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Mathematics Brain Teasers.

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Jeez, Dan. Your cracking a nut with an atom bomb!

Reverse the alphabet writing it Z to A. Starting with Z = 1, Y = 2, X = 3, .... assigning a numer to each letter ... B = 25, A = 26.

ALERT!!! Just checked my solution and I've made a mistake in the sequence of numbers originally provided. The last number should be a 6 not a 5!!!!

It should have read: 4, 22, 9, 7, 2, 6 ...

I must be getting old!

Edited by Ozymandias

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2 hours ago, Ozymandias said:

Jeez, Dan. Your cracking a nut with an atom bomb!

Reverse the alphabet writing it Z to A. Starting with Z = 1, Y = 2, X = 3, .... assigning a numer to each letter ... B = 25, A = 26.

ALERT!!! Just checked my solution and I've made a mistake in the sequence of numbers originally provided. The last number should be a 6 not a 5!!!!

It should have read: 4, 22, 9, 7, 2, 6 ...

I must be getting old!

Im not getting anything, maybe we are using different formulas?

F(x)=a^-1(x-b) mod m.

Where a is a prime number ?

I give up lol. Someone else can solve it.

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3 hours ago, danydandan said:

Im not getting anything, maybe we are using different formulas?

F(x)=a^-1(x-b) mod m.

Where a is a prime number ?

I give up lol. Someone else can solve it.

Dan, you're being too hard on yourself. You had it cracked! The answer was 18.

...............................

A=25,B=23    UCPRWT
A=25,B=24    VDQSXU
A=25,B=25    WERTYV
A=25,B=3    AIVXCZ
A=25,B=4    BJWYDA

......................................

The correct letter sequence is highlighted above except that  I gave the last number as 5 instead of 6 so the last letter should have been U, not V

The letter sequence WERTYU is from the top QWERTY row of buttons on a standard keyboard. The next letter is I giving the next number in the sequence as 18 .....

Q   W    E   R    T    Y    U    I    O   P

10   4    22   9    7    2    6   18  12  11   .........

I think you should set another teaser yourself.

Edited by Ozymandias
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1 hour ago, Ozymandias said:

Dan, you're being too hard on yourself. You had it cracked! The answer was 18.

...............................

A=25,B=23    UCPRWT
A=25,B=24    VDQSXU
A=25,B=25    WERTYV
A=25,B=3    AIVXCZ
A=25,B=4    BJWYDA

......................................

The correct letter sequence is highlighted above except that  I gave the last number as 5 instead of 6 so the last letter should have been U, not V

The letter sequence WERTYU is from the top QWERTY row of buttons on a standard keyboard. The next letter is I giving the next number in the sequence as 18 .....

Q   W    E   R    T    Y    U    I    O   P

10   4    22   9    7    2    6   18  12  11   .........

I think you should set another teaser yourself.

That never would have occurred to me. Lol. I was so fixed on it being a word that it just eluded me, and as usual when someone points something out, it now looks so simple. I'll get one in a few minutes I'll have a look at my old Texts and Tests books.

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This might be easy or difficult depending on your outlook I suppose.

Can anyone write a mathematical expression that equals exactly 5, using two number 2s and any symbols and operations. You can use multiple operators and symbols. There are probably multiple solutions I can think of only one.

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2^2 - e^[(pi)i] = 5

2 + 2 - e^[(pi)i] = 5

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7 hours ago, Ozymandias said:

2^2 - e^[(pi)i] = 5

2 + 2 - e^[(pi)i] = 5

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3 hours ago, danydandan said:

Oh, very nice, Dan. Yours is simpler than mine and probably truer to the solution criteria you set. How do you get the sqrt sign using a keypad on a phone?

My question: a room measures 4 metres wide by 12 metres long and is 3 metres high. How far is the room's exact centre from any of the its corners?

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12 minutes ago, Ozymandias said:

Oh, very nice, Dan. Yours is simpler than mine and probably truer to the solution criteria you set. How do you get the sqrt sign using a keypad on a phone?

My question: a room measures 4 metres wide by 12 metres long and is 3 metres high. How far is the room's exact centre from any of the its corners?

I have a one plus 3T phone the key pad/key board has a good amount of mathematical symbols.

