Question 1
Answer the following questions.
1(a):
Prove that 4(2x - 3) - 2(2x + 9) \equiv 4x - 30
Line 1: 4(2x - 3) - 2(2x + 9) \equiv 4x - 30
Line 2: [ Missing Line ]
Line 3: 4x - 30 \equiv 4x - 30
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: 8x - 12 - 4x - 18 \equiv 4x - 30
B: 6x + 1 + 7 \equiv 4x - 30
C: 2x + 7 - 4x - 11 \equiv 4x - 30
D: 2x - 1 - x \equiv4x - 30
Answer: A
Workings:
Line 1: 4(2x - 3) - 2(2x + 9) \equiv 4x - 30
Line 2: 8x - 12 - 4x - 18 \equiv 4x - 30
Line 3: 4x - 30 \equiv 4x - 30
Marks = 2
1(b):
Prove that (n - 1)^2 - (n - 2)^2 \equiv 2n - 3
Line 1: (n - 1)^2 - (n - 2)^2 \equiv 2n - 3
Line 2: [ Missing Line ]
Line 3: 2n - 3 \equiv 2n - 3
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: n^2 - 2n + 1 - n^2 + 4n - 4 \equiv 2n - 3
B: n^2 - 2n - 1 + n^2 - 4n - 4 \equiv 2n - 3
C: n^2 + 2n + 1 - n^2 - 4n + 4 \equiv 2n - 3
D: n^2 - 2n - 1 - n^2 - 4n - 4 \equiv 2n - 3
Answer: A
Workings:
Line 1: (n - 1)^2 - (n - 2)^2 \equiv 2n - 3
Line 2: n^2 - 2n + 1 - n^2 + 4n - 4 \equiv 2n - 3
Line 3: 2n - 3 \equiv 2n - 3
Marks = 2
1(c):
Prove that (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 1: (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 2: [ Missing Line ]
Line 3: n^2 + n - 8 \equiv (n + 4)(n - 3) + 4
Line 4: (n^2 + n -12) + 4 \equiv (n + 4)(n - 3) + 4
Line 5: (n + 4)(n - 3) + 4 \equiv (n + 4)(n - 3) + 4
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: n^2 + 4n + 4 - 3n - 12 \equiv (n + 4)(n - 3) + 4
B: 2n^2 + 6n + 4 - 4n + 1 \equiv (n + 4)(n - 3) + 4
C: -2n + 4 - 4n - 7 \equiv (n + 4)(n - 3) + 4
D: n^2 + 2n + 1 - 3n - 1 \equiv (n + 4)(n - 3) + 4
Answer: A
Workings:
Line 1: (n + 2)^2 - 3(n + 4) \equiv (n + 4)(n - 3) + 4
Line 2: n^2 + 4n + 4 - 3n - 12 \equiv (n + 4)(n - 3) + 4
Line 3: n^2 + n - 8 \equiv (n + 4)(n - 3) + 4
Line 4: (n^2 + n -12) + 4 \equiv (n + 4)(n - 3) + 4
Line 5: (n + 4)(n - 3) + 4 \equiv (n + 4)(n - 3) + 4
Marks = 2
1(d):
Prove that 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 1: 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 2: [ Missing Line ]
Line 3: 3n^2 + 6n - 9 - 3 + 3n \equiv (3n - 3)(n + 4)
Line 4: 3n^2 + 9n - 12 \equiv (3n - 3)(n + 4)
Line 5: (3n - 3)(n + 4) \equiv (3n - 3)(n + 4)
What is the missing Line 2 from the working above?
