Euclidean algorithm to obtain \(105 = p \times 24 + q\)

M1: Completes the algorithm correctly to obtain a zero remainder.

A1: All correct and concludes \(h = 3\)

Notes

M1: Begins the progress of back substitution.

M1: Completes the process.

A1: Correct expression or correct values.

Notes

M1: Attempts the remainder when 179 is divided by 11 and makes progress in establishing the remainder when \(179^5\) is divided by 11

A1: Obtains the remainder 1

Notes

M1: Starts the process of using the algorithm, with attempt at \(234 = p \times 96 + q\).

M1: Continues the process until remainder zero is reached.

A1: Deduces the correct highest common factor from correct work.

Notes

M1: Begins the process of back substitution by rearranging their equation with least positive remainder.

dM1: Completes the process.

A1: Correct expression or values of \(a\) and \(b\) identified.

Notes

B1: Deduces the correct equivalent congruence.

M1: Chooses a suitable strategy to solve the reduced congruence. May use multiplicative inverse as per scheme or see alts for some variations, must be using their value for \(b\)

M1: Applies their multiplicative inverse to reach \(x \equiv \ldots \pmod{39}\)

In the alternative it is for reaching \(x \equiv \ldots \pmod{39}\) from a succession of multiples.

A1: For 15 as a solution, need not have the \((\text{mod } 39)\). (Accept one correct solution if an alternative method is used.

A1: For stating the solution is \(15 \pmod{39}\) as the answer (not just in working) or for listing all the solutions modulo 234. Either way of expressing the answer is fine (single answer mod 39, or all 6 mod 234), but must realise it is more than just the 15, so do not accept just 15 with no indication of the modulus being considered.

ISW if they achieve the correct answer and then try to list solutions

SPECIAL CASE: B1 For one correct value of \(x\) found

(c) Way 2

B1: Deduces the multiplicative inverse of 96

M1: Multiplies through by the multiplicative inverse of 96

M1: For using reaching \(x \equiv \ldots \pmod{39}\).

A1: For 15 as a solution, need not have the \((\text{mod } 39)\). (Accept one correct solution if an alternative method is used.

A1: For stating the solution is \(15 \pmod{39}\) as the answer (not just in working) or listing all the solutions modulo 234. As per main scheme do not accept just 15 with no indication of the modulus being considered.

(corrected from the printed mark scheme: the first line of this method says “modulo 239”; the modulus is 234)

(c) Way 3

B1: Deduces the correct equivalent congruence.

M1: Tries multiplying by various numbers in an attempt to reduce the coefficient to 1. One example shown above but others are possible. Allow the M for an attempt at starting such a process.

M1: For using reaching \(x \equiv \ldots \pmod{39}\) from a succession of multiples.

A1: For 15 as a solution, need not have the \((\text{mod } 39)\). (Accept one correct solution if an alternative method is used.

A1: For stating the solution is \(15 \pmod{39}\) as the answer (not just in working) or listing all the solutions modulo 234. As per main scheme do not accept just 15 with no indication of the modulus being considered.

(c) Way 4

B1: Deduces the correct equivalent congruence.

M1: Works out the possibilities for \(16x\) to try and find one that is divisible by 16. Look for the first few evaluated.

M1: For reaching a value for \(16x\) that is a multiple of 16.

A1: For 15 as a solution, need not have the \((\text{mod } 39)\). (Accept one correct solution if an alternative method is used.

A1: For stating the solution is \(15 \pmod{39}\) as the answer (not just in working) or listing all the solutions modulo 234. As per main scheme do not accept just 15 with no indication of the modulus being considered.

NB For students who try dividing though by 12 to get \(8x \equiv 3 \pmod{234}\) a maximum of B0M1M0A0A0 is possible for attempt to reach \(x \equiv \ldots \pmod{234}\) via valid method.

Notes

B1: Correct congruence expression.

Notes

B1: Forms the equation in \(x\) and \(y\) from the given information.

M1: Substitutes for \(x\) or \(x + y\) in the congruence expression to form an expression in \(y\) only. Alternatively, they many eliminate \(y\) to get an equation in \(x\) only.

A1: Correct equation in \(y\) (or \(x\)) only.

M1: Full method to reach a residue for \(y\). If \(y\) was eliminated from the congruence they must use \(x + y = 43\) again to form an appropriate expression for \(y\) here.

A1: Interprets the situation correctly to give 16 for the minimum value for \(y\).

