A2 June 2023 Q5

EdexcelCurrent spec8 marksNumber Theory

(i) A security code is made up of 4 numerical digits followed by 3 distinct uppercase letters.

Given that the digits must be from the set {1, 2, 3, 4, 5} and the letters from the set {A, B, C, D}

(a) determine the total number of possible codes using this system.

To enable more codes to be generated, the system is adapted so that the 3 letters can appear anywhere in the code but no letter can be next to another letter.

(b) Determine the increase in the number of codes using this adapted system. (4)

(ii) A combination lock code consists of four distinct digits that can be read as a positive integer, \(N = abcd\), satisfying

all the digits are odd
\(N\) is divisible by 9
the digits appear in either ascending or descending order
\(N \equiv e \pmod{ab}\) where \(ab\) is read as a two-digit number and \(e\) is the odd digit that is not used in the code

(a) Use the first two properties to determine the four digits used in the code.

(b) Hence determine the code on the lock. (4)

Mark scheme (i)

Scheme	Marks	AO
(a) \(5^4 \times \ldots\) or \(\ldots \times 4 \times 3 \times 2\) or \(\ldots \times {}^{4}P_{3}\)	M1	1.1b
\(5^4 \times 4 \times 3 \times 2 = 15000\)	A1	1.1b
(b) The structure is _N_N_N_N_, where three _’s are letters, so have \({}^{5}C_{3}\) choices for the letters, hence new number of combinations is \({}^{5}C_{3} \times 15000\) or \(10 \times 15000\)	M1	3.1a
So increase in number of codes is 135000	A1ft	1.1b
	(4)

Notes

(i)(a)

M1: \(5^4 \times \ldots\) or \(\ldots \times 4 \times 3 \times 2\) or \(\ldots \times {}^{4}P_{3}\)

A1: For 15000

(b)

M1: Correct strategy to find the new number of codes seen or implied.

A1ft: Increase in codes is 135000. Follow through \(9 \times\) their 15000

Mark scheme (ii)

Scheme	Marks	AO
(a) Divisible by \(9 \Rightarrow a + b + c + d = 9k\) or e.g. attempt \(1 + 3 + 5 + 7 = 16\), not a multiple of 9, so reject. E.g. \(1 + 3 + 5 + 7 + 9 = 25\) so \(k = 1\) or 2, but \(1 + 3 + 5 + 7 = 16 > 9\), so \(k = 2\) hence \(\Rightarrow 25 - (a + b + c + d) = 25 - 18 = 7\) so 7 missing. Or considers options in turn and finds only \(1 + 3 + 5 + 9 = 18\) works	M1	3.1a
So the digits are 1, 3, 5 and 9	A1	1.1b
(b) Combination is either 1359 or 9531 and \(1359 \pmod{13} \equiv 59 \pmod{13} \equiv 7,\quad 9531 \pmod{95} \equiv 31\)	M1	1.1b
Combination is 1359	A1	2.2a
	(4)
	(8 marks)

Notes

(ii)(a)

M1: Full strategy to deduce the correct set of digits.

A1: Digits are 1,3, 5 and 9

Writes down the correct values scores M1A1

(b)

M1: Checks which of the two possible combinations satisfies the final property.

A1: Combination is 1359

Writes down the correct code scores M1A1