Algebraic Proof: past paper questions & worked solutions

Let \(x=0.5\overline{39}=0.5393939\ldots\). Then \(10x=5.393939\ldots\) and \(1000x=539.393939\ldots\).

Shift past the non-recurring digit and then one complete two-digit recurring block.

Subtract: \[1000x-10x=539.393939\ldots-5.393939\ldots=534.\] \[990x=534\quad\Rightarrow\quad x=\frac{534}{990}=\frac{89}{165}.\]

Identical infinite decimal tails cancel exactly; reduce the resulting fraction by 6.

Then \[ 10x=3.787878\ldots \]

and \[ 1000x=378.787878\ldots \]

Subtract: \[ 990x=375 \]

Then \[ 10x=9.5454545\ldots \]

and \[ 1000x=954.5454545\ldots \]

Subtract: \[ 990x=945 \]

Then \[ 1000x=702.\dot7\dot0\dot2 \]

Subtract: \[ 999x=702 \]

Then \[ 10x=7.636363\ldots \]

and \[ 1000x=763.636363\ldots \]

Subtract: \[ 990x=756 \]

Then \[ 1000x=306.\dot3\dot0\dot6 \]

Subtract: \[ 1000x-x=306 \]

Then \[ 10x=1.767676\ldots \]

and \[ 1000x=176.767676\ldots \]

Subtract: \[ 990x=175 \]

Then \[ 10x=3.\dot4\dot5 \]

and \[ 1000x=345.\dot4\dot5 \]

Subtract: \[ 1000x-10x=345.\dot4\dot5-3.\dot4\dot5 \]

Use denominator 21: \[ \frac37=\frac9{21},\qquad \frac13=\frac7{21} \]

Since \(4\sqrt7=\sqrt{112}\), \[ -10-4\sqrt7=-10-\sqrt{112} \]

Since \(3\sqrt5=\sqrt{45}\), \[ 7-3\sqrt5=7-\sqrt{45} \]

Let the three consecutive multiples of 4 be \[ 4n,\quad 4n+4,\quad 4n+8 \]

The difference between the square of the largest and the square of the smallest is \[ (4n+8)^2-(4n)^2 \]

This is always a multiple of 64.

Then

So \(b^2=ac+4\).

This is in the form \(a+\sqrt a\), where \(a=3\).

So \[ (a-b)(a+b)=8\times4\sqrt2=32\sqrt2 \]

The question says this equals \(y\sqrt{4y}\). Try \(y=8\): \[ 8\sqrt{32}=8\times4\sqrt2=32\sqrt2 \]

Let the three consecutive whole numbers be \[ n,\quad n+1,\quad n+2 \]

The sum of the squares of the smallest and largest is \[ n^2+(n+2)^2=n^2+n^2+4n+4=2n^2+4n+4 \]

Twice the square of the middle number is \[ 2(n+1)^2=2(n^2+2n+1)=2n^2+4n+2 \]

So \[ 2n^2+4n+4=(2n^2+4n+2)+2 \]

Multiply by the conjugate: \[ \frac{2\sqrt3}{\sqrt3+2}\times \frac{2-\sqrt3}{2-\sqrt3} =\frac{2\sqrt3(2-\sqrt3)}{4-3} \]

Since \(4\sqrt3=\sqrt{48}\), \[ 4\sqrt3-6=-6+\sqrt{48} \]

Let the two consecutive odd numbers be \[ 2n+1 \quad \text{and} \quad 2n+3 \]

The sum of their squares is \[ (2n+1)^2+(2n+3)^2 \]

This is a multiple of 8 plus 2, so the remainder is always 2.

Let the three consecutive even numbers be \[ 2n-2,\quad 2n,\quad 2n+2 \]

The middle number is \(2n\).

The difference between the square of the largest and the square of the smallest is \[ (2n+2)^2-(2n-2)^2 \]

Since \(8\times\text{middle number}=8(2n)=16n\), the result is 8 times the middle number.

Let the two consecutive square numbers be \(n^2\) and \((n+1)^2\).

\(2n+1\) is one more than an even number, so it is always odd.

Since \(N\) is a multiple of 5, let

where \(k\) is an integer.

So

Expand:

Therefore

Substitute \(N=5k\):

Since \(k\) is an integer, \(20k\) is a multiple of 20.

So \(k=9\).

Since \(3\sqrt2=\sqrt{18}\), \[ 9+3\sqrt2=9+\sqrt{18} \]

So \(n=5\).

Factorise:

So

So \(n=-19\), which is an integer.

Let the two odd numbers be \[ 2m+1 \quad \text{and} \quad 2n+1 \]

\[ 2k+1 \] where \(k\) is an integer.

So the product is odd.

Then \[ 1000x=572.727272\ldots \]

and \[ 10x=5.727272\ldots \]

Subtract: \[ 990x=567 \]

So \[ 0.5\dot7\dot2=\frac{63}{110} \]

Then \[ 10x=3.727272\ldots \]

and \[ 1000x=372.727272\ldots \]

Subtract: \[ 990x=369 \]

Rationalise: \[ \frac{9\sqrt5}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}=\frac{9\sqrt{15}}{3}=3\sqrt{15} \]

Then \[ 100x=457.\dot5\dot7 \]

Subtract: \[ 100x-x=457.\dot5\dot7-4.\dot5\dot7 \]

So this is in the form \(\dfrac{a+\sqrt a}{b}\), with \(a=2\) and \(b=3\).

The perimeter is made from three sides of length \(x\) and two sides of length \(y\): \[ 3x+2y=100 \]

So \[ y=\frac{100-3x}{2} \]

The area is the area of the rectangle plus the area of the equilateral triangle: \[ R=xy+\frac{\sqrt3}{4}x^2 \]

Substitute \(y=\dfrac{100-3x}{2}\): \[ R=x\left(\frac{100-3x}{2}\right)+\frac{\sqrt3}{4}x^2 \]

For the maximum, write \[ R=-\frac{6-\sqrt3}{4}x^2+50x \]

The coefficient of \(x^2\) is negative, so the quadratic graph opens downwards and this vertex gives a maximum.

First expand the two brackets: \[ (2y+5)(y+7)=2y^2+14y+5y+35=2y^2+19y+35 \]

Now multiply by \(3y\): \[ 3y(2y^2+19y+35)=6y^3+57y^2+105y \]

So \(a=6\), \(b=57\), and \(c=105\).

Multiply by 100: \[ 20(2x+3)+25(6x-5)=163 \]