ANSWER: Multiple Choice (Type 1)
A: 3(n^2 + 2n - 3) - 3(1-n) \equiv (3n - 3)(n + 4)
B: 3(2n^2 + 4n + 2) - 3(1-n) \equiv (3n - 3)(n + 4)
C: 3(4n + 4) - 3(1-n) \equiv (3n - 3)(n + 4)
D: 3(n^2 - 2n + 3) - 3(1-n) \equiv (3n - 3)(n + 4)
Answer: A
Workings:
Line 1: 3(n + 3)(n - 1) - 3(1 - n) \equiv (3n - 3)(n + 4)
Line 2: 3(n^2 + 2n - 3) - 3(1-n) \equiv (3n - 3)(n + 4)
Line 3: 3n^2 + 6n - 9 - 3 + 3n \equiv (3n - 3)(n + 4)
Line 4: 3n^2 + 9n - 12 \equiv (3n - 3)(n + 4)
Line 5: (3n - 3)(n + 4) \equiv (3n - 3)(n + 4)
Marks = 2
Question 2:
Show that the following statements are true:
Question 2(a): [2 marks]
(3n+1)(n+3) - n(3n+7) \equiv 3(n+1)
Line 1: 3n^2 + 9n + n + 3 - 3n^2 - 7n \equiv 3(n+1)
Line 2: [ Missing line ]
Line 3: 3(n+1) \equiv 3(n+1)
What is the missing Line 2 from the working above?
Answer type: Multiple choice type 1
A: 3n + 3 \equiv 3(n+1)
B: 3n^2 + 3 \equiv 3(n+1)
C: 3n+1 \equiv 3(n+1)
D: 3n^2 + 3n + 1 \equiv 3(n+1)
ANSWER: A
WORKING:
Line 1: 3n^2 + 9n + n + 3 - 3n^2 - 7n \equiv 3(n+1)
Line 2: 3n + 3 \equiv 3(n+1)
Line 3: 3(n+1) \equiv 3(n+1)
Question 2(b): [2 marks]
(n+3)^2 - (3n+4) \equiv (n+1)(n+2) + 3
Line 1: [ Missing line ]
Line 2: n^2 + 3n + 5 \equiv (n+1)(n+2) + 3
Line 2: (n-3)^2 - (2n+1) \equiv (n+1)(n+2) + 3
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: n^2 + 6n + 9 - 3n - 4 \equiv (n+1)(n+2) + 3
B: n^2 + 9n + 9 - 3n + 4 \equiv (n+1)(n+2) + 3
C: n^2 + 6n + 9 - 3n - 4 \equiv (n+1)(n+2) + 3
D: n^2 + 9n + 9 - 3n - 4 \equiv (n+1)(n+2) + 3
ANSWER: A
WORKING:
Line 1: n^2 + 6n + 9 - 3n - 4 \equiv (n+1)(n+2) + 3
Line 2: n^2 + 3n + 5 \equiv (n+1)(n+2) + 3
Line 2: (n-3)^2 - (2n+1) \equiv (n+1)(n+2) + 3
Question 2(c): [2 marks]
(n-3)^2 - (2n+1) \equiv (n-4)^2 - 8
Line 1: [ Missing line ]
Line 2: n^2 - 8n + 8 \equiv (n-4)^2 - 8
Line 3: (n-4)^2 - 8 \equiv (n-4)^2 - 8
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: n^2 - 6n + 9 - 2n - 1 \equiv (n-4)^2 - 8
B: n^2 - 6n + 9 - 2n + 1 \equiv (n-4)^2 - 8
C: n^2 - 6n - 9 - 2n - 1 \equiv (n-4)^2 - 8
D: n^2 - 6n - 9 - 2n + 1 \equiv (n-4)^2 - 8
ANSWER: A
WORKING:
Line 1: n^2 - 6n + 9 - 2n - 1 \equiv (n-4)^2 - 8
Line 2: n^2 - 8n + 8 \equiv (n-4)^2 - 8
Line 3: (n-4)^2 - 8 \equiv (n-4)^2 - 8
Question 3:
Answer the following questions.
Question 3(a): [1 mark]
Prove the product of two even numbers is always even.
Choose the correct proof.
Answer type: Multiple choice type 1
A: 2n \times 2m = 4nm = 2(2nm) which is always even
B: 2n + 2m = 4nm = 2(2nm) which is always even
C: 2n + 2m = 2(n+m) which is always even
D: 2n \times m = 2nm which is always even
ANSWER: A
Question 3(b): [1 mark]
Prove that the product of two odd numbers is always odd.
