[jazzship]

Emily can click a mouse ten times in 10 seconds. Buzzy can click a mouse twenty times in 20 seconds. Anthony can click a mouse five times in 5 seconds. Assume that the timing period begins with the first mouse click and ends with the final click. Which one of these computer users would be the first to complete forty clicks?

"It's Anthony. The actual period is 1 second less than the time given. Emily completes ten clicks in 9 seconds. Buzzy completes twenty clicks in 19 seconds. Anthony completes five clicks in 4 seconds. This gives us the approximate rates: Anthony 1.25 clicks/second Emily: 1.1 clicks/second, Buzzy 1.05 clicks /second.

The answer makes no sense to me. Explanation?

[Guest]

Emily can click a mouse ten times in 10 seconds. Buzzy can click a mouse twenty times in 20 seconds. Anthony can click a mouse five times in 5 seconds. Assume that the timing period begins with the first mouse click and ends with the final click. Which one of these computer users would be the first to complete forty clicks?

"It's Anthony. The actual period is 1 second less than the time given. Emily completes ten clicks in 9 seconds. Buzzy completes twenty clicks in 19 seconds. Anthony completes five clicks in 4 seconds. This gives us the approximate rates: Anthony 1.25 clicks/second Emily: 1.1 clicks/second, Buzzy 1.05 clicks /second.

The answer makes no sense to me. Explanation?

To be honest it is a little ambiguous. Antony can make 5 clicks in 5 seconds. Generally we assume that this means a rate of once per second (which is could). However, as it is phrased, Antony could click at a rate anywhere from once every second and once every 1.25 seconds. Anything in this interval will mean that after 5 seconds from the first click he will have made a total of 5 clicks. (eg at 0, 1.2, 2.4, 3.6, 4.8). The same applies for the other two.

So if we are being pedantic (as that answer seems to be) then the answer is we don't know.

[jazzship]

I was thinking along those lines too psychic, but it still seemed strange. And like you mentioned, if indeed that is the correct interpretation, there really is no concrete answer,

I need to stop buying cheap puzzle books full of bad puzzles and typos.

[Guest]

Emily can click a mouse ten times in 10 seconds. Buzzy can click a mouse twenty times in 20 seconds. Anthony can click a mouse five times in 5 seconds. Assume that the timing period begins with the first mouse click and ends with the final click. Which one of these computer users would be the first to complete forty clicks?

"It's Anthony. The actual period is 1 second less than the time given. Emily completes ten clicks in 9 seconds. Buzzy completes twenty clicks in 19 seconds. Anthony completes five clicks in 4 seconds. This gives us the approximate rates: Anthony 1.25 clicks/second Emily: 1.1 clicks/second, Buzzy 1.05 clicks /second.

The answer makes no sense to me. Explanation?

I think the answer and the explanation you gave are not correct and they should actually be the other way round.

If a person makes X clicks in Y seconds, and the time starts after the first click, then the person is actually making X-1 clicks in Y seconds.

Accordingly, Emily makes 9 clicks in 10 seconds, Buzzy makes 19 clicks in 20 seconds and Anthony makes 4 clicks in 5 seconds.

So Buzzy should be the fastest, hence the first to complete 40 clicks.

[roolstar]

Totally agree with DeeGee here.

If the first click STARTS the timer, Buzzy is the fastest clicker with 19 clicks in 20 seconds (19/20), followed by Emily with 9/10 then by Anthony with 4/5.

To better illustrate we need to consider the time between clicks and not the number of clicks in a period of time.

