[discia] GCSE Maths › Algebra › Proof › Practise
  1. Either paper 3 marks

    \(k = n^2 + 7n + 23\)

    Idris says, "\(k\) will be a prime number for all integer values of \(n\) from 1 to 8."

    Show that Idris is wrong.

    You must show that your value of \(k\) is not prime.

    Show answer
    for example \(n = 7\) gives \(k = 121 = 11 \times 11\), so \(k\) is not prime (any \(n\) from 1 to 8 with a non-prime \(k\) shown by a factor pair is also correct)

    Substitute n = 1, 2, 3, ... into \(k = n^2 + 7n + 23\) until k is not prime.

    \(n = 1\): \(k = 1 + 7 + 23 = 31\), which is prime.

    \(n = 2\): \(k = 4 + 14 + 23 = 41\), which is prime.

    Carrying on in the same way, every value of k up to \(n = 6\) is prime; at \(n = 6\), \(k = 101\).

    \(n = 7\): \(k = 49 + 49 + 23 = 121\).

    \(121 = 11 \times 11\), so \(121\) has factors other than 1 and itself.

    So \(k\) is not prime when \(n = 7\), and Idris is wrong.

    • M1substitutes values of n from 1 to 8 into k, in any order, e.g. n = 1 gives k = 31
    • M1reaches a composite value of k for some n from 1 to 8, e.g. n = 7 gives k = 121
    • A1a factor pair for that value, so k is not prime, e.g. \(121 = 11 \times 11\)
  2. Either paper 3 marks

    Here is the term-to-term rule for a sequence.

    Multiply by 2, then add 10

    The first three terms of the sequence are

    \(a + 2\) \(2a + 14\) \(4a + 38\)

    Show that the sum of the first four terms is a multiple of 5.

    Show answer
    The fourth term is \(8a + 86\). The sum of the four terms is \(15a + 140\), which is \(5(3a + 28)\), so the sum is a multiple of 5.

    Use the rule to find the fourth term: multiply the third term by 2, then add 10.

    \(2(4a + 38) + 10 = 8a + 86\)

    Add the four terms and collect like terms.

    \((a + 2) + (2a + 14) + (4a + 38) + (8a + 86) = 15a + 140\)

    Both terms are multiples of 5, so take 5 out as a factor.

    \(15a + 140 = 5(3a + 28)\)

    The sum is 5 multiplied by an expression that is a whole number whenever \(a\) is, so it is a multiple of 5.

    • M1fourth term \(8a + 86\)
    • M1sum of the four terms \(15a + 140\)
    • A1factorises as \(5(3a + 28)\) and concludes it is a multiple of 5
  3. Either paper 4 marks
    (a)

    The smallest of eleven consecutive integers is \(k\).

    Write an expression, in terms of \(k\), for the sum of the eleven integers. Give your answer in its simplest form.

    (b)

    Prove that the sum of any eleven consecutive integers is always a multiple of 11.

    Show answer
    (a) \(11k + 55\)
    (b) \(11k + 55 = 11(k + 5)\). \(k + 5\) is a whole number, so the sum is 11 times a whole number, which is what being a multiple of 11 means.

    The eleven integers are \(k\), \(k + 1\), \(k + 2\) and so on, up to \(k + 10\).

    Adding them gives eleven lots of \(k\), plus the numbers 1 to \(10\), which add to \(55\):

    \(11k + 55\)

    Take the expression from part (a) and factorise the 11 out.

    \(11k + 55 = 11(k + 5)\)

    \(k\) is a whole number, so \(k + 5\) is a whole number too. The sum is 11 multiplied by a whole number, and that is exactly what "a multiple of 11" means.

    The factorised line alone is not the proof. Saying why it proves the claim — that the bracket is a whole number — is what earns the second mark.

    • (a) M1adding the integers \(k\) to \(k + 10\)
    • (a) A1\(11k + 55\)
    • (b) M1factorising \(11k + 55 = 11(k + 5)\)
    • (b) C1bracket is a whole number, so the sum is 11 times a whole number
  4. Either paper 4 marks
    (a)

    Two whole numbers differ by 11. The smaller of them is \(k\).

    Multiply out the brackets and simplify \((k + 11)^2 - k^2\).

    (b)

    Prove that the difference between the squares of any two whole numbers that differ by 11 is always a multiple of 11.

    Show answer
    (a) \(22k + 121\)
    (b) \(22k + 121 = 11(2k + 11)\). \(2k + 11\) is a whole number, so the difference is 11 times a whole number.

    Expand the bracket first.