Line 1: (2n+1)(2m+1)
Line 2: = 4nm + 2n + 2m + 1
Line 3: [ Missing line ]
Line 4: Which is odd
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: = 2(2nm + n + m) + 1
B: = 2(2nm) + 1
C: = 2(2nm + n + m)
D: = 2(2nm)
ANSWER: A
WORKING:
Line 1: (2n+1)(2m+1)
Line 2: = 4nm + 2n + 2m + 1
Line 3: = 2(2nm + n + m) + 1
Line 4: Which is odd
Question 3(c): [2 marks]
Prove the product of three consecutive odd numbers is odd.
Line 1: [ Missing line ]
Line 2: = (4n^2 + 2n +6n + 3)(2n+5)
Line 3: = 8n^3 + 36n^2 + 46n + 15
Line 4: = 2(4n^3 + 18n^2 + 23n + 7) + 1
Line 5: Simplifies to 2(n) + 1
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (2n+1)(2n+3)(2n+5)
B: (2n+1)(2n+2)(2n+3)
C: 2n(2n+2)(2n+4)
D: n(n+1)(n+2)
ANSWER: A
WORKING:
Line 1: (2n+1)(2n+3)(2n+5)
Line 2: = (4n^2 + 2n +6n + 3)(2n+5)
Line 3: = 8n^3 + 36n^2 + 46n + 15
Line 4: = 2(4n^3 + 18n^2 + 23n + 7) + 1
Line 5: Simplifies to 2(n) + 1
Question 4:
Answer the following questions.
Question 4(a): [2 marks]
Prove algebraically that the sum of any three odd numbers is odd.
Line 1: [ Missing line ]
Line 2: = 2(n + m + p + 1) + 1
Line 3: = 2(x) + 1
Line 4: Which is odd.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (2n+1) + (2m+1) + (2p+1)
B: (2n) + (2m+1) + (2p+2)
C: (2n+1) + (2m + 2) + (2p + 3)
D: (n+1) + (m+1) + (p+1)
ANSWER: A
WORKING:
Line 1: (2n+1) + (2m+1) + (2p+1)
Line 2: = 2(n + m + p + 1) + 1
Line 3: = 2(x) + 1
Line 4: Which is odd.
Question 4(b): [2 marks]
Prove algebraically that the sum of the squares of two odd integers is always even.
Line 1: (2n+1)^2 + (2m+1)^2
Line 2: = 4n^2 + 4n + 1 + 4m^2 + 4m + 1
Line 3: [ Missing line ]
Line 4: Which is even
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: =2(2n^2 + 2m^2 + 2n + 2m + 1)
B: =4(n^2 + m^2 + n + m + 1)
C: =2(n^2 + m^2 + n + m + 1)
D: =4(n^2 + m^2 + n + m + 2)
ANSWER: A
WORKING:
Line 1: (2n+1)^2 + (2m+1)^2
Line 2: = 4n^2 + 4n + 1 + 4m^2 + 4m + 1
Line 3: = 2(2n^2 + 2m^2 + 2n + 2m + 1)
Line 4: Which is even
Question 4(c): [2 marks]
Prove that when two consecutive integers are squared, that the difference is equal to the sum of the two consecutive integers.
Line 1: [ Missing line ]
Line 2: = n^2 + 2n + 1 - n^2
Line 3: = 2n + 1
Line 4: Which is equal to the sum of the two consecutive integers.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (n+1)^2 - n^2
B: (n+1)^2 - n
C: (n^2 +1) - n^2
D: (2n+1)^2 - n^2
ANSWER: A
WORKING:
Line 1: (n+1)^2 - n^2
Line 2: = n^2 + 2n + 1 - n^2
Line 3: = 2n + 1
Line 4: Which is equal to the sum of the two consecutive integers.
Question 5:
Answer the following questions.
Question 5(a): [2 marks]
Prove that (n+3)^2 + n(3-n) -3(n+4) is a multiple of 3 for all integer values of n.