C = Click (no time)

T = Time period

For Anthony, he's clicking by following the series below

C T C T C T C T C in 5 seconds

4T = 5 seconds => T = 1.25

Anthony will then complete 40 clicks in 39xT => 39x1.25 = 48.75 seconds

Emily:

C T C T C T C T C T C T C T C T C T C in 10 seconds

9T = 10 seconds => T = 1.11 seconds

Emily will then complete 40 clicks in 39xT => 39x1.11 = 43.33 seconds

Buzzy:

C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C T C in 20 seconds

19T = 20 seconds => T = 1.05 seconds

Buzzy will then complete 40 clicks in 39xT => 39x1.05 = 41.05 seconds (THE FIRST TO GET TO 40 CLICKS)

Edited February 18, 2010 by roolstar

[jazzship]

When I read it, I interpreted it differently, perhaps incorrectly.

Assume that the timing period begins with the first mouse click and ends with the final click. Which one of these computer users would be the first to complete forty clicks?

I read it as though they were competing, this is they way they would be timed. Their initial times weren't necessarily recorded this way.

[Guest]

roolstar has it right, but all the C and T was killing me, lol. i simply did this:

assuming that it takes exactly 5 seconds for Anthony to click 5 times, time starting with first click,

it will be 4 clicks left in 5 seconds. the formula would be seconds divided by total clicks - 1 or

5/4 = 1.25 seconds average time to click all 5 times.

this logic is extended to:

Emily at 10/9 = 1.11 seconds average per click

Buzzy at 20/19 = 1.05 seconds average per click

so Buzzy is fastest with 40.95 seconds(rounded) to complete the 39 clicks after the first click.

Anthony would have taken 48.75 seconds

Emily would have taken 43.29 seconds (rounded)

[Guest]

Everyone can be correct based on their own assumptions. For example, the actual rate of clicking for Anthony can be anywhere from 1.25 clicks per second to .8 clicks per second while still achieving the 5 clicks in 5 seconds. Each person has a different range of clicks per second, which can either be on the high side or low side. Any conclusion cannot be definitive unless all assumptions are stated and depending on the assumptions the answer can be any one of the people.

[Guest]

No one can click 40 times, because they can only click 5, 10, or 20. if they can click more then it would be the one with 5 seconds, because he only waits 5 seconds. I have a feeling this is a wiseguy answer...

[Guest]

T = time between successive clicks

t0 = time at which first click occurs

X = number of clicks completed by Y seconds

After the first click occurs, it will take a time (X-1)*T until all X clicks have been made.

Thus the time at which the Xth click is made is t0 + (X-1)*T

By Y seconds we want exactly X clicks to have been made.

This sets some restrictions on our variables.

t0 cannot exceed T or else there would be another click before the click that we defined to be first.

t0 cannot be negative either because then it is not counted as a click (we begin observing clicks after t=0).

The total time required for X clicks cannot exceed Y seconds, otherwise we would have too few clicks.

The total time required for X clicks cannot be less than Y - T either or else we will have too many clicks.

We therefore have

0 <= t0 <= T

Y-T < t0 + (X-1)T < Y

With some algebra, we find that

(Y - t0)/X < T < (Y-t0)/(X-1)

remember also that 0 <= t0 <= T

In each case for this problem, Y=X

so we have

1 - t0/X < T < 1 - (t0 - 1)/(X-1)

If t0 = 0 like DeeGee suggested, then

1 < T < 1 + 1/(X-1)

Taking the highest period,

T = 1 + 1/(X-1) means that the person with the highest X will have the smallest period and the fastest rate.

However this required two assumptions: t0 = 0, and T = highest period allowed by my above inequalities.

With different assumptions you will get different situations.

For example, setting t0 = T gives

X/(X+1) < T < 1

In such a case, taking the lowest value: T = X/(X+1) means that the person with the highest X will have the highest period and the smallest rate and will lose.

Edited February 18, 2010 by mmiguel1

[Guest]

I don't know if t0 is allowed to be equal to T.

If t0 = T, then there will be another click at t0-T = 0 meaning that the first click does not occur at t0 (since the period of the clicker is a constant).

However t0 can be T- or a value infinitesimally smaller than T. Which in principle I believe yield the same results as if we just set t0=T

[Guest]

I don't completely agree with DEGEE or ROOLSTAR.

