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## Maths challenge

AuthorMessage
JQPublic

Posts: 1911

 Posted: 04:04AM Dec 10, 2012

Mods: I'm not sure which one this should go, so I put it here. If you think this belongs somewhere else, feel free to move it there without PMing me.

Here's what I suggest we do. Braingle is about our brains, and a good way to train our brains is to solve tough maths problems, but there hasn't been much about that here. Therefore, I suggest we have a thread about those.

Rules:
1. We make up maths questions (or get some from books). Do NOT use problems you find on the Internet as it's easy to find the answers for those.
2. The first person to answer a question correctly gets to post the next one.
3. If there are no correct answers within one week, the author of the current teaser should reveal the answer and post a different one.

I'll start. This one is actually a teaser which got rejected because it was considered to be maths rather than a teaser.

You've probably read or heard of teasers that go like this:

'The area of a rectangle is 60 square units and the difference between its length and its width is 4 units. What are the length and the width?'

Obviously, the dimensions of the rectangle are 6 units * 10 units.

Here's the challenge: is it always possible to find the length and width of a rectangle when we're given the area and the difference between the length and the width? (Assume that the given numbers are valid.)

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 12:02AM Dec 12, 2012

Yes it is. Let x and y be the length and width of the rectangle, x>=y, and a the given area and d the difference. Then:

(1) x * y = a
(2) x - y = d

Two simultaneous equations with two variables will always have a unique solution.

Multiply both sides of (2) by x: x^2 - xy = dx
Substitute (1) and rearrange: x^2 - dx - a = 0
Solve the quadratic equation for x: x = [d + sqrt(d^2 + 4a)] / 2 (clearly the positive solution will be required)
y=x-d, so: y=[sqrt(d^2 + 4a) - d] / 2

Using this formula to solve the original example: a=60, d=4

x = [4 + sqrt(16 + 240)] / 2 = (4 + 16) / 2 = 10
y = (16 - 4) / 2 = 6
JQPublic

Posts: 1911

 Posted: 03:36AM Dec 12, 2012

Congrats, spike, you've just won Challenge 1.

My solution was a bit different:

Yes.

Let D units be the difference between the length and the width, w units be the width and A be the area.

w(w+D) = A
w^2 + Dw = A
w^2 + Dw - A = 0

It may seem to you that the equation can return two distinct values of w, so it is not always possible to find w. However, that is untrue.

A must be positive because the area cannot possibly be zero or negative. Therefore, -A must be negative. The coefficient of w^2 is 1.

In the general form ax^2 + bx + c, the product of the roots of a quadratic equation is equal to c/a, so the product of the roots = -A / 1 = -A, which, as we've seen above, must be negative. In the real number system, if the product of two numbers is negative, then one number is positive and the other number is negative. Widths can't be negative, so the negative value must be invalid, leaving only one solution.

Adding the D to w gives the length, so it is always possible to find the length as well.

Anyhow, it's your turn. Looking forward to yours.

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 08:33PM Dec 12, 2012

Which of the following is correct (and why)?

a) There are infinitely many irrational numbers between any two distinct rational numbers
b) There are infinitely many rational numbers between any two distinct irrational numbers
c) Both of the above
d) Neither of the above

---This message was edited on 08:35PM Dec 13, 2012---
JQPublic

Posts: 1911

 Posted: 03:45AM Dec 13, 2012

I'm pretty sure b is true. I'll try to figure out a later.

However, if this is a trick question, then it's d. 'Any two rational numbers' and 'any two irrational numbers' =/= 'Any two DIFFERENT rational numbers' and 'any two DIFFERENT irrational numbers'

Edit: Now I've changed my mind and I'm not so sure that b is true any more. (I guess I said that based on my gut reaction rather than real proofs.)

---This message was edited on 03:51AM Dec 13, 2012---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 08:36PM Dec 13, 2012

Good point, JQ. No trick was intended, so I've modified the question to clarify that.
JQPublic

Posts: 1911

 Posted: 02:42AM Dec 14, 2012

Um, OK, I *think* I've got a.

