Mathematics · Oct / Nov 2024 · Paper 21 · Question 11

E.g.

n =

5

3

5

+

2

=

245

245 is not a prime number

or e.g.

245 is divisible by 5 so not true

n +

M1:

Attempts to evaluate 3

2

with n a prime number.

May be seen as a list of attempts with various numbers even if some are nonprime

e.g. n = 1 as long as at least one prime number is attempted.

Condone a slip in evaluating provided the intention was to substitute in a valid

value for n.

A1:

A correct calculation for the value and a conclusion.

There must be some reference to it not being prime or they show that the number is

divisible by e.g. 5 and state that it is false/not true.

The value they have chosen must be clearly identified as their counter example and

other considerations of n can be ignored.

Examples:

53

which is not prime is M1A1

+

2

=

245

n +

when

n =

11

then 3

2

177149

and 177149 7

÷=

25307

so false is M1A1

=

53

2

245

so false is M1A0

+

=

53

2

245

which is divisible by 25 so not prime is M1A0

+

=

Most candidates are likely to choose n = 5 as their counter example but other numbers are

possible e.g.:

n +

3

2

n

7

2189

11

177149

13

1594325

17

129140165

Scores: M1A0 (we would need e.g. “not prime” or 245/5 = 49 so false) but note that we

would have ignored the consideration of any nonprimes e.g. n = 1

=

3

+

1

(

)

2

m

k

m

2

−=

1

3

k

+

1

−=

1

9

k

2

+

6

k

+ −

1 1

or

=

3

+

2

2

1

(

3

2

)

2

1

9

2

12

4 1

m

k

m

−=

k

+

−=

k

+

k

+

−

(

)

2

2

2

m

−=

1

9

k

+

6

k

=

3 3

k

+

2

k

or

(

)

2

−=

1

9

2

+

12

+

3

=

3 3

2

+

4

+

1

m

k

k

k

k

3

1

(

)

2

m

=

k

+

2

−=

1

3

+

1

−=

1

9

2

+

6

+ −

1 1

m

k

k

k

and

=

3

+

2

(

)

2

m

k

m

2

−=

1

3

k

+

2

−=

1

9

k

2

+

12

k

+

4 1

−

(

)

and (

)

3 3

k

2

+

2

k

3 3

k

2

+

4

k

+

1

are both multiples of 3 so

m −

2

1

must be divisible by 3 when m is not divisible by 3

For the accuracy marks there should be no errors in the algebra but allow e.g. invisible

brackets to be “recovered”.

Withhold the final mark if m is used instead of their k if the work is otherwise

correct.

M1:

Starts the proof by considering at least one algebraic expression not divisible by 3 e.g.

m = 3k + 1 and attempts

m − by expanding the brackets.

2

1

Alternatively, writes

2

1

(

1

)(

1

)

and substitutes at least one algebraic

m

−=

m

−

m

+

expression not divisible by 3.

Allow equivalent representations of numbers not divisible by 3 e.g. m = 3k + 2 or

m = 3k − 1

Condone arithmetical slips and condone the use of e.g. m = 3m + 2 or

m

=

A1:

They must

obtain at least one correct expression

show that their expression is a multiple of 3 by factoring out 3 or e.g. 3k e.g.

(

)

2

(

)

or e.g. (

)(

)

3 3

k

+

2

k

or e.g.

3

k

3

k +

2

3

k

+

1 3

k

+

1

or

9

2

+

6

k

k

show that their expression is divisible by 3 e.g.

=

3

k

2

+

2

3

or

make a reasoned argument as to why the expression is divisible by 3 e.g.

obtains

9

2

6

and states e.g. is divisible by 3 as both 9 and 6 are divisible

k

+

k

by 3 (or equivalent reasoning)

Note that this may be seen in the final conclusion.

Condone a spurious “= 0” e.g. (

)

3 3

2

+

2

=

0

for this mark.

k

k

dM1: Considers both distinct cases where m is not divisible by 3 (see above)

See table below – they must have one of case A and one of case B.

A1:

Fully correct proof. They must

achieve correct expressions for

m − for both cases

2

1

show that each expression is a multiple of 3 as above

make a concluding overall statement. “Hence is a multiple of 3”. Accept

“hence proven”, “statement proved”, “QED” if they have shown for each

separate case that the expression is a multiple of 3. Ignore any reference to

what m is e.g. “for all integers m”

Do not condone any spurious “= 0” e.g. (

)

3 3

2

+

2

=

0

for this mark.

k

k

You can ignore any cases considered where m is divisible by 3 e.g. m = 3k provided this does

not clearly form part of their proof.

For reference:

Case

2

m

m −

1

3

k +

1

A

9

k

2

+

6

k

B

3

k +

2

9

2

+

12

+

3

k

k

3

k −

1

B

9

k

2

−

6

k

A

3k – 2

9

2

−

12

+

3

k

k

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