An Alternative To The Quiz Questions.

[Kimmy]

A confused bank teller transposed the dollars and cents when he cashed a check for Ms Smith, giving her dollars instead of cents and cents instead of dollars. After buying a newspaper for 50 cents, Ms Smith noticed that she had left exactly three times as much as the original check. What was the amount of the check? (Note: 1 dollar = 100 cents.)

[Nofinepix]

I've just Asked Jeeves, the explanation on how to get the answer has made my head hurt.

[Kimmy]

And mine.

[banger]

It Left Google Gasping
But here are 2 answers take your pick
Solution by Diophantine Equation
Let x be the number of dollars in the check, and y be the number of cents.
Then 100y + x − 50 = 3(100x + y).
Therefore 97y − 299x = 50.

A standard solution to this type of linear Diophantine equation uses Euclid's algorithm.

The steps of the Euclidean algorithm for calculating the greatest common divisor (gcd) of 97 and 299 are as follows:

299 = 3 × 97 + 8
97 = 12 × 8 + 1

This shows that gcd(97,299) = 1.

To solve 97y − 299x = gcd(97,299) = 1, we can proceed backwards, retracing the steps of the algorithm as follows:

1 = 97 − 8 × 12
= 97 − (299 − 3 × 97) × 12
= 37 × 97 − 12 × 299

Therefore a solution to 97y − 299x = 1 is y = 37, x = 12.
Hence a solution to 97y − 299x = 50 is y = 50 × 37 = 1850, x = 50 × 12 = 600.
It can be shown that all integer solutions of 97y − 299x = 50 are of the form y = 1850 + 299k, x = 600 + 97k, where k is any integer.

In this case, because x and y must be between 0 and 99, we choose k = −6.
This gives y = 56, x = 18.
So the check was for $18.56.
Solution by Simultaneous Equations
Let x be the number of dollars in the check, and y be the number of cents. Consider the numbers of dollars and cents Ms Smith holds at various times. The original check is for x dollars and y cents. The bank teller gave her y dollars and x cents. After buying the newspaper she has y dollars and x − 50 cents. We are also told that after buying the newspaper she has three times the amount of the original check; that is, 3x dollars and 3y cents.

Clearly (y dollars plus x − 50 cents) equals (3x dollars plus 3y cents). Then, bearing in mind that x and y must both be less than 100 (for the teller's error to make sense), we equate dollars and cents.

As −50 (x − 50) 49 and 0 3y 297, there is a relatively small number of ways in which we can equate dollars and cents. (If there were many different ways, this whole approach would not be viable.) Clearly, 3y − (x − 50) must be divisible by 100. Further, by the above inequalities, −49 3y − (x − 50) 347, giving us four multiples of 100 to check.

•If 3y − (x − 50) = 0, then we must have 3x = y, giving x = −25/4, y = −75/4
•If 3y − (x − 50) = 100, then (to balance) we must have 3x − y = −1, giving x = 47/8, y = 149/8
•If 3y − (x − 50) = 200, then we must have 3x − y = −2, giving x = 18, y = 56
•If 3y − (x − 50) = 300, then we must have 3x − y = −3, giving x = 241/8, y = 747/8
There is only one integer solution; so the check was for $18.56.

[BC]

These are tough, but I like them. Well done Banger.

[banger]

Nov 30, 2009 21:07:51 GMT BC said:

These are tough, but I like them. Well done Banger.

Confusion Between Me Google And Euciid
My solution to the puzzle is
They can keep the dollars and cents
Just pass the Paracetamol

[Kimmy]

PLEASE don't ;D the messenger.

Solution to puzzle 5: Confused bank teller
Solution by Diophantine Equation

Let x be the number of dollars in the check, and y be the number of cents.
Then 100y + x − 50 = 3(100x + y).
Therefore 97y − 299x = 50.

A standard solution to this type of linear Diophantine equation uses Euclid's algorithm.

The steps of the Euclidean algorithm for calculating the greatest common divisor (gcd) of 97 and 299 are as follows:

299 = 3 × 97 + 8
97 = 12 × 8 + 1

This shows that gcd(97,299) = 1.