    \((k + 11)^2 = k^2 + 22k + 121\)

    Now subtract \(k^2\). The two squared terms cancel:

    \(k^2 + 22k + 121 - k^2 = 22k + 121\)

    Factorise the answer to part (a).

    \(22k + 121 = 11(2k + 11)\)

    \(k\) is a whole number, so \(2k + 11\) is one as well. The difference is therefore 11 multiplied by a whole number, which makes it a multiple of 11.

    • (a) M1expanding \((k + 11)^2 = k^2 + 22k + 121\)
    • (a) A1\(22k + 121\)
    • (b) M1factorising \(22k + 121 = 11(2k + 11)\)
    • (b) C1\(2k + 11\) is a whole number, so the difference is 11 times a whole number
  5. Either paper 4 marks
    (a)

    Expand and simplify \(r(r + 16) + 64\).

    (b)

    Hence prove that \(r(r + 16) + 64\) is a square number for every whole number \(r\).

    Show answer
    (a) \(r^2 + 16r + 64\)
    (b) \(r^2 + 16r + 64 = (r + 8)^2\), and \(r + 8\) is a whole number, so the expression is the square of a whole number.

    Multiply out the bracket, then add the \(64\).

    \(r(r + 16) = r^2 + 16r\)

    \(r^2 + 16r + 64\)

    Factorise the answer to part (a). It is a perfect square trinomial: the constant \(64\) is \(8^2\), and the middle coefficient \(16\) is \(2 \times 8\).

    \(r^2 + 16r + 64 = (r + 8)^2\)

    \(r\) is a whole number, so \(r + 8\) is too, and the expression is that whole number squared — a square number.

    • (a) M1multiplying out \(r^2 + 16r\)
    • (a) A1\(r^2 + 16r + 64\)
    • (b) M1factorising to \((r + 8)^2\)
    • (b) C1\(r + 8\) is a whole number, so the expression is a whole number squared
  6. Either paper 4 marks
    (a)

    Factorise \(m^2 + 3m\).

    (b)

    Prove that \(m^2 + 3m\) is always even when \(m\) is a whole number.

    Show answer
    (a) \(m(m + 3)\)
    (b) \(m^2 + 3m = m(m + 3)\). The two factors differ by 3, which is odd, so one of them is even and the other is odd. A product with an even factor is even.

    \(m\) is a factor of both terms.

    \(m^2 + 3m = m(m + 3)\)

    Use the factorised form from part (a): \(m(m + 3)\).

    The two factors are \(m\) and \(m + 3\), and they differ by 3, which is an odd number. Adding an odd number to a whole number changes it from even to odd, or from odd to even — so whichever \(m\) happens to be, exactly one of the two factors is even.

    A product with an even factor is even, so \(m^2 + 3m\) is always even.

    • (b) M1uses the factorised form \(m(m + 3)\)
    • (b) M1the factors differ by 3, an odd number, so one of them is even
    • (b) A1a product with an even factor is even
  7. Either paper 4 marks

    \(m\) and \(n\) are integers.

    (a)

    Prove that \((2m+1)^2 - (2n-1)^2 = 4(m+n)(m-n+1)\)

    (b)

    Marcus says the result in part (a) shows that the difference of the squares of any two odd numbers must be a multiple of 16. Is Marcus correct? You must give reasons for your answer.

    Show answer
    (a) \((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n = 4(m+n)(m-n+1)\)
    (b) No, Marcus is not correct: \(3\) and \(1\) are both odd, and \(3^2 - 1^2 = 9 - 1 = 8\), which is not a multiple of 16.

    Expand each square:

    \((2m+1)^2 = 4m^2 + 4m + 1\)

    \((2n-1)^2 = 4n^2 - 4n + 1\)

    Subtract:

    \((2m+1)^2 - (2n-1)^2 = 4m^2 + 4m + 1 - (4n^2 - 4n + 1) = 4m^2 + 4m - 4n^2 + 4n\)

    Factorise fully:

    \(4m^2 + 4m - 4n^2 + 4n = 4(m^2 + m - n^2 + n) = 4(m+n)(m-n+1)\)

    Part (a) shows that \((2m+1)^2 - (2n-1)^2 = 4(m+n)(m-n+1)\) for any integers \(m\) and \(n\).

    Taking \(m=1\) and \(n=1\) gives the odd numbers \(2m+1=3\) and \(2n-1=1\), and \(3^2 - 1^2 = 9 - 1 = 8\).

    But \(8\) is not a multiple of 16, and Marcus is not correct.