Alternative

B1: Forms the equation in \(x\) and \(y\) from the given information.

M1: uses their congruence express, sets = 0 and forms an equation in \(x\), \(y\) and a third variable. Solve simultaneous equations to find \(y\) in terms of the third variable.

A1: Correct equation for \(y\)

M1: Uses value for the third variable to find a value for \(y\)

A1: Interprets the situation correctly to give 16 for the minimum value for \(y\).

Notes

B1: Accurately recalls Fermat’s Little Theorem. Allow if stated correctly in general form, \(a^{p-1} \equiv 1 \pmod{p}\). May be implied.

M1: Attempts to write the index in a suitable form and then applies the theorem, look for \(80 = 22k + r\) or equivalent leading to a reduced index.

dM1: For a full method to reduce to a least residue. There are many ways that they can split up their value, including use of a calculator to work with a higher power as long as an attempt at FLT to reduce initially has been seen. Condone a sign slip e.g using \(21^2 \equiv 19 \pmod{23}\) instead of \(21^2 \equiv -19 \pmod{23}\) as long as there is a complete method and there is only one sign slip

A1: Correct answer.

Notes

(i)(a)

M1: \(5^4 \times \ldots\) or \(\ldots \times 4 \times 3 \times 2\) or \(\ldots \times {}^{4}P_{3}\)

A1: For 15000

(b)

M1: Correct strategy to find the new number of codes seen or implied.

A1ft: Increase in codes is 135000. Follow through \(9 \times\) their 15000

Notes

(ii)(a)

M1: Full strategy to deduce the correct set of digits.

A1: Digits are 1,3, 5 and 9

Writes down the correct values scores M1A1

(b)

M1: Checks which of the two possible combinations satisfies the final property.

A1: Combination is 1359

Writes down the correct code scores M1A1

Notes

B1: States \(214 \equiv 6 \pmod{8}\)

M1: Uses modulo arithmetic to find \((214)^6 \equiv a \pmod{8}\)

A1: Achieves \((214)^6 \equiv 0 \pmod{8}\) from correct modulo arithmetic and draws the conclusion that \(214^6\) is divisible by 8

Notes

M1: Uses the division rule for 11 to find a value for \(a\).

A1: \(a = 1\)

Notes

M1: Uses the division rule for 3 to find a value for \(b\).

A1: Any two correct values for \(b\).

A1: All four correct values for \(b\).

Notes

M1: Attempt at the Euclidean algorithm to find the gcd of 168 and 66, condone a numerical slip.

A1*: A fully correct application of the Euclidean algorithm and explains that the highest common factor is 6.

Notes

M1: Attempts to apply back substitution to find suitable values.

A1: Correct expression in just the numbers 168 and 66 reached (all remainders eliminated). Need not be fully simplified.

A1: Deduces the correct values of \(a\) and \(b\). Allow if only seen in the expression.

Notes

B1: Correct explanation. Must refer to 10 not being a multiple of 6.

Notes

B1: Uses the result in (b) and divides through by 6 to form a correct statement between 11 and 28. Alternatively uses Euclidean algorithm for 28 and 11 to form an equation for 1 as multiples of 28 and 11. Allow this mark for a correct unsimplified equation.

M1: Multiplies through in their equation by 8 or the original equation through by \(-5\) (or other multiplicative inverse).

A1: Deduces the correct solutions. Accept either \(-40 \pmod{28}\) or \(16 \pmod{28}\), or any other correct congruence.

Alternative 1

Note the order of marks has changed

M1: forms an equation for 8 in terms of multiples of 28 and 11

B1: Correct equation, allow this mark for a correct unsimplified equation.

A1: Correct solution

Alternative 2

B1: Correct equation

M1: Uses Fermat’s little theorem

A1: Correct solution

Trial and error approach

(d) – likely to score B0M1A1.

B1: Gives some kind of proof that the solution is unique. E.g. by reference to hcf(11,28) = 1 or trial of every possible value.

M1: Applies trial and improvement method to find one solution.

A1: Correct solution found.