"If the first click STARTS the timer, Buzzy is the fastest clicker with 19 clicks in 20 seconds (19/20), followed by Emily with 9/10 then by Anthony with 4/5."

So, if the first click STARTS the timer, then each of them may click on the 2nd second. However, the total number of seconds is the same. Think of it this way. I could be driving 60 miles per hour, 30 miles per half hour, or 15 miles per quarter hour. What is my pace? They're all equal.

Since the riddle doesn't address stamina as a factor (i.e. usually a sprinter can't run a 400 meter race at the exact same pace as they can run 200 meters), I believe they all finish at the same time regardless of when the clock starts:

Scenario A (1st click is AFTER the clock starts): Buzzy might get 19 clicks in 20 seconds, followed by Emily with 9/10 and Anthony with 4/5.....BUT.....once they're out of the gates so to speak, they'll continue on at the pace that was stated. In this case, they would all finish in 40 seconds (since they're all clicking at a pace of 1 click per second).

Scenario B (clock starts AFTER the first click) : In this scenario, they all finish in 39 seconds assuming they all continue at their respective pace.

Edited February 19, 2010 by kk11

[Guest]

DeeGee and RoolStar are right.

"So, if the first click STARTS the timer, then each of them may click on the 2nd second. However, the total number of seconds is the same. Think of it this way. I could be driving 60 miles per hour, 30 miles per half hour, or 15 miles per quarter hour. What is my pace? They're all equal."

If you are trying to compare it this way then look at it like this. You could go 60 miles in an hour or 30 miles in half an hour or 15 miles in 15 minutes. BUT if the timer doesn't start until you have already gone a mile, THEN you are actually going 59 miles an hour or 29 miles in 30 minutes or 14 miles in 15 minutes.

WHICH MEANS the person going 59 mph is going faster than the person going 58 mph (29*3) or the person going 56 mph (14*4).

"Scenario B (clock starts AFTER the first click) : In this scenario, they all finish in 39 seconds assuming they all continue at their respective pace."

You are assuming they are clicking at a consistent rate of 1 per second which is exactly the trick to this problem. If the clock starts after the first click (as stated in the OP) they are all clicking at a rate slightly slower than 1 per second.

[Guest]

I don't completely agree with DEGEE or ROOLSTAR.

"If the first click STARTS the timer, Buzzy is the fastest clicker with 19 clicks in 20 seconds (19/20), followed by Emily with 9/10 then by Anthony with 4/5."

So, if the first click STARTS the timer, then each of them may click on the 2nd second. However, the total number of seconds is the same. Think of it this way. I could be driving 60 miles per hour, 30 miles per half hour, or 15 miles per quarter hour. What is my pace? They're all equal.

Since the riddle doesn't address stamina as a factor (i.e. usually a sprinter can't run a 400 meter race at the exact same pace as they can run 200 meters), I believe they all finish at the same time regardless of when the clock starts:

Scenario A (1st click is AFTER the clock starts): Buzzy might get 19 clicks in 20 seconds, followed by Emily with 9/10 and Anthony with 4/5.....BUT.....once they're out of the gates so to speak, they'll continue on at the pace that was stated. In this case, they would all finish in 40 seconds (since they're all clicking at a pace of 1 click per second).

Scenario B (clock starts AFTER the first click) : In this scenario, they all finish in 39 seconds assuming they all continue at their respective pace.

"60 miles per hour = 30 miles per half our = 15 miles per quarter hour"

Yes, this is the most common way that wording of the type in the problem is supposed to be interpreted.

Are there other ways to interpret it though?

"Buzzy can click a mouse twenty times in 20 seconds"

What if Buzzy actually clicks at a rate of 21/20 = 1.05 clicks/second?

The period between clicks is 20/21 seconds/click.

This is faster than the rate one would expect from the typical interpretation.

Can he get away with making only 20 clicks in a span of 20 seconds?

