Year 7 Mock Landmark – Autumn 2
Sequences (Calculator Allowed)
1. A sequence is:
6, 10, 14, 18, …
a. Find the expression for the 𝑛th term of the sequence.
b. Find the 50th term of the sequence.
c. Is 648 in this sequence? Explain your answer.
2. A sequence is:
11, 8, 5, 2, …
Find the expression for the nth term of the sequence.
3. A sequence starts:
2, 7, …
Which of the following could be possible nth term formulae for the sequence:
(i) 5𝑛 − 3
(ii) 𝑛3 + 1
(iii) 𝑛2 + 2𝑛 − 1
4. (a) Find the values of 21 , 22 , 23 , 24 and 25 .
(b) Hence or otherwise, determine the units digit of 2250
5. A sequence starts with 31. Thereafter, the following rule is used to get the next term in the
sequence:
“If the last number in the sequence is odd, add 3 to get the next number. If it is even, divide
it by 2 to get the next”.
a. Determine the first 12 terms of the sequence (starting with 31).
b. (Difficult) Determine the 100th term of the sequence.
Number Theory (Calculator Not Allowed)
1. (a) Express 504 as a product of its prime factors.
(b) Hence determine how many factors 504 has.
2. Express 100 as the sum of:
a. three different primes
b. two squares
3. For the numbers 24 and 30, determine:
a. The Lowest Common Multiple
b. The Highest Common Factor
4. (a) Find the prime factorisation of 180.
(b) Given also that 378 = 2 × 33 × 7, determine the lowest common multiple and highest
common factor of 180 and 378. You may leave your answers in prime factorised form.
5. A four-digit numbers consists of the digits "457𝑎". Give the possible values of the digit 𝑎 if:
a. The number is divisible by 3.
b. The number is divisible by 4.
c. The number is divisible by 6.
d. The number is divisible by 11.
6. A palindromic number is a number that reads the same when the order of its digits is
reversed. What is the difference between the largest and smallest five-digit palindromic
numbers that are both multiples of 45?
7. Given that 27 × 5 × 72 = 31360, determine the smallest number we need to multiply
31360 by to obtain a square number.
8. (Difficult) An integer has 8 factors and is only allowed to have 2 and/or 3 as its prime factors.
How many integers have this property?
Area/Perimeter (Calculator Allowed)
1. Find the following areas, giving your answers to 2 decimal places where appropriate4 (and
ensuring you give units).
2. A farmer wishes to plant crops on his field, which is in the shape indicated in the below
diagram. Each packet of seed costs £3.75 and each box will covers 50m2 of land. Calculate
the cost to seed his land. (Note: he can only buy a whole number of boxes)
3. Calculate the perimeter of the following shapes, giving your answer in exact form.
4. Calculate the fraction of the square shaded.
5. (Difficult) The diagram consists of a circle and four quarter circles. Find the area of the
shaded region, in terms of 𝜋.
Solutions:
Sequences
1. a) 4𝑛 + 2
b) 202
c)
648−2
4
= 161.5 which is not a whole number, so 648 is not in the sequence.
2. 14 − 3𝑛 or −3𝑛 + 14
3. (i) Yes, as 5 × 1 − 3 = 2 and 5 × 2 − 3 = 7
(ii) No, as 13 + 1 = 2 but 23 + 1 = 9 which is not 7.
(iii) Yes, as 12 + (2 × 1) − 1 = 2 and 22 + (2 × 2) − 1 = 7
4. 21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32
We can see units digits go in repeating sequence 2, 4, 8, 6. Remainder when 250 is divided
by 4 is 2, so units digit must be 4.
5. a) 31, 34, 17, 20, 10, 5, 8, 4, 2, 1, 4, 2, …
b) Starts repeating from 4, 2, 1, … The 8th term is 4, so we can go a multiple of 3 forward and
have the same term. 98th term is also 4, so 100th term is 1.
Number Theory
1. a) 504 = 23 × 32 × 7
b) 4 × 3 × 2 = 24 factors
2. a) One of the numbers has to be 2 (as primes can’t all be odd as they’d add to an odd
number). One possibility 79 + 19 + 2
b) 64 + 36 = 100
3. a) 120
b) 6
4. a) 180 = 22 × 32 × 5
b) 378 = 2 × 33
×7
Thus 𝐻𝐶𝐹(180,378) = 2 × 32 = 18, 𝐿𝐶𝑀(180,378) = 22 × 33 × 5 × 7
5. a) 𝑎 = 2, 5, 8 (So that digits add to multiple of 3)
b) 𝑎 = 2, 6 (So that last two digits divisible by 4)
c) 𝑎 = 2, 8 (same as part (a), but last digit must also be even)
d) 𝑎 = 6 (so that (4 + 7) − (5 + 6) = 0 which is a multiple of 11)
6. If divisible by 45, then divisible by 9 and 5. If divisible by 5, each must end with 5 or 0. But
given it’s a palindromic number, if it ends in a 0, it has to start with a 0 (which is not
permitted). So we must have "5𝑎𝑏𝑎5" for both numbers.
Largest number is where 𝑎 = 9. Given that 5 + 9 + 9 + 5 = 28, middle digit 𝑏 must be 8 in
order to make the number divisible by 9. So 59895 is the largest number.
Similarly smallest is 50𝑏05. 𝑏 must be 8 to make the digits add to 9.
Difference: 59895 − 50905 = 8990
7. Powers need to be even for number to be a square, so we need an extra 2 and 5. i.e. we
need to multiply be 10.
8. Suppose Number is of form 2𝑎 × 3𝑏 . There are 8 factors, thus adding 1 to each of 𝑎 and 𝑏
and then multiplying these gives 8. The only numbers that multiple to give 8 are 2 × 4 and
4 × 2 (there’s also 8 × 1 and 1 × 8, but if we then subtracted one, one of the powers would
be 0).
Thus possibilities are 21 × 33 = 54 and 23 × 31 = 24.
However we could also just have 37 or just have 27 , each of which would have 8 factors.
Area/Perimeter
1. a) 27.5 cm2
b) 56 cm2
c) 21 cm2
d) 25𝜋 = 78.54 cm2
e) 25 −
25𝜋
4
= 5.37 cm2
f) 48 m2
2. Area = (80 × 60) +
𝜋×302
2
− 5506.858 m2
5506.858 ÷ 50 = 110.1 so 111 boxes required.
111 × 3.75 = £416.25
3. a) 12 × 𝜋 = 12𝜋
b)
4.
3
16
2×𝜋×4
2
+ 4 + 12 = 4𝜋 + 16
(You can add appropriate lines to divide into 16 congruent triangles)
5. Each of the four shaded areas can be split into 2.
Each of these 8 shapes are a quarter circle with a triangle cut out.
Therefore area of each shape:
𝜋×22
4
−
2×2
2
= 𝜋−2
Therefore shaded area:
8𝜋 − 16