Proof(aqa math) for rag rating, Exercises of Mathematics

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A - Proof MS

Time: 96:00 Marks: 80

R < 38 < A < 58 < G

Grade | Minimum Marks

A* | 69

A | 58

B | 48

C | 38

D | 28

E | 18

Topics covered:

• Proof (Y1)
• Proof (Y2)

Proof (Y1) MS

Time: 48:00 Marks: 40

R < 19 < A < 29 < G

Grade | Minimum Marks

A* | 34

A | 29

B | 24

C | 19

D | 14

E | 9

AQA AS Maths: Problem solving 1 © MEI 12/03/ Exercise level 3 solutions page 2 of 3 integralmaths.org

Then + + + =

• = =

n n n n

so 7 + 8 + 9 + 10 + 11 = 45

Suppose 45 can be written as the sum of 6 consecutive integers: Then + + + =

• = =

n n n n

so 5 + 6 + 7 + 8 + 9 + 10 = 45

Suppose 45 can be written as the sum of 7 consecutive integers: Then + + + =

• =

n n n This does not have an integer solution. Suppose 45 can be written as the sum of 8 consecutive integers: Then +^ +^ +^ =

• =

n n n This does not have an integer solution. Suppose 45 can be written as the sum of 9 consecutive integers: Then + + + =

• = =

n n n n

so 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

45 can’t be written as a sum of 10 consecutive integers as the lowest possible sum is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 55

So, there are 5 different ways of writing 45 as the sum of consecutive integers.

The two triangles are similar as they are both right-angled and they share a common angle. Similar triangles have corresponding sides in the same ratio.

So, =  =  =

x (^) x

x

AQA AS Maths: Problem solving 1 © MEI 12/03/ Exercise level 3 solutions page 3 of 3 integralmaths.org

1. Suppose n is a positive integer which leaves a remainder of 11 when divided into 1000.

Then kn + 11 = 1000 where k is an integer.

So, kn = 989

The only factors of 989 are 1, 23, 43 and 989

so 989 can be written as 1  989 or 23  43

n cannot be 1 as 1000  1 = 1000 with no remainder.

n can be 23, 43 or 989 So there are three possible positive integers.

AQA AS Maths: Problem solving 2 © MEI 12/03/ Exercise level 3 solutions page 2 of 2 integralmaths.org

If b = 6, b ² = 36 so a can only be zero which means it is not a 2 digit number. If b = 7, b ² = 49 so a must be 8. 87 does not work as 8 + 7² is not 87 If b = 8, b ² = 64 so a must be 4. 48 does not work as 4 + 8² is not 48 If b = 9, b ² = 81 so a must be 8. 89 works as 8 + 9² = 89 So the only number that satisfies the criteria is 89.

1. (a)( )

( )

(^2 2 ) 2

b c b bc c b c b c (b) Any positive integer can be written as 10 b + c for some integers b and c , and the expression for the square becomes 10 k + c^2 so the final digit of the square number is determined only by c^2. For different values of c , you get: 12 = 1, 2^2 = 4, 3^2 = 9, 4^2 = 10 + 6, 5^2 = 20 + 5, 6^2 = 30 + 6, 72 = 40 + 9, 8^2 = 60 + 4, 9^2 = 80 + 1, 0^2 = 0 So square numbers end in 1, 4, 5, 6, or 9, and never in 2, 3, 7, or 8.

Proof (Y2) MS

Time: 48:00 Marks: 40

R < 19 < A < 29 < G

Grade | Minimum Marks

A* | 34

A | 29

B | 24

C | 19

D | 14

E | 9

AQA A level Proof topic assessment solutions

3 of 4 03/08/21 © MEI integralmaths.org

1. Method 1: All odd numbers can be written in the form 2 n + 1, n 

( ) ( )

n n n n n

n and n + 1 are two consecutive numbers, therefore one of them divides by 2

Therefore, (^4) n n ( + (^1) )divides by 8 Therefore, 4 n n ( + (^1) ) + 1 has remainder 1 when divided by 8. The square of an odd number has remainder 1 when divided by 8.

Method 2: All odd number can be written either in the form 4 n + 1, n  or the form 4 n − 1, n  ( ) ( )

(^2 ) 2

n n n n n 8 2( n^2 + n )+ 1 has remainder 1 when divided by 8.

( ) ( )

(^2 ) 2

n n n n n 8 2( n^2 − n )+ 1 has remainder 1 when divided by 8. The square of an odd number has remainder 1 when divided by 8. [5]

6. Any integer n either has one digit or it can be written in the form 10 a + b where

a b , are integers and b is the final digit of n.

( +^ ) =^ ( +^ )+

(^2 ) 10 a b 10 10 a 2 ab b

Therefore, the final digit of n^2 is the final digit of b^2 where b is the final digit of

n.

The possible values of b are 0,1,2,3, 4,5,6,7,8,9.

2 2 2 2 2 2 2 2 2 2 0 0 1 1 2 4 3 9 4 16 5 25 6 36 7 49 8 64 9 81

AQA A level Proof topic assessment solutions

4 of 4 03/08/21 © MEI integralmaths.org

The only possible final digits of n^2 are 0,1, 4,5,6,9. Therefore, the final digit of a square number cannot be 3. [6]

1. Suppose that log 2 3 is rational. Then it can be written as a fraction m n where m n , are integers. =

=

log 2 3

3 2 3 2

m n m n

m n

But this is a contradiction because 3 m is an odd number and 2 n is an even number. Therefore, log 2 3 is irrational. [6]

1. (a) Assume 5 is rational. Then it can be written as a fraction m n where m n , are integers with no common factor. =

=

2 2 2 2

m n m n n m m^2 has a factor of 5. Since 5 is a prime number and it is a factor of^ m^2 , 5 must

be a factor of m. We can therefore write m as 5 q where q is an integer

=( )

=

2 2 2 2 2 2

n q n q n q n^2 has a factor of 5. Since 5 is a prime number and it is a factor of^ n^2 , 5 must

be a factor of n.

m and n have a common factor of 5, which contradicts the assumption that

they have no common factors. Therefore, 5 is irrational. [6]

(b) Following the same method of proof gives 4 n^2^ = m^2 , so m^2 has a factor of 4.

However, this does not imply that m has a factor of 4, only that m has a factor

of 2 because 4 = 22. Therefore the proof fails and we cannot conclude that 4 is irrational. [2]

Total 40 marks