Mathematics · Oct / Nov 2022 · Paper 21 · Question 1
YMA01
As printed in the paper
The question as text
1. Given that a, b and c are integers greater than 0 such that • c = b + 2 • a + b + c = 10 Prove, by exhaustion, that the product of a, b and c is always even. You may use the table below to illustrate your answer. (3) You may not need to use all rows of this table. a b c 1 2
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Show the answer and mark scheme
a
b
c
(abc)
6
1
3
4
2
4
2
3
5
Any one correct row for b = 1, b = 2 or b = 3.
Products do not need to be found for this mark.
Attempts the product abc for at least 2 valid combinations.
Finds all three valid combinations with correct products seen and
somewhere/shows why this is exhaustive and concludes. *
Note that in most cases the M1 can only follow B1 but there may be some exceptions.
Numerical approach using the table:
B1: Any one correct row for b = 1, b = 2 or b = 3. Products do not need to be found for this mark.
M1: Attempts the product abc for at least 2 valid combinations.
A1*: Requires:
the row or are discounted in some way
is even, each product stated as even, etc.
Algebraic/logic approach:
B1: Uses the information to obtain a correct equation connecting a and b e.g. a + 2b = 8, a = 8 − 2b
M1: States a must be even and considers the product abc in some way
A1*: States e.g. abc is even with a reason e.g. “even × anything is even”
Pure Algebraic approach:
B1: Uses the information to obtain a correct equation connecting a and b e.g. a + 2b = 8, a = 8 − 2b
(
) (
)
M1:
abc
=
8
−
2
b b b
+
2
Attempts the product of a, b and c in terms of b (or some other letter)
A1*:
abc
2 4
(
b b b
) (
2
)
which is even, hence proven, QED etc.
=
−
+
Concludes abc is even and makes a (minimal) conclusion. There must be no algebraic errors.
NB using this approach “
abc
= −
2
b
3
+
4
b
2
+
16
b
which is even hence proven” is not sufficient –
they would need to say e.g. which is even + even + even or factor out the 2.
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