Question 6:
6 | a | u = u + 3 0 , u = u + 3 0 0 ,
2 1 3 2
u = u + 3 0 0 0 , …
4 3
u =u +310n (n  1) with u =7
n + 1 n 1 | M1
A1
[2] | 1.1a
2.2b | Any sensible method (may be
implied by a correct, or nearly
correct, answer)
Must include first term. Condone
missing “range” of n. | OR u = 1 0 u − 3 3 ,
2 1
u =10u −33, …
3 2
OR u =10u −33
n + 1 n
with u = 7
1
b | CS is u = C
1
For PS try u = a  1 0 n
n
a = 13
u = 110n +C and u =7 used …
n 3 1
u = 13  1 0 n + 1 13
n | B1
M1
A1
M1
A1
[5] | 1.1
1.1
1.1
1.1
1.1 | Including substn. to determine a
Their GS = CS + PS and attempt to
find a numerical value for C
AG
Since 111 = 3  37, it follows that if 37 | u
k
then 37 | u also, and proof follows by
k + 3
induction | M1dep
*
A1 | 2.1
2.4 | Inductive reasoning attempted from
their 𝑢
𝑘+3
All correctly explained (for final A,
must note that this implies that {u }
n
contains infinitely many multiples of
37)
Alternative method 1
mod 37, { u } = {7, 0, 4, 7, 0 …}
n
Baseline case: u is a multiple of 37
2 | B1
B1 | Noting that every third term is (or
appears to be) a multiple of 37
Noted or used
Assume 𝑢 is a multiple of 37 for some k
3𝑘−1 | M1
Then 𝑢 = 1 (103𝑘+2 +11) = 𝑢 +
3𝑘+2 3𝑘−1
3
111×3×103𝑘−1 | M1* | Attempt to write their 𝑢 in
3𝑘+2
terms of 𝑢
3𝑘−1
A1
Since 111 = 3  37, it follows that if 37 | u
k
then 37 | u also, and proof follows by
k + 3
induction | M1dep
*
A1 | Inductive reasoning attempted from
their 𝑢
𝑘+3
All correctly explained (for final A,
must note that this implies that { u }
n
contains infinitely many multiples of
37)
Alternative method 1
mod 37, { u } = {7, 0, 4, 7, 0 …}
n
Baseline case: u is a multiple of 37
2
B1
B1
Then 𝑢 = 1 (103𝑘+2 +11) = 𝑢 +
3𝑘+2 3𝑘−1
3
111×3×103𝑘−1
M1dep
*
A1
Since 111 = 3  37, it follows that if 37 | u
k
then 37 | u also, and proof follows by
k + 3
induction
Alternative method 2
 
The sequence 10n +11 runs | B1 | Working with the sequence (or with
the original one)
 .
2 1 , 1 1 1 , 1 0 1 1 , 1 0 0 1 1 , 1 0 0 0 1 1 . .
Noting that 111 = 3  37 and 100011 = | B1 | Must make the conclusion that { u }
n
2703  37 and noting that every third term | contains infinitely many multiples of
is (appears to be) a multiple of 37, so … | 37 at some stage
Testing 103n + 2 +11 | M1 | Directly or inductively | Or 103𝑛−1 +11
(1 ) n
= 0 3  1 0 2 + 1 1 | M1 | Use of the laws of indices
= 1 0 0  1 0 0 0 n + 1 1 | A1
Since 999 is a multiple of 37, | M1 | Use of modular arithmetic
1000  1 (mod 37)
1 0 3 n + 2 + 1 1  1001n +11=111
= 3 37  0 (mod 37) | A1 | All correctly concluded
[7]
Working with the sequence (or with
the original one)
a | i = C a = B b = A
A B A C C B
c = C p = B q = A
B A C A B C | B1
B1
B1
[3] | 1.1
1.1
1.1 | 1st B for correct i and  1 of a, b,
c
2nd B for all of i, a, b, c correct
3rd B for p, q correct
b | B1
B1
B1
[3] | 1.1
1.1
1.1 | First row and first column are
given
Correct composition of
reflections (33 square)
Correct composition of rotations
(22 square)
All the rest correct (white cells) | NB The main diagonal
& composition of
rotations are obvious;
one composition of
reflections gives the
others by the LSP;
similarly for each of the
white rectangles.
c | {i, a}, {i, b}, {i, c} and {i, p, q} | B1
B1
B1
[3] | 1.1 1.1
1.1 | Any one subgroup of order 2
The subgroup of order 3
All four correct with no extras | Condone inclusion of
{i} and/or D
3
d | i | NOT cyclic, as D has no element of order 6
3
or IF they first explain that D is non-
3
abelian (with valid reason) then allow the
claim “non-abelian  non-cyclic” | B1
[1] | 2.4 | Or since elements have orders:
a, b, c … 2 and p, q … 3 | Just ‘no generator’ not
sufficient
ii | NOT abelian, as (e.g.) ab = p but ba = q | B1
[1] | 2.4 | Pull the mark forward from d i if
they have done this there | Accept no symmetry in
the leading diagonal in
the Cayley table
e | H is cyclic (while D isn’t)
3
H is abelian (while D isn’t)
3 | B1
B1
[2] | 2.5
2.2a | Any two differences clearly
stated or described on group
structure or order of elements or
subgroup order
PMT
Need to get in touch?
If you ever have any questions about OCR qualifications or services (including administration, logistics and teaching) please feel free to get in
touch with our customer support centre.
Call us on
01223 553998
Alternatively, you can email us on
support@ocr.org.uk
For more information visit
ocr.org.uk/qualifications/resource-finder
ocr.org.uk
Twitter/ocrexams
/ocrexams
/company/ocr
/ocrexams
OCR is part of Cambridge University Press & Assessment, a department of the University of Cambridge.
For staff training purposes and as part of our quality assurance programme your call may be recorded or monitored. © OCR
2022 Oxford Cambridge and RSA Examinations is a Company Limited by Guarantee. Registered in England. Registered office
The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA.
Registered company number 3484466. OCR is an exempt charity.
OCR operates academic and vocational qualifications regulated by Ofqual, Qualifications Wales and CCEA as listed in their
qualifications registers including A Levels, GCSEs, Cambridge Technicals and Cambridge Nationals.
OCR provides resources to help you deliver our qualifications. These resources do not represent any particular teaching method
we expect you to use. We update our resources regularly and aim to make sure content is accurate but please check the OCR
website so that you have the most up-to-date version. OCR cannot be held responsible for any errors or omissions in these
resources.
Though we make every effort to check our resources, there may be contradictions between published support and the
specification, so it is important that you always use information in the latest specification. We indicate any specification changes
within the document itself, change the version number and provide a summary of the changes. If you do notice a discrepancy
between the specification and a resource, please contact us.
Whether you already offer OCR qualifications, are new to OCR or are thinking about switching, you can request more
information using our Expression of Interest form.
Please get in touch if you want to discuss the accessibility of resources we offer to support you in delivering our qualifications.