I'll give the answer to that in a bit, see if anyone else can get it.

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21 hours ago, Ozymandias said:

My question: a room measures 4 metres wide by 12 metres long and is 3 metres high. How far is the room's exact centre from any of the its corners?

Easiest solution is,

D=√(3^2+12^2+4^2)

D=13

13/2. 6.5 meters. Which is half the space diagonal of rectangular prism, or the room. That 6.5 should be the same for any corner to the exact centre of the room.

Eir, was hacked last night. I think a DOD attack so broadband and mobile data was down, could not answer sooner. If your an Eir user change all your passwords associated with them and monitor your account, or put a restriction for a while via whatever bank you use.

Edited by danydandan
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3 hours ago, danydandan said:

Easiest solution is,

D=√(3^2+12^2+4^2)

D=13

13/2. 6.5 meters. Which is half the space diagonal of rectangular prism, or the room. That 6.5 should be the same for any corner to the exact centre of the room.

Eir, was hacked last night. I think a DOD attack so broadband and mobile data was down, could not answer sooner. If your an Eir user change all your passwords associated with them and monitor your account, or put a restriction for a while via whatever bank you use.

Of course, Dan. The answer is 6.5m. A simple application of Pythagoras' Theorem. The diagonals from opposite corners all intersect each other at the centre of the room.

I can't understand the total lack of interest but it's your turn to set a teaser. Maybe try to make it more amenable to a wider audience.

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9 minutes ago, Ozymandias said:

Of course, Dan. The answer is 6.5m. A simple application of Pythagoras' Theorem. The diagonals from opposite corners all intersect each other at the centre of the room.

I can't understand the total lack of interest but it's your turn to set a teaser. Maybe try to make it more amenable to a wider audience.

Simple one now so. I was going to post an efficiency type one, kinda engineery.

I am given a choice of three boxes. One of the boxes holds a prize. So I choose one, before I'm given the box to look inside, I'm asked if I'm sure about my choice or would I like to choose another.

Do I have better odds if I stick to my choice or if I choose again?

Edited by danydandan
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17 hours ago, danydandan said:

Simple one now so. I was going to post an efficiency type one, kinda engineery.

I am given a choice of three boxes. One of the boxes holds a prize. So I choose one, before I'm given the box to look inside, I'm asked if I'm sure about my choice or would I like to choose another.

Do I have better odds if I stick to my choice or if I choose again?

Is this meant to be the Monty Hall problem?

With no extra information, don't think the odds change.

Edited by Golden Duck
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2 hours ago, Golden Duck said:

Is this meant to be the Monty Hall problem?

With no extra information, don't think the odds change.

Not sure who Monty Hall is.

Edit: nope it's not the Monty Hall problem, according to Wikipedia the chooser gets to see what's in the box, so to speak. And then is given a second chance. It's similar to my question but statistically I think they are the same. If you consider it one event the odds change, if you consider it two different events the odds change. Either way your either left with 1/2 or 2/3 from starting with 1/3. I'm not sure if allowing the chooser to see his initial choice and allowing them to then change it leads to a second event. I think it does.

You can set the next question GD.

Edited by danydandan
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A ball is thrown vertically upwards with a velocity of 20m/sec. What's it's height at 12m/sec?

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4 hours ago, danydandan said:

A ball is thrown vertically upwards with a velocity of 20m/sec. What's it's height at 12m/sec?

I've only just seen this now. I sent Golden Duck a private message asking him to set the next teaser as requested by you. He has not responded.

Ignoring any aerodynamic drag on the ball the answer is 2.854 metre to the nearest millimetre.

Initial Velocity (U) = 20m/s

Subsequent Velocity (V) = 12m/s

Deceleration Rate (a) = - 9.81m/s^2

Distance Moved (S) = ???

Assuming uniform acceleration, no aerodynamic effects acting on the ball and that it follows a straight-line path upwards ...

From V^2 = U^2 + 2aS

S = (V^2 - U^2) / 2a = (12^2 - 20^2) / 2(-9.81) = (144 - 400) / (-19.62) = - 56 / -19.62 = 2.8542... = 2.854m

Edited by Ozymandias
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16 hours ago, Ozymandias said:

I've only just seen this now. I sent Golden Duck a private message asking him to set the next teaser as requested by you. He has not responded.