Correct answer stated with no supportive working scores B0M1A1

Notes

B1: See scheme

Notes

B1: See scheme

Notes

M1: Uses the divisibility rule for 11 and the restrictions on the values \(a\), \(b\) and \(c\) leading to \(a - b + c = 0\) or 11 only

A1*: Uses the information that the sum of the digits is even and that \(a - b + c = 0\) or 11 from the divisibility rule for 11 to show that \(a - b + c = 0\) cso

Notes

B1: States \(a + b + c \equiv 8 \bmod 9\) or \(a + b + c + 1 \equiv 0 \bmod 9\)

M1: Uses their equations to find a value for \(a + c = \ldots\) or a value for \(b\)

M1: Uses their values to find at least two correct values for \(N\)

A1: Deduces all four correct values of \(N\) and no extra values

Trial and error

B1: For identifying \(a + c = b\)

M1: For one correct value, this implies B1

M1: For two correct values

A1: Deduces all four correct values of \(N\) and no extra values

Alternative

B1: States \((N) = 11n \equiv 8 \bmod 9\)

M1: Finds the multiplicative inverse of 11 and uses this to find an expression for \(N\)

M1: Uses their expression for \(N\) to find at least two values for \(N\)

A1: Deduces all four correct values of \(N\) and no extra values

Notes

M1: Complete attempt at the Euclidean algorithm to find the gcd (hcf) of 124 and 17, condone a numerical slip.

A1: A fully correct application of the Euclidean algorithm and draws the conclusion that since the gcd/hcf = 1 therefore 124 and 17 are relatively prime (coprime).

Notes

M1: Attempts to find the Bezout’s identity, condone sign slips

A1: Correct Bezout’s identity.

A1: Deduces a set of correct values of \(x\) and \(y\).

Alternatively

M1: Uses the solution to (a) to find an identity of the form \(a \times 124 + b \times 17 = c\)

A1: Correct identity

A1: Deduces a set of correct values of \(x\) and \(y\).

Alternative

Alternative

Notes

M1: A complete method to reach using modulo arithmetic to achieve \(x \equiv \ldots \bmod 17\)

For example

Multiplies through by their multiplicative inverse of 124 and reaches \(x \equiv \ldots \bmod 17\). Bezout’s identity used in part (b) follow through on their multiple of 124

A1: \(x \equiv 8 \bmod 17\) o.e.

Notes

M1: Starts the process of using the algorithm, with attempt at \(416 = p \times 72 + q\).

M1: Continues the process until remainder zero is reached.

A1: Deduces the correct highest common factor from correct work.

Notes

M1: Begins the process of back substitution by rearranging their equation with least positive remainder.

M1: Completes the process.

A1: Correct expression or values of \(a\) and \(b\) identified.

Notes

M1: Uses the answer to (b) and reduces modulo 416 to reach −8 as a congruent number. Or a full process to multiply \(23 \times 72 = 1656\) and reduce by subtracting multiples of 416

A1: For 408.

Notes

M1: Uses modulo 13 to reduce the equation in some way to reduce a power of 5, e.g. using \(5^2 = 25 \equiv -1\) or 12. There will be lots of approaches that can be used here (e.g. \(5^3 \equiv 125 \equiv -5 \pmod{13}\) is another possibility.

Note the question instructs calculator solutions are not acceptable so finding \(5^{10}\) itself is not acceptable.

dM1: Completes to a smallest (positive or negative) residue – allow if slips, but look for reducing to a number in \(\{-6, \ldots 6\}\). Again other routes than the one shown are possible. E.g.
\(5^{10} \equiv (5^3)^3 \times 5 \equiv (-5)^3 \times 5 \equiv -125 \times 5 \equiv 5 \times 5 \equiv 25 \equiv 12 \pmod{13}\)

A1: For 12.

Notes

M1: Sets up the proof by stating or implying that if there is a subgroup of order 11 then by Lagrange’s Theorem 11 must divide 5291848. May not mention Lagrange’s Theorem at this stage. A formal assumption is not required as long as it is implicit.

M1: Applies the divisibility test for 11. Look for an attempt at the alternating sum being used.

A1: Alternating sum is 23, so derives a contradiction as 11 does not divide \(|G|\), and conclusion made. Use of Lagrange’s Theorem must be clear, though it need not be named.

Notes

M1: Begins process of completing the table by filling in at least one row or column correctly.

A1: Five or more rows or columns completed correctly.

A1: Completely correct table.

(ii)(a) and (b) are marked together: (6) in total.

Notes

B1: Identifies 16 as the identity element. No reason needed.

B1: Identifies all inverses or gives reason why each element has an inverse (may refer to each row and column containing the identity once only and symmetrically about the diagonal).

B1: Refers to closure and associativity to deduce \((X, \times_{30})\) is a group.