To solve 97y − 299x = gcd(97,299) = 1, we can proceed backwards, retracing the steps of the algorithm as follows:1 = 97 − 8 × 12
= 97 − (299 − 3 × 97) × 12
= 37 × 97 − 12 × 299


Therefore a solution to 97y − 299x = 1 is y = 37, x = 12.
Hence a solution to 97y − 299x = 50 is y = 50 × 37 = 1850, x = 50 × 12 = 600.
It can be shown that all integer solutions of 97y − 299x = 50 are of the form y = 1850 + 299k, x = 600 + 97k, where k is any integer.

In this case, because x and y must be between 0 and 99, we choose k = −6.
This gives y = 56, x = 18.
So the check was for $18.56.

[Kimmy]

7. Five men, a monkey, and some coconuts

Five men crash-land their airplane on a deserted island in the South Pacific. On their first day they gather as many coconuts as they can find into one big pile. They decide that, since it is getting dark, they will wait until the next day to divide the coconuts.

That night each man took a turn watching for rescue searchers while the others slept. The first watcher got bored so he decided to divide the coconuts into five equal piles. When he did this, he found he had one remaining coconut. He gave this coconut to a monkey, took one of the piles, and hid it for himself. Then he jumbled up the four other piles into one big pile again.

To cut a long story short, each of the five men ended up doing exactly the same thing. They each divided the coconuts into five equal piles and had one extra coconut left over, which they gave to the monkey. They each took one of the five piles and hid those coconuts. They each came back and jumbled up the remaining four piles into one big pile.

What is the smallest number of coconuts there could have been in the original pile?

[banger]

After a little mental arithmatic I found that the smallest number of coconuts there could have been in the original pile is 3121

[banger]

I must hold my hand up and confess that I sought help from Google for my answer but admit it does any of us really understand the answers
(Or is Kimmy trying to make monkeys of us)
ANSWER
Let the original pile have n coconuts. Let a be the number of coconuts in each of the five piles made by the first man, b the number of coconuts in each of the five piles made by the second man, and so on.

Writing a Diophantine equation to represent the actions of each man, we have

n = 5a + 1 n + 4 = 5(a + 1)
4a = 5b + 1 4(a + 1) = 5(b + 1)
4b = 5c + 1 4(b + 1) = 5(c + 1)
4c = 5d + 1 4(c + 1) = 5(d + 1)
4d = 5e + 1 4(d + 1) = 5(e + 1)

Hence n + 4 = 5 × (5/4)4 (e + 1), and so n = (55/44) (e + 1) − 4.

Note that, since 5 and 4 are relatively prime, 55/44 = 3125/256 is a fraction in its lowest terms. Hence the only integer solutions of the above equation are where e + 1 is a multiple of 44, whereupon d + 1, c + 1, b + 1, and a + 1 are all integers.

So the general solution is n = 3125r − 4, where r is a positive integer, giving a smallest solution of 3121 coconuts in the original pile.

If I have underestimated anyone's intelligence I apologise

[Kimmy]

Solution to puzzle 7: Five men, a monkey, and some coconuts

Let the original pile have n coconuts. Let a be the number of coconuts in each of the five piles made by the first man, b the number of coconuts in each of the five piles made by the second man, and so on.

Writing a Diophantine equation to represent the actions of each man, we haven = 5a + 1 n + 4 = 5(a + 1)
4a = 5b + 1 4(a + 1) = 5(b + 1)
4b = 5c + 1 4(b + 1) = 5(c + 1)
4c = 5d + 1 4(c + 1) = 5(d + 1)
4d = 5e + 1 4(d + 1) = 5(e + 1)


Hence n + 4 = 5 × (5/4)4 (e + 1), and so n = (55/44) (e + 1) − 4.

Note that, since 5 and 4 are relatively prime, 55/44 = 3125/256 is a fraction in its lowest terms. Hence the only integer solutions of the above equation are where e + 1 is a multiple of 44, whereupon d + 1, c + 1, b + 1, and a + 1 are all integers.

So the general solution is n = 3125r − 4, where r is a positive integer, giving a smallest solution of 3121 coconuts in the original pile.
Remarks

It's clear from the form of the equations that we can generalize this result. Given m > 2 men in the airplane, the smallest solution would be mm − (m−1) coconuts. This follows because, for m > 2, m and m−1 are always relatively prime. (Any divisor of m and m−1 must also divide their difference.)

[Kimmy]

11. Dice game

Two students play a game based on the total roll of two standard dice. Student A says that a 12 will be rolled first. Student B says that two consecutive 7s will be rolled first. The students keep rolling until one of them wins. What is the probability that A will win?