If he makes his first click at 1-d seconds where d > 0 and d is approaching 0 in a limit,

then here will be the times for his clicks:

First: 1-d

Second: 1-d + 20/21

Third: 1-d + 2*20/21

...

...

Nineteenth: 1-d + 18*20/21

Twentieth: 1-d + 19*20/21

Twenty First: 1-d + 20*20/21

Let us approximate the constants in these last two terms to get an idea of what is going on:

Twentieth: 19.095-d

Twenty First: 20.048-d

Therefore it is clear that at the completion of the 20th second since starting, Buzzy has completed exactly 20 clicks, and that his 21st click occurs sometime after the 20th second (around 20.048).

Therefore one might say Buzzy can make 20 clicks in 20 seconds.

Yet, he shall not make 40 clicks in 40 seconds if we continue our counting:

Fortieth: 1-d + 39*20/21

Forty First: 1-d + 40*20/21

Approximate again:

Fortieth: 38.143-d

Forty First: 39.095-d

Forty Second:40.048-d

We see that Buzzy makes his 41st click before 40 seconds is over.

We therefore say that Buzzy can make 41 clicks in 40 seconds.

As I have shown, this does not contradict the statement that Buzzy makes 20 clicks in 20 seconds

even though 41/40 does not equal 20/20.

In such a case, Buzzy will make his 40th click sometime before 40 seconds have elapsed and you cannot so easily say that each player will tie in the competition.

The complete way to describe all possibilities is given by the inequalities I posted earlier.

[Guest]

I made a mistake in my last post:

1-d should be 20/21-d

Starting with 1-d doesn't make sense.

In that case, Buzzy will make 21 clicks in the first 20 seconds, like I should have realized.

This also doesn't satisfy the original constraints in my earlier post.

In particular T > (Y-t0)/X which I originally wrote as >=, but now realize should be strictly >.

That equation translates to 20/21 > 20/21 + d/20 which is not true for any d >= 0.

Yet I stated that d > 0 earlier, which comes from the assertion that t0 < T.

Perhaps the problem is that I should never have set T to 20/21.

I think that the following are correct, even though my example above is wrong.

(Y-t0)/X < T <= (Y-t0)/(X-1)

0 <= t0 < T

I realize though that my bad example may have taken some of my believability.

I'm not really sure anymore.

Edited February 19, 2010 by mmiguel1

[Guest]

I know I've written too many posts, but I don't think I will write anything after this one.

T = time between clicks, assumed constant

t = time after clock starts when the first click occurs

X = number of clicks person is said to click in Y seconds.

Any of these variables followed by a, e, or b (Anthony, Emily, Buzzy) means that the variable describes the quantity stated above for that person.

Since T is constant, for all intents and purposes the players click in a purely periodic fashion before and after the clock starts for all time.

This means that should we examine a situation where a player begins at a particular t, then this situation is completely equivalent to other situations where the player starts at t + k*T where k is an integer.

Therefore t = 0 and t = T are equivalent situations, and we should not consider both.

Let us limit ourselves to the half-interval [0,T) when considering the values of t.

0 <= t < T

X clicks must be complete by the time the Yth second has elapsed since the clock began.

The time at which the nth click occurs is t + (n-1)*T

The Xth click must occur before or at the moment that Y seconds have elapsed: t+(X-1)*T <= Y

Additionally, the X+1th click must occur strictly after Y seconds have elapsed: t + X*T > Y

We therefore have that t <= Y - (X-1)*T and that t > Y - XT

Note that in our problem for each player X=Y.

We therefore have X(1-T) < t <= T(1-X) + X

when comparing this to 0 <= t < T

we see that it must be true that

0 <= -T(X-1) + X

and

X(1-T) < T

Solving for T in each case gives

X/(X+1) < T <= X/(X-1)

For X=5, 5/6 < Ta <= 5/4, 0 <= ta < Ta

For X=10, 10/11 < Te <= 10/9, 0 <= te < Te

For X=20, 20/21 < Tb <= 20/19, 0 <= tb < Tb

also,

X(1-T) < t <= T(1-X) + X must be satisfied.

