AS June 2024 Q2

EdexcelCurrent spec6 marksNumber Theory

2. Tiles are sold in boxes with 21 tiles in each box.

The tiles are laid out in \(x\) rows of 5 tiles and \(y\) rows of 6 tiles.

All the tiles from a box are used before the next box is opened.

When all the rows of tiles have been laid, there are \(n\) tiles left in the last opened box.

(a) Write down a congruence expression for \(n\) in the form\[ax + by \pmod{c}\]where \(a\), \(b\) and \(c\) are integers. (1)

Given that

exactly 43 rows of tiles are laid
there are no tiles left in the last opened box

(b) use your congruence expression to determine the minimum number of rows of 6 tiles laid. (5)

Scheme	Marks	AO
\((n \equiv)\, 5x + 6y \pmod{21}\)	B1	1.1b
	(1)

Mark scheme (b)

Scheme	Marks	AO
\(x + y = 43\)	B1	1.1b
\(\Rightarrow n \equiv 5(x + y) + y \equiv 5 \times 43 + y \pmod{21}\)	M1	3.1a
\(\equiv 5 \times 1 + y \equiv 5 + y \pmod{21}\)	A1	1.1b
\(n \equiv 0 \pmod{21} \Rightarrow 5 + y \equiv 0 \pmod{21} \Rightarrow y \equiv -5 \pmod{21}\)	M1	1.1b
So (as \(y \geqslant 0\)) minimum for \(y\) is 16	A1	2.3
	(5)
	(6 marks)

Notes

B1: Forms the equation in \(x\) and \(y\) from the given information.

M1: Substitutes for \(x\) or \(x + y\) in the congruence expression to form an expression in \(y\) only. Alternatively, they many eliminate \(y\) to get an equation in \(x\) only.

A1: Correct equation in \(y\) (or \(x\)) only.

M1: Full method to reach a residue for \(y\). If \(y\) was eliminated from the congruence they must use \(x + y = 43\) again to form an appropriate expression for \(y\) here.

A1: Interprets the situation correctly to give 16 for the minimum value for \(y\).

Alternative

Scheme	Marks	AO
\(x + y = 43\)	B1	1.1b
\(5x + 6y \pmod{21} = 0 \Rightarrow \ldots\ 5x + 6y = 21N\) Solve simultaneous \(x + y = 43\) and \(5x + 6y = 21N\)	M1	3.1a
\(y = 21N - 215\)	A1	1.1b
\(N = 11 \Rightarrow y = 21 \times 11 - 215 = \ldots\)	M1	1.1b
minimum for \(y\) is 16	A1	2.3
	(5)

B1: Forms the equation in \(x\) and \(y\) from the given information.

M1: uses their congruence express, sets = 0 and forms an equation in \(x\), \(y\) and a third variable. Solve simultaneous equations to find \(y\) in terms of the third variable.

A1: Correct equation for \(y\)

M1: Uses value for the third variable to find a value for \(y\)

A1: Interprets the situation correctly to give 16 for the minimum value for \(y\).