Ignoring any aerodynamic drag on the ball the answer is 2.854 metre to the nearest millimetre.

Initial Velocity (U) = 20m/s

Subsequent Velocity (V) = 12m/s

Deceleration Rate (a) = - 9.81m/s^2

Distance Moved (S) = ???

Assuming uniform acceleration, no aerodynamic effects acting on the ball and that it follows a straight-line path upwards ...

From V^2 = U^2 + 2aS

S = (V^2 - U^2) / 2a = (12^2 - 20^2) / 2(-9.81) = (144 - 400) / (-19.62) = - 56 / -19.62 = 2.8542... = 2.854m

S=(20^2-12^2)/2x(9.81).

S=(256)/(19.56)

S=13.08m

Method is correct.

Using the above, find the time between the instants at which the ball is at this height.

Edited by danydandan
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1 hour ago, danydandan said:

S=(20^2-12^2)/2x(9.81).

S=(256)/(19.56)

S=13.08m

Method is correct.

Using the above, find the time between the instants at which the ball is at this height.

No, Dan. I've been lecturing on this stuff for over 40 years so I ought to know what I'm doing.

The ball begins its upward motion with an initial velocity (u = 20m/s) and is not accelerating under gravity (g = 9.81m/s^2) but is decelerating (negative acceleration) under the pull of gravity (g = -9.81m/s^2). At some point in its upward travel it will have slowed to a subsequent velocity (v) of 12m/s.

Assuming no aerodynamic effects the formulae applicable here are:  (1) v = u + at ;   (2) v^2 = u^2 + 2aS ;   and (3) S = ut + (1/2)at^2

You are using the correct formula but you have either transposed it incorrectly when making S the subject or you have mixed up the value of the initial velocity (u = 20 m/s) with the later (v = 12m/s)

Initial Velocity (u)            = 20m/s

Subsequent Velocity (v) = 12m/s

Deceleration Rate (a)     = - 9.81m/s^2

Distance Moved (S)       =     ?

Time (t)                          =     ?

From V^2 = U^2 + 2aS

S = (V^2 - U^2) / 2a = (12^2 - 20^2) / 2(-9.81) = (144 - 400) / (-19.62) = - 56 / -19.62 = 2.8542... = 2.854m

EDIT @ 10 mins later. You should always check a post that is being challenged before replying on your high horse!!! My original calculation was in error because of a typo!!!!

I subtracted 400 from 144 and should have written -256 NOT -56. Therefore -256/-19.62 = 13.0479.. = 13.048m to nearest mm.

You are correct Dan!!

Edited by Ozymandias
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Now I have to redeem myself.

'... find the time between the instants at which the ball is at this height [13.048m]'.

Assuming no aerodynamic effects the same formulae are applicable here:  (1) v = u + at ;   (2) v^2 = u^2 + 2aS ;   and (3) S = ut + (1/2)at^2

Initial Velocity (u)            = 20m/s

Subsequent Velocity (v) = 12m/s

Deceleration Rate (a)     = - 9.81m/s^2

Distance Moved (S)       =     13.048m

Time (t)                          =     ?

Using S = ut + (1/2)at^2

13.048 = 20t + (1/2)(-9.81)t^2

13.048 = 20t - 4.905t^2  (a quadratic equation in t)

4.905t^2 - 20t + 13.048 = 0 (solve for t using the formulae t = {-(-b) +/- sqrt[b^2 - 4ac]}/2a

So,    t = {-(-20)+/- sqrt[(-20)^2 - 4(4.905)(13.048)}/2(4.905)

t =  {20  +/-  sqrt(400 - 256)} / 9.81

t =  {20 +/- 12}/9.81

t = 32/9.81  or  8/9.81  =  3.2619  or 0.8155

Therefore, the interval or time between these two times is 3.2619 - 0.8155  =  2.4465 seconds

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41 minutes ago, Ozymandias said:

No, Dan. I've been lecturing on this stuff for over 40 years so I ought to know what I'm doing.

The ball begins its upward motion with an initial velocity (u = 20m/s) and is not accelerating under gravity (g = 9.81m/s^2) but is decelerating (negative acceleration) under the pull of gravity (g = -9.81m/s^2). At some point in its upward travel it will have slowed to a subsequent velocity (v) of 12m/s.