The winner is decided by whoever has the least amount of time elapsed at their 40th click.

Min of

ta + 39*Ta,

tb + 39*Tb,

te + 39*Te

Any person can win depending on how you choose Ta, ta, Tb, tb, Te, and te.

You may choose any values for these that satisfy the inequalities above.

Example,

Anthony wins if

(Ta,ta,Tb,tb,Te,te) = (1,0,20/19,0,10/9,0)

Emily wins if

(Ta,ta,Tb,tb,Te,te) = (5/4,0,20/19,0,1,0)

Buzzy wins if

(Ta,ta,Tb,tb,Te,te) = (5/4,0,1,0,10/9,0)

Edited February 19, 2010 by mmiguel1

[Guest]

Found more explicit bounds for t:

Anthony:

5/6 < Ta <= 5/4

If T >= 1, 0 <= ta <= 5-4*Ta

else, 5(1-Ta) < ta < Ta

Emily:

10/11 < Te <= 10/9

If T >= 1, 0 <= te <= 10-9*Te

else, 10*(1-Te) < te < Te

Buzzy:

20/21 < Tb <= 20/19

If T >= 1, 0 <= tb <= 20-19*Tb

else, 20*(1-Tb) < tb < Tb

[Guest]

This is amazing... the details of discussions which have come about in solving a problem that Jazzship claims to have come from a no-good cheap book with printing errors!! Take it easy guys... may be we are all over-thinking it.

[Guest]

This is amazing... the details of discussions which have come about in solving a problem that Jazzship claims to have come from a no-good cheap book with printing errors!! Take it easy guys... may be we are all over-thinking it.

But we are having fun!

[jazzship]

... Jazzship claims to have come from a no-good cheap book with printing errors!! Take it easy guys... may be we are all over-thinking it.

Maybe I jumped to conclusions about this particular puzzle, but not completely without reason. It was/is a cheap book - literally - and it is full of typographical errors. The answer they give to this problem is Anthony, which no one seems to agree with.

[Guest]

The answer they give to this problem is Anthony, which no one seems to agree with.

Well, the book doesn't always have to be right ;-)

[Guest]

Just to keep it going. Now calculate for non constant intervals between clicks. eg geometric progression. So Anthony's clicks then a quarter of a second later clicks again then half a second later. Until he has clicked 5 times at 3.75 seconds, but the next click wont come for another 4 seconds. This could simulate clicker-finger fatigue. You could designate a common ratio very near 1 to be more realistic.

Edit: It seems like any way you slice it Buzzy will always be fastest. Is there a way to make it so he isn't?

Edited February 19, 2010 by Semper Rideo

[Guest]

To get 5 clicks in 5 seconds, one would have to click at a rate of one click per 1.25 seconds, 10 clicks in 10 seconds is a rate of one click per 1.11 seconds, and 20 clicks in 20 seconds is a rate of one click every 0.95 seconds. The fastest rate will be the one to reach 40 clicks first, so Buzzy wins.

[Guest]

The answer to this question is indeterminate because it is dependent upon the length of time between "click sets". There is no reason to assume that the time between "slick sets" is exactly one second or any other interval.

[Guest]

I think the answer and the explanation you gave are not correct and they should actually be the other way round.

If a person makes X clicks in Y seconds, and the time starts after the first click, then the person is actually making X-1 clicks in Y seconds.

Accordingly, Emily makes 9 clicks in 10 seconds, Buzzy makes 19 clicks in 20 seconds and Anthony makes 4 clicks in 5 seconds.

So Buzzy should be the fastest, hence the first to complete 40 clicks.

This is the correct computation. Buzzy wins.

[Guest]

I would have to say that none of them would even reach 40. Their click rates are so slow that they must have brain damage and couldn't count that high anyways. Less than one click per second?