Let y = x^1/2 and the two rational numbers be a and b where b > a. This function is non-linear, so I'll put it through logarithmic transformation and get log y=1/2*log x. The resultant graph is a straight line with a slope of 1/2 and log y-intercept of 0, i.e. passing through the origin. The points on the graph where log x = log a and log x = log b are the rational numbers, and the line segment with these points as the endpoints is the range of log x. Since straight lines are made of an infinite number of points, log x has an infinite number of values. There's no limit on the number of rational numbers or irrational numbers on a line segment, so there's an infinite set of irrational values of log x satisfying log a < log x < log b. When log a < log x < log b, a < x < b, so there's an infinite set of irrational values of x as well.

[Note: where I live, we use log without a base for the common logarithm, not the natural logarithm. 'lg' is also seldom used.]

PS I probably didn't need the logarithmic transformation part, but I'm not familiar with curves so I just decided to play safe.

Edit: WAIT ... I forgot x doesn't have to be irrational when log x is irrational ... Darn, back to square one.

Edit edit: NO WAIT .... Hey, this could have been much easier, exponentless and logless, and probably correct.

Let a and b be the two rational numbers on the number line. Now, there's an infinite set of numbers between a and b. There's no limit on the number of points between a and b. Therefore, there should be an infinite number of both rational numbers and irrational numbers between a and b. Therefore, (a) is true.

Now just to figure out (b). I think this is much harder because I can't think in terms of a Cartesian plane or a number line because seriously, how do you plot pi on a those? If only irrational numbers were limited to unsimplifyable surds ...

PS Is there an algebraic definition of an irrational number other than being able to be represented by a ratio a:b? If there is, I'd probably sound much less layman ...

PPS I changed the colour of my wrong answer to white.

---This message was edited on 05:02AM Dec 14, 2012---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
JQPublic

Posts: 1911

 Posted: 05:24AM Dec 15, 2012

OK ... I think the answer is C.

Let's say there are two irrational numbers, a and b, which start differing after X digits. The difference between the two numbers must therefore be able to be divided into an infinite number of rational numbers. Therefore, b is also correct.

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 12:35AM Dec 16, 2012

JQPublic wrote:
OK ... I think the answer is C.

Let's say there are two irrational numbers, a and b, which start differing after X digits. The difference between the two numbers must therefore be able to be divided into an infinite number of rational numbers. Therefore, b is also correct.

That's the one! Though it seems counter-intuitive for both a and b to be true simultaneously, once you wrap your head around the fact that there really is no concept of consecutive rational or irrational numbers, then it becomes almost trivial. Infinity moves in mysterious ways...
JQPublic

Posts: 1911

 Posted: 03:44AM Dec 16, 2012

Thanks spike

Firstly, just a little nitpick on the answer to the first question. You wrote 'two simultaneous equations with two variables will always have a unique solution.' That is not always true (although it is true in this case). For example, 2x+y=0 and 4x+2y=0 (consistent equations) have an infinite number of solutions, while 2x+y+3=0 and 4x+2y=0 (inconsistent equations) have no real solution.

Anyway, here's an algebra one:

Given this quadratic equation: 2x^2 - 3 = 3x + 9

Let α and β be its two roots. Find 3α + 2β^2.

Catch: You are not allowed to find α and β first. Also, the quadratic formula is strictly prohibited if you want to substitute it in.

---This message was edited on 06:16AM Dec 16, 2012---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
JQPublic

Posts: 1911

 Posted: 07:05PM Dec 28, 2012

OK, time to reveal the answer. Sorry for being late and pity nobody got it.

2x^2 - 3 = 3x + 9
2x^2 - 3x - 12 = 0

α + β = -(-3)/2 = 3/2
αβ = -12/2 = -6

2x^2 - 3x - 12 = 0
3x = 2x^2 - 12
3α = 2α^2 - 12

3α + 2β^2
= 2α^2 - 12 + 2β^2
= 2(α^2 + β^2) - 12
= 2(α^2 + β^2) - 12
= 2(α^2 + 2αβ + β^2 - 2αβ) - 12
= 2(α + β)^2 - 4αβ - 12
= 2(-6)^2 - 4(-6) - 12
= 84

Now for the new question.

Given the function f(x) = 4x^2 + 6x + k, where k is a constant, and that f(x) is positive for all values of x. Find the range of values of k.

John looked at the problem and was quick to scribble out an answer.

f(x) > 0, so the discriminant > 0
36 - 16k > 0
-16k > -36
k < 9/4

However, he was wrong. Why? What is the correct answer?

