Mathematics · May / June 2025 · Paper 21 · Question 10
YMA01
As printed in the paper
The question as text
10. In this question you must show detailed reasoning. Use algebra to prove by exhaustion that, for all positive integers m that are not multiples of 3, the value of m 2 + 3m + 2 is always a multiple of 3 (4)
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
Sets
=
3
+ or
1
=
3
+
2
( p > 0) o.e. and attempts
m
2
+
3
m
+
2
m
p
m
p
(
)
(
)
(
)
2
2
2
+
3
+
2
=
3
+
1
2
+
3 3
+
1
+
2
=
9
+
15
+
6
=
3 3
+
5
+
2
E.g.
m
m
p
p
p
p
p
p
A1
Sets
+ and
( p > 0) and attempts
2
+
3
+
2
m
=
3
p
1
m
=
3
p
+
2
m
m
(
)
2
(
)
2
(
)
2
2
m
+
3
m
+
2
=
3
p
+
2
+
3 3
p
+
2
+
2
=
9
p
+
21
p
+
12
=
3 3
p
+
7
p
+
4
E.g.
And gives a minimal conclusion *
3
1
3
2
Main scheme algebraic method using e.g.
m
=
p
+ and
m
=
p
+
You will need to look at both cases and mark the one which is fully correct first.
Allow a different variable to p and may be different letters for the two cases.
Condone use of m as a variable for the first three marks.
There should be no errors in the algebra for the A marks including invisible brackets.
Note that there are other allowable pairs of combinations covering the two distinct cases
e.g.
=
3
− and
1
=
3
+ or e.g.
1
=
3
−
2
and
=
3
+
2
m
p
m
p
m
p
m
p
but not e.g.
m
=
3
p
−and
1
m
=
3
p
+
2
Note: An attempt substituting in values rather than using algebra will score 0 marks.
+ o.e or
2
M1: Sets
m
=
3
p
1
m
=
3
p
+
2
o.e., expands
m
+
3
m
+
2
and achieves a quadratic expression
which may be unsimplified
A1: Correct quadratic expression with constant terms collected for
2
+
3
+
2
for either of the two
m
m
cases and shows or gives a reason why the expression is a multiple of 3. They must have fully
multiplied out or the quadratic expression must have a factor of 3 taken out.
Do not isw if they simplify their quadratic incorrectly.
2
9
+
15
+
6
e.g.
p
p
and states each term is a multiple of 3 or alternatively, correctly divides by 3 and
concludes that it is divisible by 3
e.g. (
)
2
2
3 3
p
+
5
p
+
2
or e.g.
3(3
p
+
2 )
p
+
3(3
p
+
1)
+
3
2
e.g.
9
p
+
21
p
+
12
and states each term is a multiple of 3 or alternatively correctly divides by 3 and
concludes that it is divisible by 3
e.g. (
)
2
3 3
+
7
+
4
p
p
dM1: Sets
=
3
+ o.e and
1
=
3
+
2
o.e. and expands
2
+
3
+
2
and achieves two quadratic
m
p
m
p
m
m
expressions which may be unsimplified. See first M1 for guidance. It is dependent on the previous
method mark.
A1*: Requires
correct quadratic expression for
2
+
3
+
2
for both cases which is not in terms of m
m
m
shows or gives a reason for why the expressions are multiples of 3 (see first A1 for guidance)
makes a concluding overall statement. “Hence
2
+
3
+
2
is always a multiple of 3 for all m
m
m
that are not multiples of 3” (or equivalent) Allow generic conclusions e.g. “proven”, “hence
true”
Withhold the final mark if they have a spurious = 0 on the end of their quadratic expressions.
Below is a table for the different possible cases and the quadratic expressions:
m
2
+
3
m
+
2
3
p −
2
9
p
2
−
3
p
=
3(3
p
2
−
p
)
3
p −
1
9
p
2
+
3
p
=
3(3
p
2
+
p
)
3
p +
1
9
2
15
6
3(3
2
5
2)
p
+
p
+
=
p
+
p
+
3
p +
2
9
2
+
21
+
12
=
3(3
2
+
7
+
4)
p
p
p
p
Factorising first then using logic
Note the same marking principle can be applied as in the main scheme and notes
2
3
2
(
1
)(
2
)
m
+
m
+
=
m
+
m
+
(
)(
)
(
)(
2
+
3
+
2
=
3
+ +
1 1 3
+ +
1 2
=
3 3
+
2
Sets e.g.
=
3
+ leading to
1
m
m
p
p
p
m
p
(
)(
)
(
2
+
3
+
2
=
3
−+
1 1 3
−+
1 2
=
3
3
+
1
Sets e.g.
=
3
− leading to
1
m
m
p
p
p
p
m
p
Alternative using algebra with logic
2
3
2
(
1
)(
2
)
m
+
m
+
=
m
+
m
+
represents two consecutive numbers and if m is not a multiple of 3 then one of (
e.g. (
m
+
1)(
m
+
2)
or (
)
2
m+
2
must be a multiple of 3 hence
m
+
3
m
+
2
is a multiple of 3.
If you see this approach or other attempts that you think may be creditworthy then send to review.
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