Brain Teasers ? or 50 Years On........... ?

So, can anybody recall how to convert a recurring decimal back to its proper fraction ?

Hi Jan-Erik,

Nice link and I'm sure that many people will have had horrible exciting flash-backs to school days

My own method is very similar, but possibly more consistent regardless of the type of recurrence. So using two of the examples in the above link:-

0.004545...

• Let r= 0.0045...
• 10000r = 45.4545...     (1)
•     100r = 0.4545...        (2)
• (1) - (2) 9900r = 45.0
• r = 45/9900

which reduces to 1/220

  0.3777...

• Let r= 0.3777...
• 100r = 37.777...     (1)
• 10r = 3.777...          (2)
• (1) - (2) 90r = 34
• r = 34/90

which reduces to 17/45

The basic principle is to:-

• Let r = the recurrent decimal
• generate TWO multiples of "r" eg 10,000r and 100r such that the first multiple puts one (and only one) whole recurrence in front of the decimal point and the second multiple starts the whole recurrence immediately after the decimal point.
• Subtract the second from the first then divide and cancel as far as possible.

Apologies for not using the "dot above" to denote the recurrence. Let's just say "Hoopla"

The diagram shows a mosaic tile layout in its first four stages of sequential development.

Task 1. Find an expression for the number of RED tiles in the "n"th layout.

4(n - 1) + 1

Well done sj.

You could simplify your formula just a little to read 4n-3, well done.

Anybody want to provide an expression for the TOTAL number of squares in the "n"th Layout ?

Don

Thought it was about time I had another go at one of your teasers.

I think the total number of blue tiles  is given by: 2* (n-1) * (n-1)

So the total number of tiles is given by: (4n-3) + 2 (n-1) (n-1)

Regards,

Peter

x = 2n∧2 - 1

i.e. the total squares for a complete square mosaic = (2n - 1)∧2

Less areas of the 2 blank sides = 2(n - 1)∧2

SJ - I think we agree. But I do prefer your method of subtraction, rather than my method of addition. If I simplify my expression, I also get the result of 2* n^2 -1

Hi Peter,

I expected that they should result in the same expression as the number of blue blocks would be the same as the number of blocks which make up the blank area.

Steve

You guys are too good !

My (simple) way (for the first one) was:-

          Layout         n=   1    2    3     4

The RED numbers increase      1    5    9    13

the common differences are      4     4    4    so 4n is in the formula and it is a linear sequence

          4n=   4    8   12   16

                                     term is     1    5     9    13  so you have to subtract 3 (from the 4n values) to get the term

so the required expression is 4n - 3

Hope this makes sense. I've tried to line the numbers up vertically to show what I am doing, but Hoopla won't let me !

My solution for the TOTAL number of tiles is:-

Layout Number      n = 1     2     3     4
No. of tiles          1     7    17   31
Difference          6    10    14
Second Difference           4      4
Second differences are constant hence a Quadratic n²
And the co-efficient of “n” is "Second Difference"÷2 ie 4÷2 = 2 hence 2n²

No of tiles          1     7     17     31
2n²               2     8     18     32

(2n² – 1 gives the TOTAL number of tiles in each succeeding layout

Oh dear, I can't get the numbers to line up again !

And finally, (on this specific task !)

I have just created two consecutive tile layouts. I used 414 red tiles in total.

Which two layouts have I produced ?

Hopefully the numbers line up and make more sense !

Hopefully a bit more clear !

If I understand the question correctly then n = 52 and n = 53 are the 2 consecutive layouts.

i.e. 4 * 52 = 208 - 3 = 205

and 4 * 53 = 212 - 3 = 209

205 + 209 = 414 red tiles

Let n = the first layout

As there is a difference of 4 blocks between the 2 layouts,  4n - 3 = (414 - 4) /2

4n - 3 = 205 so n = 202/4 = 52 and the second layout is n + 1 = 53.

Hi SJ,

Very neat solution you have there. I like it !

Mine is set out above although it is a bit boring because it's all algebra !

x³ - 3x - 1 = 0

x = ?

Don Atkinson posted:

x³ - 3x - 1 = 0

x = ?

The method I used is called irritation iteration 

Don

I started with the iteration method - but then thought there must be another way. Google to the aid !

By substituting x=w+1/w you end up with solving a quadratic. Only "problem" is that the quadratic has imaginary roots ! But I managed to follow through the method and got the following 3 solutions:

x = 2 Cos 20 ( = approx. 1.87938)

x = 2 Cos 140 ( = approx. -1.532)

x = 2 Cos 260 ( = approx. -0.3473)

Maybe something easier next time

Regards,

Peter

sophiebear0_0 posted:

Don

I started with the iteration method - but then thought there must be another way. Google to the aid !

By substituting x=w+1/w you end up with solving a quadratic. Only "problem" is that the quadratic has imaginary roots ! But I managed to follow through the method and got the following 3 solutions:

x = 2 Cos 20 ( = approx. 1.87938)

x = 2 Cos 140 ( = approx. -1.532)

x = 2 Cos 260 ( = approx. -0.3473)

Maybe something easier next time

Regards,

Peter

 

Peter, you did really well. In fact, that was brilliant.

if you had followed the iteration method you would probably have only found the -1.532 solution. Well done.

These are only meant to be current GCSE questions, so they shouldn't be too hard, as I said earlier, kids these days have things easy.............or do they?

easier one next time

Oh, on second thoughts you would also have found the other two solutions as well because there is a change of sign when substituting for values of "x" between -2 and -1; -1 and 0; and between 1 and 2.

I cheated and used Excel with values of x between -5 and +5 to find the approximate locations of changes from positive to negative, and then to iterate in those regions.

Naim Audio moved into Direct Sales for one week only leading up to Xmas. In preparation, they manufactured “X” 250DR power amps and “Y” 300DR power amps.

Over the first day they sold 5 of the 250DR's and 3 of the 300DR's. The ratio of the remaining Sale stock at the end of Day 1 of 250DR's to 300DR's was 5:8.

Naim decided to manufacture overnight a further 10 of each power amp thus raising the sales stock level for the start of Day 2 and the ratio became 5:7.

At the end of the Sales Week, they had sold all the power amps.

How many of each power amp did they sell in their Direct Sales.