[Hint: There's an alternative solution if you can't figure out why this is wrong.]

---This message was edited on 07:27PM Dec 28, 2012---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 08:38PM Jan 1, 2013

JQPublic wrote:

α + β = -(-3)/2 = 3/2
αβ = -12/2 = -6

---This message was edited on 08:39PM Jan 1, 2013---
spikethru4

Posts: 17

 Posted: 09:13PM Jan 1, 2013

JQPublic wrote:

Now for the new question.

Given the function f(x) = 4x^2 + 6x + k, where k is a constant, and that f(x) is positive for all values of x. Find the range of values of k.

John looked at the problem and was quick to scribble out an answer.

f(x) > 0, so the discriminant > 0
36 - 16k > 0
-16k > -36
k < 9/4

However, he was wrong. Why? What is the correct answer?

[Hint: There's an alternative solution if you can't figure out why this is wrong.]

John switched his signs twice at the end: -16k > -36 => k > -36/-16 > 9/4.

Let a be the value of x for which f(x) is minimum. At this point, f(a)>0, so 4a^2 + 6a + k > 0, or k > -4a^2 - 6a.

To find a, solve the derivative f'(x)=0:

f'(x) = 8x + 6. f'(a) = 0 => a = -6/8 = -3/4

Therefore k > 9/4

---This message was edited on 09:39PM Jan 1, 2013---
JQPublic

Posts: 1911

 Posted: 04:00AM Jan 2, 2013

spikethru4 wrote:
JQPublic wrote:

α + β = -(-3)/2 = 3/2
αβ = -12/2 = -6

Actually, that's possible.

To form a quadratic equation with its roots, (α-x)(β-x)=0 will do, but I'll have to add in a non-zero constant A as well (since that can appear in the last step of the equation-solving, but is unrelated to the roots since it can't equal zero.

ax^2+bx+c≡A(α-x)(β-x)

R.H.S.
=A(α-x)(β-x)
=A(x^2-αx-βx+αβ)
=Ax^2-Aαx-Aβx+Aαβ
=Ax^2-A(α+β)x+Aαβ

Therefore A = a, -A(α+β) = b, Aαβ = c

-b/a = -[-A(α+β)]/A = α+β
c/a = Aαβ/A = αβ

*puts on immature 'I told you so' face*

---This message was edited on 08:14AM Jan 2, 2013---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
JQPublic

Posts: 1911

 Posted: 04:09AM Jan 2, 2013

spikethru4 wrote:
JQPublic wrote:

Now for the new question.

Given the function f(x) = 4x^2 + 6x + k, where k is a constant, and that f(x) is positive for all values of x. Find the range of values of k.

John looked at the problem and was quick to scribble out an answer.

f(x) > 0, so the discriminant > 0
36 - 16k > 0
-16k > -36
k < 9/4

However, he was wrong. Why? What is the correct answer?

[Hint: There's an alternative solution if you can't figure out why this is wrong.]

John switched his signs twice at the end: -16k > -36 => k > -36/-16 > 9/4.

Let a be the value of x for which f(x) is minimum. At this point, f(a)>0, so 4a^2 + 6a + k > 0, or k > -4a^2 - 6a.

To find a, solve the derivative f'(x)=0:

f'(x) = 8x + 6. f'(a) = 0 => a = -6/8 = -3/4

Therefore k > 9/4

I've never learnt (and never will) learn calculus, so I really have no idea what you're talking about. You did find the right answer, but not the correct explanation of John's mistake (since the inequality sign switches directions when both sides are divided by a negative number.) How about trying again?

---This message was edited on 04:11AM Jan 2, 2013---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 09:44PM Jan 2, 2013

*puts on immature 'I told you so' face*

Fair dos.

(since the inequality sign switches directions when both sides are divided by a negative number.)

Whoops!

*Light-bulb switches on*

If f(x)>0 for all x, then f(x)=0 has no real solutions, therefore the discriminant must be negative, not positive as John thought.

Calculus in a nutshell
The rate of change of a curve, or the slope of the tangent at any point, is a function known as the first derivative. For polynomials, the derivative of ax^n is nax^(n-1) and the derivative of the sum of such terms is the sum of the derivatives. So the first derivative of f(x), f'(x), is 8x + 6.