Line 1: = n^2 + 3n + 3n + 9 + 3n - n^2 - 3n - 12
Line 2: = 6n - 3
Line 3: [ Missing line ]
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: = 3(2n+1)
B: = 3(n+1)
C: =3
D: = 2(3n+1)
ANSWER: A
WORKING:
Line 1: = n^2 + 3n + 3n + 9 + 3n - n^2 - 3n - 12
Line 2: = 6n - 3
Line 3: = 3(2n+1)
Question 5(b): [2 marks]
Prove algebraically that the sum of two consecutive numbers is odd.
Choose the correct proof.
Answer type: Multiple choice type 1
A: If a number is n, then the next number is n+1. Therefore the sum is n + (n+1) = 2n + 1 which is odd.
B: If a number is n, then the next number is 2n+1. Therefore the sum is n + (2n+1) = 3n + 1 which is odd.
C: If a number is n, then the next number is 2n. Therefore the sum is n + (2n) = 3n which is odd.
D: If a number is n, then the next number is n+2. Therefore the sum is n + (n+2) = 2n + 2 = 2(n+1) which is odd.
ANSWER: A
Question 5(c): [2 marks]
Prove algebraically that the sum of the squares of two consecutive multiples of 5 is not a multiple of 10.
Line 1: [ Missing line ]
Line 2: = 25n^2 + 25n^2 + 25n + 25n + 25
Line 3: = 50n^2 + 50n + 25
Line 4: Which is not a multiple of 10.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (5n)^2 + (5n+5)^2
B: n^2 + (n+5)^2
C: (5n)^2 + (5n+1)^2
D: ((5n) + (5n+5))^2
ANSWER: A
WORKING:
Line 1: (5n)^2 + (5n+5)^2
Line 2: = 25n^2 + 25n^2 + 25n + 25n + 25
Line 3: = 50n^2 + 50n + 25
Line 4: Which is not a multiple of 10.
Question 6:
Answer the following questions.
Question 6(a):
Show algebraically that the sum of any 3 consecutive even numbers is always a divisible by 6.
Line 1: [ Missing line ]
Line 2: = 6n + 6
Line 3: = 6(n+1)
Line 4: Which is divisible by 6.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: 2n + (2n+2) + (2n+4)
B: 2n + (2n+1) + (2n+2)
C: n + (n+2) + (n+4)
D: (2n+1) + (2n+3) + (2n+5)
ANSWER: A
WORKING:
Line 1: 2n + (2n+2) + (2n+4)
Line 2: = 6n + 6
Line 3: = 6(n+1)
Line 4: Which is divisible by 6.
Question 6(b): [2 marks]
Prove algebraically that (4n+2)^2 - (2n+2)^2 is a multiple of 4 for all positive integers.
Line 1: = 16n^2 + 16n + 4 - 4n^2 - 8n - 4
Line 2: = 12n^2 + 8n
Line 3: [ Missing line ]
Line 4: Which is a multiple of 4.
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: =4(3n^2+2n)
B: =4(3n + 2)
C: =2(6n^2 + 3n)
D: =4(3n^2 - 2n)
ANSWER: A
WORKING:
Line 1: = 16n^2 + 16n + 4 - 4n^2 - 8n - 4
Line 2: = 12n^2 + 8n
Line 3: =4(3n^2+2n)
Line 4: Which is a multiple of 4.
Question 6(c): [2 marks]
Prove algebraically that (2n+3)^2 - (2n-3)^2 is a multiple of 4 for all positive integers.
Line 1: =4n^2+12n+9 - 4n^2+12n-9
Line 2: =24n
Line 3: [ Missing line ]
Line 4: Which is a multiple of 8
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: = 8(3n)
B: = 4(2n+1)
C: = 3(8n-1)
D: = 4(6n-1)
ANSWER: A
WORKING:
Line 1: =4n^2+12n+9 - 4n^2+12n-9
Line 2: =24n
Line 3: = 8(3n)
Line 4: Which is a multiple of 8
Question 7:
Answer the following questions.
