# IB DP Maths Topic 10.2 a|b⇒b=na for some n∈Z . HL Paper 3

## Question

Show that $${\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = {\text{gcd}}\left( {k – 1,\,2} \right)$$, where $$k \in {\mathbb{Z}^ + },\,k > 1$$.

[4]
a.

State the value of $${\text{gcd}}\left( {4k + 2,\,3k + 1} \right)$$ for odd positive integers $$k$$.

[1]
b.i.

State the value of $${\text{gcd}}\left( {4k + 2,\,3k + 1} \right)$$ for even positive integers $$k$$.

[1]
b.ii.

## Markscheme

METHOD 1

attempting to use the Euclidean algorithm       M1

$$4k + 2 = 1\left( {3k + 1} \right) + \left( {k + 1} \right)$$      A1

$$3k + 1 = 2\left( {k + 1} \right) + \left( {k – 1} \right)$$      A1

$$K + 1 = \left( {k – 1} \right) + 2$$      A1

$$= {\text{gcd}}\left( {k – 1,\,2} \right)$$     AG

METHOD 2

$${\text{gcd}}\left( {4k + 2,\,3k + 1} \right)$$

$$= {\text{gcd}}\left( {4k + 2 – \left( {3k + 1} \right),\,3k + 1} \right)$$     M1

$$= {\text{gcd}}\left( {3k + 1,\,k + 1} \right)\,\,\left( { = {\text{gcd}}\left( {{\text{k + 1,}}\,{\text{3k + 1}}} \right)} \right)$$     A1

$$= {\text{gcd}}\left( {3k + 1 – 2\left( {k + 1} \right),\,k + 1} \right)\,\,\left( { = {\text{gcd}}\left( {k – 1{\text{,}}\,k + {\text{1}}} \right)} \right)$$     A1

$$= {\text{gcd}}\left( {k + 1 – \left( {k – 1} \right),\,k – 1} \right)\,\,\left( { = {\text{gcd}}\left( {{\text{2,}}\,k – {\text{1}}} \right)} \right)$$     A1

$$= {\text{gcd}}\left( {k – 1,\,2} \right)$$     AG

[4 marks]

a.

(for $$k$$ odd), $${\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = 2$$     A1

[1 mark]

b.i.

(for $$k$$ even), $${\text{gcd}}\left( {4k + 2,\,3k + 1} \right) = 1$$     A1

[1 mark]

b.ii.

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