Question difficulty level:
Basic
Moderate
Challenging
Difficult
Genius :)
Basic
Moderate
Challenging
Difficult
Genius :)
Paper 1 – No Calculator
[Maximum mark: 4]
Show that the difference of the squares of two consecutive positive integers is equal to the sum of the integers.
Let’s represent the integers as n and n + 1
For the difference of the squares of the integers we get:
(n + 1)^{2} - n^{2} = (n^{2} + 2n + 1) - n^{2} =
= n^{2} + 2n + 1 - n^{2} = 2n + 1
For the sum of the integers we get:
n + (n + 1) = 2n + 1
Since 2n + 1 = 2n + 1, the difference of the squares of two consecutive positive integers is equal to the sum of the integers.
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Paper 1 – No Calculator
[Maximum mark: 6]
Consider three consecutive positive integers.
Show that
(a) the sum of these integers is divisible by 3.
[2]
(b) the sum of the squares of these integers is never divisible by 3.
[4]
Let’s represent the integers as n, n + 1, and n + 2
(a)
The sum of the integers is:
n + (n + 1) + (n + 2) = 3n + 3
Factorising, we get that the sum is 3(n + 1), therefore the sum of three consecutive integers is divisible by 3.
(b)
The sum of the squares of the integers is:
n^{2} + (n + 1)^{2} + (n + 2)^{2} =
= n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4)
= 3n^{2} + 6n + 5
Factorising 3 from the first two terms we get:
3(n^{2} + 2n) + 5
Since 3(n^{2} + 2n) is divisible by 3 while 5 is not divisible by 3, the sum of the squares of three consecutive integers is never divisible by 3.
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Paper 1 – No Calculator
[Maximum mark: 4]
Consider two consecutive integers, n and n + 1.
Show that the sum of the squares of these integers is odd.
The sum of the squares of the integers is:
n^{2} + (n + 1)^{2} = n^{2} + n^{2} + 2n + 1 =
= 2n^{2} + 2n + 1 = 2(n^{2} + n) + 1
Therefore the sum of these integers is odd.
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Paper 1 – No Calculator
[Maximum mark: 6]
Four consecutive integers are n, n+1, n+2 and n+3.
(a) Prove that the sum of these integers is an even number.
[2]
(b) Prove that the sum of the squares of these integers is not divisible by 4.
[4]
(a)
Finding the sum of the integers:
n + n + 1 + n + 2 + n + 3 = 4n + 6
Factorising we get 2(2n + 3), therefore the sum of these four integers is divisible by 2, so it is an even number.
(b)
Finding the sum of the squares:
n^{2} + (n + 1)^{2} + (n + 2)^{2} + (n + 3)^{2}
n^{2} + (n^{2} + 2n + 1) + (n^{2} + 4n + 4) + (n^{2} + 6n + 9)
4n^{2} + 12n + 14
4n^{2} + 12n + 12 + 2 = 4(n^{2} + 3n + 3) + 2
Since 4(n^{2} + 3n + 3) is divisible by 4, 4(n^{2} + 3n + 3) + 2 is not divisible by 4.
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Paper 1 – No Calculator
[Maximum mark: 3]
Consider two positive integers, k and k + 2.
Show that the sum of these integers is even.
The sum of the integers is:
k + (k + 2) = 2k + 2 = 2(k + 1)
Therefore the sum of these integers is even.
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Paper 1 – No Calculator
[Maximum mark: 5]
Prove that the mean of four consecutive positive integers is equal to the median of these integers.
Let’s represent the integers as n, n + 1, n + 2, and n + 3
The mean of the integers can be written as:
\frac{n +(n + 1)+(n + 2)+(n + 3)}{4} = \frac{4n + 6}{4} = \frac{2n + 3}{2}
The median of the four integers is:
\frac{(n + 1)+(n + 2)}{2} = \frac{2n + 3}{2}
Since \frac{2n + 3}{2} = \frac{2n + 3}{2}, the mean of four consecutive positive integers is equal to the median of these integers.
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