Brain Teaser No 1
Posted by: Don Atkinson on 16 November 2001
THE EXPLORER
An explorer set off on a journey. He walked a mile south, a mile east and a mile north. At this point he was back at his start. Where on earth was his starting point? OK, other than the North Pole, which is pretty obvious, where else could he have started this journey?
Cheers
Don
Posted on: 05 May 2003 by Don Atkinson
Hopefully, a more legible tree
Cheers
Don
Cheers
Don
Posted on: 06 May 2003 by Matthew T
A bit late to this but I managed to get 15 as well. I just counted them.
Matthew
Matthew
Posted on: 06 May 2003 by Don Atkinson
Matthew,
A bit late to this but I managed to get 15 as well. I just counted them.
You didn't need to count them, I had numbered them down the right hand side.............
Cheers
Don
A bit late to this but I managed to get 15 as well. I just counted them.
You didn't need to count them, I had numbered them down the right hand side.............
Cheers
Don
Posted on: 06 May 2003 by Don Atkinson
A 50% upgrade....
What is the smallest whole number that increases by exactly 50% when you transpose the last digit (ie the unit value at the right) and place it at the front ie at the left.
An example might help make the question clear. Suppose we want to try 1234. We would move the 4 from the RHS to the LHS to make the new number 4123. Test to see whether 4123 is exactly 50% bigger than 1234......well it isn't, so try a few more numbers......
Got the idea??
Cheers
Don
What is the smallest whole number that increases by exactly 50% when you transpose the last digit (ie the unit value at the right) and place it at the front ie at the left.
An example might help make the question clear. Suppose we want to try 1234. We would move the 4 from the RHS to the LHS to make the new number 4123. Test to see whether 4123 is exactly 50% bigger than 1234......well it isn't, so try a few more numbers......
Got the idea??
Cheers
Don
Posted on: 07 May 2003 by Matthew T
Don,
As far as I can tell, if there is a number that fits your requirements it is over a million.
Matthew
As far as I can tell, if there is a number that fits your requirements it is over a million.
Matthew
Posted on: 07 May 2003 by steved
Don,
There is no number that fulfils the criterion.
Assume the last digit of the original number is "x".
Simplifying the resulting equation always results in a fraction multiplied by "x" which cannot be an integer.
Steve D
There is no number that fulfils the criterion.
Assume the last digit of the original number is "x".
Simplifying the resulting equation always results in a fraction multiplied by "x" which cannot be an integer.
Steve D
Posted on: 07 May 2003 by steved
Just to explain further:
For a 2 digit number, the fraction is 17/28
For a 3 digit number, the fraction is 197/28
For a 4 digit number, the fraction is 1997/28
For a "n" digit number, the fraction is:
((2 x (10 to the power of n-1)) - 3)/28
The fraction can never be a whole number.
Steve D
For a 2 digit number, the fraction is 17/28
For a 3 digit number, the fraction is 197/28
For a 4 digit number, the fraction is 1997/28
For a "n" digit number, the fraction is:
((2 x (10 to the power of n-1)) - 3)/28
The fraction can never be a whole number.
Steve D
Posted on: 07 May 2003 by Matthew T
Steved,
I'm not sure your proof is completely valid because if any of the initial integers are greater then 7 multiplying by 1.5 will play havoc. I started with that route but realised that it might not work. Could be wrong, but need to find an alternative proof to be completely statisfied.
cheers
Matthew
I'm not sure your proof is completely valid because if any of the initial integers are greater then 7 multiplying by 1.5 will play havoc. I started with that route but realised that it might not work. Could be wrong, but need to find an alternative proof to be completely statisfied.
cheers
Matthew
Posted on: 07 May 2003 by Matthew T
Don,
Is 0 a whole number? (sneaky, sneaky)
Matthew
Is 0 a whole number? (sneaky, sneaky)
Matthew
Posted on: 07 May 2003 by Matthew T
OK my proof that this is impossible.
The intial number starts with A and ends with Z
A*10^n + B*10 + Z
multiplying by 1.5 gives
A*1.5*10^n + 1.5*B + Z*1.5
from Don premise we can deduce that
Z = A*1.5 (1 s.f.) and where both Z and A are positive integers less then 10
We can also conclude that Z must be even as multiplying by 1.5 must give a integer
Possibilities
Z= 2,4,6,8
A= 3,6,9,12
for Z=2,4,6 multiplying A by 1.5 will result in a number for all B that is greater then Z
for Z=8 A=12 which is not a possibility as this will change the number of digits in the number.