Hint to puzzle 11: Dice game

Let p be the probability that student A wins. Consider the possible outcomes of the first two rolls. (Recall that each roll consists of the throw of two dice.) Hence express p as a weighted mean, and solve for p.

[banger]

Wrestling with this problem after having two falls this is a submission

[BC]

I didn't understand the hint at all!

But my guess is 50% or 1 in 2.

My thinking as follows:

A - Chances of throwing a 12 with 2 dice is 1:36

B - Chances of throwing a 7 is 1:6

(1&6 2&5 3&4 4&3 5&2 6&1 = 6:36 or 1:6)

So to throw 7 twice is 1:6 x 1:6 = 1:36

OK, I'm probably wrong, but I do try. Actually, Mrs BC says I can be very trying. Which is nice to know.

[banger]

You Can't Beat A Trier

[Kimmy]

Answer to puzzle 11: Dice game

The probability that student A will win is 7/13.

Solution to puzzle 11: Dice game

Let p be the probability that student A wins. We consider the possible outcomes of the first two rolls. (Recall that each roll consists of the throw of two dice.) Consider the following mutually exclusive cases, which encompass all possibilities.
If the first roll is a 12 (probability 1/36), A wins immediately.
If the first roll is a 7 and the second roll is a 12 (probability 1/6 · 1/36 = 1/216), A wins immediately.
If the first and second rolls are both 7 (probability 1/6 · 1/6 = 1/36), A cannot win. (That is, B wins immediately.)
If the first roll is a 7 and the second roll is neither a 7 nor a 12 (probability 1/6 · 29/36 = 29/216), A wins with probability p.
If the first roll is neither a 7 nor a 12 (probability 29/36), A wins with probability p.

Note that in the last two cases we are effectively back at square one; hence the probability that A subsequently wins is p.

Probability p is the weighted mean of all of the above possibilities.

Hence p = 1/36 + 1/216 + (29/216)p + (29/36)p.

Therefore p = 7/13.

[Kimmy]

17. Three children

On the first day of a new job, a colleague invites you around for a barbecue. As the two of you arrive at his home, a young boy throws open the door to welcome his father. “My other two kids will be home soon!” remarks your colleague.

Waiting in the kitchen while your colleague gets some drinks from the basement, you notice a letter from the principal of the local school tacked to the noticeboard. “Dear Parents,” it begins, “This is the time of year when I write to all parents, such as yourselves, who have a girl or girls in the school, asking you to volunteer your time to help the girls' soccer team.” “Hmmm,” you think to yourself, “clearly they have at least one of each!”

This, of course, leaves two possibilities: two boys and a girl, or two girls and a boy. Are these two possibilities equally likely, or is one more likely than the other?

Note: This is not a trick puzzle. You should assume all things that it seems you're meant to assume, and not assume things that you aren't told to assume. If things can easily be imagined in either of two ways, you should assume that they are equally likely. For example, you may be able to imagine a reason that a colleague with two boys and a girl would be more likely to have invited you to dinner than one with two girls and a boy. If so, this would affect the probabilities of the two possibilities. But if your imagination is that good, you can probably imagine the opposite as well. You should assume that any such extra information not mentioned in the story is not available.

[BC]

Dec 3, 2009 8:59:35 GMT Kimmy said:

Answer to puzzle 11: Dice game

The probability that student A will win is 7/13.

Solution to puzzle 11: Dice game

Let p be the probability that student A wins. We consider the possible outcomes of the first two rolls. (Recall that each roll consists of the throw of two dice.) Consider the following mutually exclusive cases, which encompass all possibilities.
If the first roll is a 12 (probability 1/36), A wins immediately.
If the first roll is a 7 and the second roll is a 12 (probability 1/6 · 1/36 = 1/216), A wins immediately.
If the first and second rolls are both 7 (probability 1/6 · 1/6 = 1/36), A cannot win. (That is, B wins immediately.)
If the first roll is a 7 and the second roll is neither a 7 nor a 12 (probability 1/6 · 29/36 = 29/216), A wins with probability p.
If the first roll is neither a 7 nor a 12 (probability 29/36), A wins with probability p.

Note that in the last two cases we are effectively back at square one; hence the probability that A subsequently wins is p.

Probability p is the weighted mean of all of the above possibilities.

Hence p = 1/36 + 1/216 + (29/216)p + (29/36)p.

Therefore p = 7/13.

Easy for you to say.