Assuming no aerodynamic effects the formulae applicable here are:  (1) v = u + at ;   (2) v^2 = u^2 + 2aS ;   and (3) S = ut + (1/2)at^2

You are using the correct formula but you have either transposed it incorrectly when making S the subject or you have mixed up the value of the initial velocity (u = 20 m/s) with the later (v = 12m/s)

Initial Velocity (u)            = 20m/s

Subsequent Velocity (v) = 12m/s

Deceleration Rate (a)     = - 9.81m/s^2

Distance Moved (S)       =     ?

Time (t)                          =     ?

From V^2 = U^2 + 2aS

S = (V^2 - U^2) / 2a = (12^2 - 20^2) / 2(-9.81) = (144 - 400) / (-19.62) = - 56 / -19.62 = 2.8542... = 2.854m

EDIT @ 10 mins later. You should always check a post that is being challenged before replying on your high horse!!! My original calculation was in error because of a typo!!!!

I subtracted 400 from 144 and should have written -256 NOT -56. Therefore -256/-19.62 = 13.0479.. = 13.048m to nearest mm.

You are correct Dan!!

We are both idiots lol.

I made loads of mistakes, your way is correct. I wrote my equation so no negatives would appear. Which is incorrect.

Man does not say much for either of us. Lol.

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2 minutes ago, Ozymandias said:

Now I have to redeem myself.

'... find the time between the instants at which the ball is at this height [13.048m]'.

Assuming no aerodynamic effects the same formulae are applicable here:  (1) v = u + at ;   (2) v^2 = u^2 + 2aS ;   and (3) S = ut + (1/2)at^2

Initial Velocity (u)            = 20m/s

Subsequent Velocity (v) = 12m/s

Deceleration Rate (a)     = - 9.81m/s^2

Distance Moved (S)       =     13.048m

Time (t)                          =     ?

Using S = ut + (1/2)at^2

13.048 = 20t + (1/2)(-9.81)t^2

13.048 = 20t - 4.905t^2  (a quadratic equation in t)

4.905t^2 - 20t + 13.048 = 0 (solve for t using the formulae t = {-(-b) +/- sqrt[b^2 - 4ac]}/2a

So,    t = {-(-20)+/- sqrt[(-20)^2 - 4(4.905)(13.048)}/2(4.905)

t =  {20  +/-  sqrt(400 - 256)} / 9.81

t =  {20 +/- 12}/9.81

t = 32/9.81  or  8/9.81  =  3.2619  or 0.8155

Therefore, the interval or time between these two times is 3.2619 - 0.8155  =  2.4465 seconds

Perfect.

Edit: Well that's what I got too. Don't make it right lol.

Edited by danydandan
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Assume that a bee flies directly back and forth at a constant 5m/s between its hive situated beside a straight road and a honey pot being carried towards the hive by a beekeeper walking at a steady 3km/hr along the same road. When the bee starts to fly the honey pot is initially 1 kilometre from its hive. How far will the bee have flown in total when the beekeeper reaches its hive? [Admittedly, the poor bee will be knackered!!!]

Edited by Ozymandias
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On 10/20/2018 at 4:11 PM, Ozymandias said:

Assume that a bee flies directly back and forth at a constant 5m/s between its hive situated beside a straight road and a honey pot being carried towards the hive by a beekeeper walking at a steady 3km/hr along the same road. When the bee starts to fly the honey pot is initially 1 kilometre from its hive. How far will the bee have flown in total when the beekeeper reaches its hive? [Admittedly, the poor bee will be knackered!!!]

Any takers? I note that there have been over 60 views of this teaser which isn't as difficult as it appears. Although there is some maths needed it doesn't require an elaborate mathematical solution.

Edited by Ozymandias
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12 minutes ago, Ozymandias said:

Any takers? I note that there have been over 60 views of this teaser which isn't as difficult as it appears. Although there is some maths needed it doesn't require an elaborate mathematical solution.

I'm waiting to give someone else a chance, wait till tomorrow.

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1 hour ago, danydandan said:

I'm waiting to give someone else a chance, wait till tomorrow.

OK, Dan. I'll do nothing.
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