Something like a proof
Matthew
The intial number starts with A and ends with Z
A*10^n + B*10 + Z
multiplying by 1.5 gives
A*1.5*10^n + 1.5*B + Z*1.5
from Don premise we can deduce that
Z = A*1.5 (1 s.f.) and where both Z and A are positive integers less then 10
We can also conclude that Z must be even as multiplying by 1.5 must give a integer
Possibilities
Z= 2,4,6,8
A= 3,6,9,12
for Z=2,4,6 multiplying A by 1.5 will result in a number for all B that is greater then Z
for Z=8 A=12 which is not a possibility as this will change the number of digits in the number.
Something like a proof
Matthew
Posted on: 07 May 2003 by steved
Matthew,
Your proof is much simpler than mine, so well done. However, I don't agree that mine isn't valid: assume for simplicity 3 digit number:-
first number abx
second number xab
equation 100x + 10a + b = 1.5(100a + 10b + x)
expand 100x + 10a + b = 150a + 15b +1.5x
simplify 98.5x = 140a + 14b or 197x = 280a + 28b
or 197x/28 = 10a + b
The right hand side must be an integer, and therefore so must the left hand side. However, the left hand side cannot be an integer, so there is no true solution. Steve D
Your proof is much simpler than mine, so well done. However, I don't agree that mine isn't valid: assume for simplicity 3 digit number:-
first number abx
second number xab
equation 100x + 10a + b = 1.5(100a + 10b + x)
expand 100x + 10a + b = 150a + 15b +1.5x
simplify 98.5x = 140a + 14b or 197x = 280a + 28b
or 197x/28 = 10a + b
The right hand side must be an integer, and therefore so must the left hand side. However, the left hand side cannot be an integer, so there is no true solution. Steve D
Posted on: 07 May 2003 by Don Atkinson
Guys, Guys,
I need to be sure that my teaser is clearly understood.
As an example, we will take any whole number (no fractions, no decimals, zero not allowed but the number could include a zero; it doesn't in fact, but that's beside the point)
So, the arbitrary example is 287654. We shift the 4 from the rhs to the lhs and form the second number, 428765. We test to see whether 428765 = 1.5 * 287654. Well, it isn't so we have to try a few more numbers. The second number is a whole number and is exactly 50% bigger than the first number.
For example, 21 is exactly 50% more than 14. Again 14 and 21 don't fill the bill either...
Both the numbers I have in mind are less than a million. The arithmetic is prescise, not approximate.
Do you feel that the task is clear? I hope you do.
Cheers
Don
I need to be sure that my teaser is clearly understood.
As an example, we will take any whole number (no fractions, no decimals, zero not allowed but the number could include a zero; it doesn't in fact, but that's beside the point)
So, the arbitrary example is 287654. We shift the 4 from the rhs to the lhs and form the second number, 428765. We test to see whether 428765 = 1.5 * 287654. Well, it isn't so we have to try a few more numbers. The second number is a whole number and is exactly 50% bigger than the first number.
For example, 21 is exactly 50% more than 14. Again 14 and 21 don't fill the bill either...
Both the numbers I have in mind are less than a million. The arithmetic is prescise, not approximate.
Do you feel that the task is clear? I hope you do.
Cheers
Don
Posted on: 08 May 2003 by steved
Don,
285714
Steve D
285714
Steve D
Posted on: 08 May 2003 by Matthew T
Ok Don, we didn't read the question correctly, I now have the answer, which I will with hold to let others have a go. The number is six digits though. Does zero still count though?
Matthew
Matthew
Posted on: 08 May 2003 by Matthew T
Thanks for jumping in there Steve!
That's the answer I got.
That's the answer I got.
Posted on: 08 May 2003 by Don Atkinson
Matthew, Steve,
Any tips on how to remember these two numbers (285714 & 428571) ?
Cheers
Don
Any tips on how to remember these two numbers (285714 & 428571) ?
Cheers
Don
Posted on: 09 May 2003 by Matthew T
Don,
Even more interesting!
Whole parts of 2,000,000/7 and 3,000,000/7.
Matthew
Even more interesting!
Whole parts of 2,000,000/7 and 3,000,000/7.