SC Allow B0B0B1ft for deducing not a group with valid reason if identity or inverse checks fail.

Notes

M1: For an attempt to apply a correct Fermat’s Little theorem at least once in the question with either \(p = 11\), \(p = 7\) or \(p = 3\) on either the \(46^{46}\) or \(47^{47}\) term.

M1: Applies FLT and congruence arithmetic fully to find the residue of \(46^{46} + 47^{47}\) modulo 11. There will be lots of different routes, so look for an attempt to apply FLT that leads to determining if 11 is a divisor or not.

A1: \(46^{46} + 47^{47} \equiv 7 \pmod{11}\) (accept equivalents as long as it is clear it is not congruent to 0) and deduces it is not a possible order for a subgroup.

Notes

M1: Applies checks for both 7 and 3 as divisors of \(46^{46} + 47^{47}\) via similar strategy.

M1: Applies FLT with \(p = 3\) to find a smaller residue modulo 3. Other routes are possible.

M1: Applies FLT with \(p = 7\) to find a smaller residue modulo 7. Other routes are possible.

A1: Shows \(46^{46} + 47^{47}\) congruent to 0 modulo 3 and modulo 7, and deduces 21 divides \(46^{46} + 47^{47}\) hence it is a possible order for a subgroup.

Alt:

M1: Reduces the bases modulo 21 and applies a power reduction technique using congruences for at least one of the power of 46 or 47

M1: Reduces fully by congruence arithmetic either the \(46^{46}\) or \(47^{47}\) term.

M1: Reduces fully by congruence arithmetic both the \(46^{46}\) and \(47^{47}\) terms

A1: Shows \(46^{46} + 47^{47}\) congruent to 0 modulo 21, and deduces 21 divides \(46^{46} + 47^{47}\) hence it is a possible order for a subgroup.

Notes

M1: Begins the process of applying the Euclidean algorithm, with attempt at the first two steps. Allow slips.

M1: Completes the process to the stage shown – if errors have been made at least three steps should have been made in reaching their final line (ending +1) to score this mark.

A1: Algorithm correctly carried out – as shown.

M1: Starts the process of back substitution – at least two substitutions made.

A1: Completes the process and finds the correct values for \(a\) and \(b\).

Notes

B1ft: Deduces correct multiplicative inverse. Accept 23 or anything congruent to 23 modulo 125 or follow through their \(b\).

Notes

M1: Multiplies 16 by their multiplicative inverse or any other full method to proceed to the solution, e.g. multiplying the identity found in (a) through by 16 and reducing modulo 125.

A1: \(x \equiv 118 \pmod{125}\). Accept 118 or anything congruent to 118 modulo 125 as long as it is part of a correct modulo statement. (Do not accept just 368 on its own.)

Notes

M1: Identifies one of the odd prime factors of 66 and proceeds to check divisibility for it.

A1: Correct method and deduction for either divisibility by 3 or by 11

A1: Correct method and deduction for both divisibility by 3 and by 11

A1: Notes also divisibility by 2 and explains why divisibility by 66 follows. The explanation may have been given in a preamble “66 = 2×3×11 so divisible by 66 if divisible by 2, 3 and 11”

The must be a correct reason for divisibility by 2, ie “it is even” or “last digit is even”. Do not accept “last digit is 0” with no reason given.

NB There is no divisibility test for 6, so attempting such will result in the loss of the last two A marks. (E.g. sum of digits being a multiple of 6 is an incorrect test, as for instance 15 satisfies 1+5 = 6 but is not a multiple of 6.)

Notes

Allow credit for any part for relevant work seen throughout.

M1: Writes \(n = abcd\) as sum of multiples of the digits, as shown in scheme.

A1*: Explains how each coefficient reduces modulo 7 to give the stated answer.

Notes

B1: Applies the divisibility test for 9, either as in scheme or stating \(a + b + c + d\) is a multiple of 9 (or similar). \(a + b + c + d \equiv 0\ (\mathrm{mod}\ 9)\) is fine.

M1: Uses the third fact in combination with the second to eliminate \(c\) and \(d\) and deduce \(k\) even. Alternatively, \(a + b + c + d \equiv 2(a + b) \equiv 0\ (\mathrm{mod}\ 9) \Rightarrow 10(a + b) \equiv 5 \times 0\ (\mathrm{mod}\ 9) \Rightarrow a + b \equiv 0\ (\mathrm{mod}\ 9)\)

B1: Eliminates the possibilities that \(k = 0\) or 4 using the fourth fact. Alt: reasons as \(b \lt a\) then \(a + b\) cannot be 0 or 18.