    • (a) M1expands both squares
    • (a) M1simplifies to \(4m^2 + 4m - 4n^2 + 4n\)
    • (a) A1factorises to \(4(m+n)(m-n+1)\)
  8. Either paper 1 mark

    \(a\) is a prime number and \(b\) is an even number.

    \(n = 5a + b + 7\)

    Which statement about \(n\) must be true? Write down the correct statement.

    A: always odd

    B: always prime

    C: always even

    D: could be even or odd

    Show answer
    could be even or odd

    The term \(b\) is even, so it never changes the parity of \(n\).

    At \(a = 2\), the only even prime: \(n = 17 + b\), and \(17\) is odd.

    At an odd prime such as \(a = 3\): \(n = 22 + b\), and \(22\) is even.

    Both parities occur, so n could be even or odd.

  9. Either paper 4 marks

    Here is a quadrilateral \(ABCD\). The diagonal \(AC\) is drawn and \(DA = DC\).

    Angle \(BAC = 54^\circ\), angle \(BCA = 49^\circ\) and angle \(CAD = 45^\circ\).

    ABCD

    Prove that \(ABCD\) is not a cyclic quadrilateral.

    You must give a reason for each angle you work out.

    Show answer
    Angle \(ABC = 77^\circ\) and angle \(ADC = 90^\circ\), which add to \(167^\circ\), less than \(180^\circ\). Opposite angles do not sum to \(180^\circ\), so \(ABCD\) is not cyclic.

    In triangle \(ABC\), the angles add up to \(180^\circ\):

    \(\angle ABC = 180 - 54 - 49 = 77^\circ\)

    Triangle \(ACD\) is isosceles with \(DA = DC\), so the base angles are equal (angles in an isosceles triangle):

    \(\angle ACD = \angle CAD = 45^\circ\)

    The angles in triangle \(ACD\) add up to \(180^\circ\):

    \(\angle ADC = 180 - 45 - 45 = 90^\circ\)

    \(B\) and \(D\) are opposite angles:

    \(\angle ABC + \angle ADC = 77 + 90 = 167^\circ\)

    The opposite angles of a cyclic quadrilateral sum to \(180^\circ\). Here the sum is less than \(180^\circ\), so \(ABCD\) is not a cyclic quadrilateral.

    (Using \(A\) and \(C\) instead: \(\angle BAD = 54 + 45 = 99^\circ\) and \(\angle BCD = 49 + 45 = 94^\circ\), which add to \(193^\circ\), also not \(180^\circ\).)

    • M1angle \(ABC = 180 - 54 - 49 = 77^\circ\), angles in a triangle sum to \(180^\circ\) (or angle \(BAD = 99^\circ\) by adding angles at \(A\))
    • M1angle \(ACD = 45^\circ\) as \(DA = DC\) (base angles of an isosceles triangle), so angle \(ADC = 180 - 2 \times 45 = 90^\circ\) (or angle \(BCD = 94^\circ\))
    • M1sum of the opposite pair, \(77 + 90 = 167^\circ\) (or \(99 + 94 = 193^\circ\)), compared with \(180^\circ\)
    • A1stated that opposite angles do not sum to \(180^\circ\) (for \(B\) and \(D\), less than \(180^\circ\)), so not cyclic
  10. Either paper 4 marks
    (a)

    \(2p - 3\) is an odd number for every whole number \(p\).

    Expand and simplify \((2p - 3)^2\).

    (b)

    Prove that the square of any odd number is odd.

    Show answer
    (a) \(4p^2 - 12p + 9\)
    (b) \(4p^2 - 12p + 9 = 2(2p^2 - 6p + 4) + 1\), which is two times a whole number plus one, so it is odd.

    Squaring a bracket means multiplying it by itself, so there are two middle terms and they are the same.

    \((2p - 3)^2 = 4p^2 - 12p + 9\)

    An even number is 2 times a whole number, and an odd number is 2 times a whole number plus 1. So the job is to write part (a) in that shape.

    \(4p^2 - 12p + 9 = 2(2p^2 - 6p + 4) + 1\)

    Everything inside the bracket is a whole number, so the whole expression is 2 times a whole number, plus 1. That is the definition of an odd number, so the square of an odd number is always odd.

    • (a) M1multiplies the bracket by itself, with a correct term \(4p^2\) or \(9\)
    • (a) A1\(4p^2 - 12p + 9\)
    • (b) M1factors out 2 from the terms: \(2(2p^2 - 6p + 4) + 1\)
    • (b) C1states it is 2 times a whole number plus 1, so it is odd