Matthew
Posted on: 09 May 2003 by Don Atkinson
Whole parts of 2,000,000/7 and 3,000,000/7.
and not just the whole parts.....the number keeps repeating as the decimal part
Cheers
Don
and not just the whole parts.....the number keeps repeating as the decimal part
Cheers
Don
Posted on: 15 May 2003 by Don Atkinson
I saw this one in an IQ booklet. Although I know what the answer is I am not convinced I have a good solution.
Giveaway
In Salisbury, a town of 66,000 people, there lived a rich man. He offered £38.00 to each male and a certain sum to each female. Of the males, only a nineteenth collected their money and of all the females only 2 out of every 10 collected their money.
He gave away £132,000.00 in total. How much did each female receive?
Of course, being in an IQ booklet, you only had 5 minutes to do the sum IN YOUR HEAD.
Cheers
Don
No cheating !!!!
Giveaway
In Salisbury, a town of 66,000 people, there lived a rich man. He offered £38.00 to each male and a certain sum to each female. Of the males, only a nineteenth collected their money and of all the females only 2 out of every 10 collected their money.
He gave away £132,000.00 in total. How much did each female receive?
Of course, being in an IQ booklet, you only had 5 minutes to do the sum IN YOUR HEAD.
Cheers
Don
No cheating !!!!
Posted on: 15 May 2003 by Paul Ranson
From the monies,
38M + xF = 132000
From the population,
19M + 5F = 66000
M & F are count of males/females that get the money, x is how much a female gets.
Anyway it's obvious that x = 10 is a solution. Probably not good enough though...
Paul
38M + xF = 132000
From the population,
19M + 5F = 66000
M & F are count of males/females that get the money, x is how much a female gets.
Anyway it's obvious that x = 10 is a solution. Probably not good enough though...
Paul
Posted on: 16 May 2003 by steved
Don
£10 is correct.
Paul,
Your equations are fine, and £10 is a unique solution.
Steve D
£10 is correct.
Paul,
Your equations are fine, and £10 is a unique solution.
Steve D
Posted on: 16 May 2003 by Matthew T
Paul,
You could write your equations as
2m+ x*f/5 = 132000
and
m + f = 66000
which can alternatively be writen as
x*f/5 - 2f = 0
divide by f
x/5 - 2 = 0
x = 10
Matthew
You could write your equations as
2m+ x*f/5 = 132000
and
m + f = 66000
which can alternatively be writen as
x*f/5 - 2f = 0
divide by f
x/5 - 2 = 0
x = 10
Matthew
Posted on: 16 May 2003 by Paul Ranson
This was supposed to be a 'head thing'.
The way I first saw it was that £132000 is an expenditure of £2 per head, but only 1/19 of males collect, so they each get £38, but 1/5 of females are smart so they each get £10.
Paul
The way I first saw it was that £132000 is an expenditure of £2 per head, but only 1/19 of males collect, so they each get £38, but 1/5 of females are smart so they each get £10.
Paul
Posted on: 16 May 2003 by Don Atkinson
Isn't it time for brain teaser number 2?
How time flies when you're enjoying yourself...........
Cheers
Don
How time flies when you're enjoying yourself...........
Cheers
Don
Posted on: 16 May 2003 by Don Atkinson
Paul was right to remind you lot (including himself i noticed) that it was a head thing......some might argue 'mental'......but let's not go down that road.
My logic (mental) was
the 'average' payout across the whole population was 132000/66000 = £2 ph
the 'average' payout across the male population was 38/19 = £2 ph
hence the 'average' payout across the female population has got to be £2 ph also (to maintain the overall average)
if only 1 in 5 of the ladies took up the offer, they had to collect 2*5 = £10 each in order to create that £2 ph average.
'fraid it took me a little longer than the allowed 5 minutes to sus..
Paul, Matthew, Steve, how long ???
Cheers
Don
My logic (mental) was
the 'average' payout across the whole population was 132000/66000 = £2 ph
the 'average' payout across the male population was 38/19 = £2 ph
hence the 'average' payout across the female population has got to be £2 ph also (to maintain the overall average)
if only 1 in 5 of the ladies took up the offer, they had to collect 2*5 = £10 each in order to create that £2 ph average.
'fraid it took me a little longer than the allowed 5 minutes to sus..
Paul, Matthew, Steve, how long ???
Cheers
Don