A1*: Completes the proof, with all steps explained. Allow if the previous B has not been earned.

Notes

M1: Combines the results of (a) and (b) to eliminate \(b\) and \(d\) from the result in (a).

A1*: Correct completion to the given statement. There will be different ways to achieve this, one example is shown. Check work carefully!

Notes

M1: There may be many approaches here. Scored for beginning a process of eliminating possibilities for one of the digits, e.g. deducing \(a\) is at least 5 as shown. (May be done by individually eliminating the lower values.)

dM1: Continues the process eliminating at least two more cases.

A1: Complete argument, leading to one solution only (and no possibility of others). Need not be expressed as eloquently as above, but should be clear all cases have been considered.

B1: For \(n = 6345\) obtained in any way.

(corrected from the printed mark scheme: the first line is printed as \(\text{“}a\,..\,5\text{”}\), a garbled symbol for \(a \geqslant 5\))

Notes

M1: Starts the process by showing 2 correct stages

A1: Completes the algorithm correctly

A1: Correct HCF

Notes

M1: Starts the reversal process by completing at least 2 stages

A1: Completes the algorithm correctly

A1: Correct values for \(a\) and \(b\)

Notes

B1: Correct answer 1562275 (no need to see calculation)

Notes

M1: Attempts the product shown.

A1: 495495

Notes

M1: Works out the different cases that give more than half adults – e.g. attempts to find combinations for 5, 6, 7 or 8 adults (or 0, 1, 2 or 3 juniors). Need not be the correct formula, look for evidence of the correct cases considered. Do not allow this mark if the case of 4 adults and 4 juniors is included.

M1: Sums the possibilities. Allow if the 4 adults/4 juniors case is included, but the binomial products must be correct but e.g. \({}^{14}C_{0}\) may not be seen (as it is 1). Must be an attempt at using the correct formulae.

A1: cao 383955

Notes

B1: Recalls Fermat’s Little Theorem correctly. May be implied in their work.

M1: Attempts 542 in the form \(12a + b\). Score if an attempt at \(\left(6^{12}\right)^{45}\) or similar is seen with attempt to combine with another term. Allow M1 for attempt \(\text{“}542 = \text{their } 12\text{”} \times a + b\)

A1: Uses their \(a\) and \(b\) to write \(6^{542}\) correctly in terms of \(6^{12}\) or \(6^{13}\) (must be one of these two)

M1: Completes the process to find the residue (more convoluted roots are possible but look for a complete process to reach a residue using their attempt at Fermat’s Little Theorem at least once).

A1: Correct residue. Allow if the “45” was incorrect so long as the remainder was 2.

Notes

B1: Correct value. Accept as 7! for this part but must be evaluated in the remaining parts.

Notes

M1: Evidence that the 4 students have been considered as one unit among many and sees the problem as permutations of 4 items. Score for \(4! \times k\) where \(k \ne 1\)

A1: Correct value

Notes

M1: Realises that the other 5 students can sit in any position – evidenced by sight of 5! (in (c))

A1: Correct value

Notes

M1: A correct strategy applied. E.g. interprets the situation as the answer to part (ii)(a) minus the ways that they can sit together So score for \(7! - \ldots\) or \(5040 - \ldots\) where \(\ldots\) is not zero.

Alternatively, considers the different positions Devindra can sit for each different position for Charles, and with 5! positions for the rest. Look for 5!\(\times\)(sum of 7 terms) oe.

A1: Correct value

Notes

M1: For an understanding of mod notation and finding a correct value for \(a = 4\), 6 or 12

A1: For all three correct values for a and no extras \(a = 4\), 6 & 12

Notes

(ii) see scheme

Notes

M1: For applying a divisibility test for dividing by 11 to £13 940 220 or 139 402 2000p

A1: Fully correct method and concludes not divisible by 11 and interprets conclusion in context

Notes

M1: Executes the correct process by adding and subtracting alternating digits or equivalent

A1: Completes correctly with a correct conclusion

Notes

M1: Uses the Euclidean algorithm showing two stages (Must be Euclidean algorithm not e.g. using prime factors)

A1: Completes the algorithm correctly

A1: All correct and concludes HCF is 6