Why is this useful? When the rate of change is zero, the curve is at a minimum or maximum value. In a quadratic curve with a>0, there is one minimum value. Find this minimum value, and that tells us how far up (or down) the y-axis we need to slide the curve (adjust the value of k) so that the minimum value is >0.
JQPublic

Posts: 1911

 Posted: 04:55AM Jan 3, 2013

Yes, that's right, it's the answer. Thanks for the quick crash course in calculus.

My alternative solution was much easier:

4x^2 + 6x + k
= 4(x^2 + 3/2x) + k
= 4(x^2 + 3/2x + 9/16 - 9/16) + k
= 4(x + 3/4)^2 - 9/4 + k

Since f(x) > 0,
- 9/4 + k > 0
k > 9/4

---This message was edited on 04:59AM Jan 3, 2013---

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens
spikethru4

Posts: 17

 Posted: 01:27AM Jan 4, 2013

Next question:

In a game of Monopoly, using standard box rules, what is the probability that a player lands in Jail during his/her first turn?

For the purposes of this question, we are using my 1976 Standard UK set, which has the following board layout and Chance/Community Chest cards:

Board layout (clockwise from GO)
Jail/Just Visiting, Pall Mall, Electric Company, Whitehall, Northumberland Avenue, Marylebone Station, Bow Street, Community Chest, Marlborough Street, Vine Street
Free Parking, Strand, Chance, Fleet Street, Trafalgar Square, Fenchurch St Station, Leicester Square, Coventry Street, Water Works, Piccadilly
Go to Jail, Regent Street, Oxford Street, Community Chest, Bond Street, Liverpool St Station, Chance, Park Lane, Super Tax, Mayfair

Relevant Chance Cards (16 total)
Go to Jail
Go back three spaces

Relevant Community Chest Cards (16 total)
Go to Jail
Go back to Old Kent Road
RGW4

Posts: 1838

 Posted: 01:29AM Jan 6, 2013

spikethru4 wrote:
Next question:

In a game of Monopoly, using standard box rules, what is the probability that a player lands in Jail during his/her first turn?

For the purposes of this question, we are using my 1976 Standard UK set, which has the following board layout and Chance/Community Chest cards:

Relevant Chance Cards (16 total)
Go to Jail
Go back three spaces

Relevant Community Chest Cards (16 total)
Go to Jail
Go back to Old Kent Road

First, it is interesting to see how an English board is set up. There are so many different boards available now, and I know my daughter and son own about 6-8 different ones between them. The game is the same, the places differ, sorry, I just don't see spending money to feel like I am in the Navy (which I was) or in New York. Mute point... again, the boards are basically the same.

Now, I am not going to try and impress anyone with math equations to show a point, because frankly, Algebra is not my strong point. However, if I was setting a bet in Vegas on this, I would give going to jail on the first round high odds... probably something like 30 to 1, maybe even 50 to 1 if I was feeling generous. So, I guess that would be my answer to your question.

Getting three doubles in a row, the odds are like 1 in 200 (I could be more accurate, but it is late and I don't want to pull out a pen and paper to do the math).

To land on community chess or chance and get a go to jail card on the first turn, is a 1 in 16 chance odd on the first card... considering there are only 16 cards in each pile.

Doubles are the key to this question. You either have to get 3 doubles in a row, a "Go to Jail" card, or hit that corner spot of going to Jail! Moving to another location on the board is incidental. Getting closer to jail with a go to property card does not put you in it!

Hitting the community chest box on the first roll of snake eyes... may get you a return to "Go" card or a "Go to Jail Card" or a "Go to Kent Road" card (Hm... I don't recall a card going to Mediterranean"?)

So, I will stick with my 30 to 1 odds...and say if I saw Vegas giving better odds,. I probably would go with it.

---This message was edited on 02:13AM Jan 6, 2013---

"That's one of the remarkable things about life. It's never so bad that it can't get worse." Calvin of Calvin and Hobbes
JQPublic

Posts: 1911

 Posted: 08:49AM Jan 6, 2013

Question: when you say 'land in jail', does just visiting count?

'An idea, like a ghost, must be spoken to a little before it will explain itself.' - Charles Dickens

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