[BC]

17 Two girls and a boy is more likely. Lots of possibilities, but the most likely scenario I can think of is that it is a girls school (why not ask for help on the boys soccer team). If it is a girls school and two kids are arriving together, then they are girls. The boy goes to a different school, which is why he's home early.

And with that, I rest my case m'laud.

[Kimmy]

[Kimmy]

Solution to puzzle 17: Three children

We assume that each birth is an independent event, for which the probability of a boy is the same as the probability of a girl. There are, then, three possibilities for your colleague's family, all equally likely:
Boy, Boy, Girl
Boy, Girl, Boy
Boy, Girl, Girl

Therefore there is a 2/3 chance that the colleague has two boys and a girl, and a 1/3 chance he has two girls and a boy.
Remarks

Note that in each of the ordered triples above, (BBG, BGB, BGG), the first letter represents the gender of the first child. The context of the word first may be chosen at our convenience. For example, it may be the eldest child (if we sort by descending age), the shortest (sort by ascending height), or perhaps the child whose forename is first alphabetically. In this case, based upon the information we are given, the context we choose is “the first child we meet.” The second and third letters in each ordered triple represent the genders of the other two children, neither of whom we have met.

Another way to arrive at the answer is to note that the boy who opens the door is essentially a red herring. Leaving him aside, the puzzle asks us to compare the probabilities that the other two children are (a) both girls, or (b) one girl and one boy. (The letter from the principal is carefully worded to leave both options open.) The second option (older sister, younger brother, or older brother, younger sister) is twice as likely as the first (elder sister, younger sister.)

As with many conditional probability questions, this result may seem surprising at first. If you are skeptical, I urge you to carry out a simulation, either manually or programatically. Setting up such a simulation forces you to analyze exactly what is happening.

The situation can be modelled by throwing three coins. Let heads represent boys, and tails girls. One coin in each throw should be set aside to be checked whether it is a boy or a girl. (This corresponds to the boy who opens the door.) It needn't be the same coin each time (though it could be), and it needn't be thrown first (though it might be), but it must be chosen independently of whether it shows heads or tails.

One convenient method would be to throw one dime and two nickels. If the dime (first child) shows heads (a boy) and one or both of the nickels shows tails (at least one girl), then we have a faithful representation of the puzzle situation. Other scenarios are discarded. Of the scenarios retained, in roughly two out of three cases the three coins will show two heads.

[Kimmy]

My BrainBashers electronic world atlas has developed a fault, I did a listing of miles from England to particular countries and here is the result:

Spain 14,000 miles
Fiji 12,000 miles
Germany 18,000 miles
Brazil 16,000 miles

How far away did it list Iceland as?

[BC]

My first, obviously incorrect from previous efforts, guess is 22,000

Based on down 2k up 6k down 2k up 6k

---

My second, probably also incorrect although more likely, guess is 18,000

Based on number of letters in the word corresponding with miles i.e. Iceland has 7 letters the same as Germany.

---

[BC]

That is:

Fiji 4 = 12,000 miles
Spain 5 = 14,000 miles
Brazil 6 = 16,000 miles
Germany 7 = 18,000 miles
Iceland 7 = 18,000 miles

[Kimmy]

ANSWER.

20,000: each vowel is worth 4,000 miles and each consonant is worth 2,000.

[Kimmy]

Puzzle 1

How many gifts, in total, did I receive during the Twelve Days of Christmas - according to the song?

[BC]

I thought at first 1 + 2 + 3 + 4 + etc.

But then, two french hens could be seen as 1 present. So I'm going to go for 12.

Might as well get the red pen out now...

X

[Kimmy]

Answer.

364: for example, on the fifth day of Christmas, my true love gave to me: five gold rings + four calling birds + three French hens + two turtle doves + a partridge in a pear tree.

[Kimmy]

Puzzle 3

A long train, half a kilometre long, is about to enter a long tunnel. The tunnel is 10k long. If the speed of the train is 35kph, how long will it take for the entire train to pass through the tunnel - from the front of the train entering to the end of the train leaving the tunnel?

[BC]

Dec 6, 2009 11:19:32 GMT Kimmy said:

Puzzle 3

A long train, half a kilometre long, is about to enter a long tunnel. The tunnel is 10k long. If the speed of the train is 35kph, how long will it take for the entire train to pass through the tunnel - from the front of the train entering to the end of the train leaving the tunnel?

18 minutes