Question 7(a): [2 marks]
If 2n is always even for all positive integer values of n, prove algebraically that the sum of the squares of any two consecutive even numbers is always a multiple of 4.
Line 1: [ Missing line ]
Line 2: = 4n^2 + 4n^2 +8n + 4
Line 3: = 4(2n^2 + 2n + 1)
Line 4: Which is always a multiple of 4.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (2n)^2 + (2n+2)^2
B: (2n)^2 + (2n+1)^2
C: (2n+1)^2 + (2n+3)^2
D: (2n) + (2n+2)
ANSWER: A
WORKING:
Line 1: (2n)^2 + (2n+2)^2
Line 2: = 4n^2 + 4n^2 +8n + 4
Line 3: = 4(2n^2 + 2n + 1)
Line 4: Which is always a multiple of 4.
Question 7(b): [2 marks]
Prove algebraically that the difference between the squares of any two consecutive numbers is always an odd number.
Line 1: [ Missing line ]
Line 2: = n^2 + 2n + 1 - n^2
Line 3: = 2n+1
Line 4: Which is odd.
What is the missing Line 1 from the working above?
Answer type: Multiple choice type 1
A: (n+1)^2 - n^2
B: (n+1)^2 - n
C: (n^2 +1) - n^2
D: (2n+1)^2 - n^2
ANSWER: A
WORKING:
Line 1: (n+1)^2 - n^2
Line 2: = n^2 + 2n + 1 - n^2
Line 3: = 2n+1
Line 4: Which is odd.
Question 8:
Answer the following questions.
Question 8(a): [1 mark]
Tom says that 7x - (2x+3)(x+2) is always negative.
Is he correct?
Answer type: Multiple choice type 1
A: Yes
B: No
ANSWER: A
WORKING:
7x-(2x+3)(x+2) = 7x - (2x^2 + 4x + 3x + 6) = -2x^2 - 6 = -(2x^2+6)(2x^2+6) is always positive, so multiplying by a negative means that the answer is always negative, so Tom is correct.
Question 8(b): [1 mark]
Changing a single number in Tom’s statement would lead to a change in your conclusion.
Choose the correct answer.
Answer type: Multiple choice type 1
A: Change +3 into a -3
B: Change 7 into a -7
ANSWER: A
Question 9:
Answer the following questions.
Question 9(a): [2 marks]
Show that the difference between 14^{20} and 21^2 is a multiple of 7.
Line 1: (7 \times 2)^{20} - (7 \times 3)^2
Line 2: = 7^{20} \times 2^{20} - 7^2 \times 3^2
Line 3: [ Missing line ]
Line 4: Therefore the answer is divisible by 7, so must be a multiple of 7.
What is the missing Line 3 from the working above?
Answer type: Multiple choice type 1
A: = 7(7^{19} \times 2^{20} - 7 \times 3^2)
B: = 7(7^{20} \times 2^{20} - 7^2 \times 3^2)
C: = 7(7^{18} \times 2^{20} - 7 \times 3^2)
D: = 7(7^{19} \times 2^{20} + 7 \times 3^2)
ANSWER: A
WORKING:
Line 1: (7 \times 2)^{20} - (7 \times 3)^2
Line 2: = 7^{20} \times 2^{20} - 7^2 \times 3^2
Line 3: = 7(7^{19} \times 2^{20} - 7 \times 3^2)
Line 4: Therefore the answer is divisible by 7, so must be a multiple of 7.
Question 9(b): [2 marks]
Determine whether 3^{60} - 25 is a prime number or not.
Answer type: Multiple choice type 1
A: It is not prime
B: It is prime
C: It is impossible to tell
ANSWER: A
WORKING:
3^{60} is always odd, because you are multiplying odd numbers. Also 25 is odd. So the difference between two odd numbers, i.e. subtracting them, is always even.
3^{60}-25 is going to be even. All even numbers are divisible by 2, so 3^{60}-25 is not prime.