A Fistful of Brain Teasers

Naming depends whether you use traditional British terminology, or the now more widely and possibly near universal formerly American system.

i would say: Twelve quadrillion, three hundred and forty five trillion, six hundred and seventy eight billion, nine hundred and eighty seven million, six hundred and fifty four thousand, three hundred and twenty one.

But it looks better written out - actually a rather lovely product,  12345678987654321  

Or with commas but doesnt look so neat: 12,345,678,987,654,321

 

12345678987654321 is quite a neat outcome.

hopefully, others will also have appreciated this.

11 x 11 = 121 and similar products yield similar results.

the given 111 111 111 example provides the largest product comprising single digit components. The symmetry continues beyond eg 1 111 111 111 But the appearance is complicated because the product includes double-digit components.

Anyway, I hope this little diversion was interesting.

0 appears eleven times (that’s a Zero to avoid ambiguity)

2 appears twenty times

3 appears twenty times

4 appears twenty times

5 appears twenty times

6 appears twenty times

7 appears twenty times

8 appears twenty times

9 appears twenty times

There is nothing sinister or obscure about the list in which all the above things occur.

How many times would the number 1 appear ?

Home to Pub Puzzle JPEG

Going to the Pub

I live in the house shown in the bottom left of the diagram and pop up to the pub, top right, most evenings. The diagram is orientated North-Up and the Pub is NE of the house.

The possible paths linking the two places are indicated by the lines in the diagram and I always walk either due N, due E or NE. That is, I always go so that every step brings me closer to the pub !

What is the greatest number of different routes I can take ?

Home to Pub Part solution JPEG

IB has ruled himself out of any further attempt at this one. (nice try IB BTW - and not too far wide of the mark !

So, to encourage "the others" ..........

I have entered the number of possible routes available to reach a dozen of the "nodes"

You might be able to detect a pattern developing and....

this might lead to a accurate solution

notnaim man posted:

Total frustration, I have a pile of envelopes with numbers scribbled on. I got 347, then 375. Then, I have been told there is a similar puzzle about getting to see a mistress without being seen en route every day of the year, so without counting again, 365?

Tantalising, Frustrating, annoying............That's what some of these Brain Teasers are all about Also a bit of fun andgreat satisfaction when you get there !

Take a careful look at the "13" Node. ie there are 13 different routes to that specific Node.

They are quite easy to follow. and en-route you will have passed through Nodes that could have been reached by 5, 3 and another 5 routes respectively. (You might have noticed, there is only 1 route to the NW corner and there is only 1 route to the SE corner).

There is a "pattern" developing, with a Pub at the end.

(but sorry, the number of routes isn't 345, 347, 365 or 375)

More Frustration ! or Great satisfaction ?

One more prompt,.....

Going "North" from the "Five" to the West of "13", the next two Nodes have 7 then 9 possible routes.

Likewise, going "East" from the "Five" to the South of "13", the next two Nodes also have 7 then 9 possible routes.

Each Node is derived simply by adding together the number of routes from the nodes leading into it. Usually there are three nodes leading into the next node.

You only have eight nodes to fill in and you should be able to do those in your head !............but pencil and paper are not prohibited

Digits JPEG

In the diagram above, you will notice that the nine digits, 1 to 9, have been arranged such that the second line is twice as big as the first and the third line is three times as big as the first.

There are three other ways of arranging the nine digits so as to produce the same result.

Anybody ?

2 7 3                         3 2 7                      2 1 9

5 4 6                          6 5 4                      4 3 8

8 1 9                          9 8 1                       6 5 7

 

On the same idea, there is just one arrangement where each row is double the one above. What is it? (This may be a quicker one)

 

Innocent Bystander posted:

2 7 3                         3 2 7                      2 1 9

5 4 6                          6 5 4                      4 3 8

8 1 9                          9 8 1                       6 5 7

 

On the same idea, there is just one arrangement where each row is double the one above. What is it? (This may be a quicker one)

 

I think i'll give others a crack at this one before I respond

(you will obviously need to read my intro to this, a couple of posts above)

The joys of train travel and cycling !

I occasionally travel up to London by train. On my return journey I have the choice of three options :-

  • I can leave the train at Newbury and cycle home.
  • I can stay on the train for another 15 miles to Overton, then again cycle home
  • Or I can change trains at Newbury and travel the 14 miles to Pewsey, before again cycling home.

Funnily enough, the distances from Newbury, Overton and Pewsey to my home are identical ! Even more remarkable, the railway lines between Newbury and Overton and Newbury and Pewsey are dead straight…..as are the cycle paths between each station and my house !

I happen to know that the distance between Overton and Pewsey is 13 miles.

How far is my home from the three stations ?

BTW, the cycle paths are all picturesque so I choose my route to suit the weather and my fancy for particular scenic views

Slide3

The story breaks down to a simple bit of geometry that you would have done at school.

The diagram above illustrates this as follows.

The triangle NOP (Newbury, Overton, Pewsey) shows the train lines (black) from London and the cyclepaths (red) from each of NOP to my Home.

Given that the stated distance from each Station to Home is identical, it follows that my Home is at the Circumcentre of Triangle NOP and the distance is the radius of the circumscribed circle.

The circumcentre is where the perpendicular bisectors of NO, OP and PN meet. (Dashed black lines)

All you need do now, is calculate the radius of the circumscribed circle to triangle NOP with sides 13,14 and 15.

Most of the formulae that you need were used in the Farmer Watt’s field puzzle, way back on Page 8, but you will also need the FULL Sine Rule formula.

Can anybody recall the FULL Sine Rule formula ?

A few min spare today - and I could see it from the cosine rule, not needing the sine rule.

So by my estimation it is just under 8.1 miles (my calculator makes it 8.097 - but in reality probably fractionally more because that doesn’t allow for the cars, pedestrians and dogs you may have to dodge en route).

Innocent Bystander posted:

A few min spare today - and I could see it from the cosine rule, not needing the sine rule.

So by my estimation it is just under 8.1 miles (my calculator makes it 8.097 - but in reality probably fractionally more because that doesn’t allow for the cars, pedestrians and dogs you may have to dodge en route).

Ha ha ! there are no cars, pedestrians or dogs en-route..........this is winky's valhalla !!!

Your estimation is very close to the correct figure.......this is a nice, but encouraging way of saying it's wrong! But what a bloody good shot !

I didn't use the Cosine Rule, but that doesn't mean your technique is wrong, you might just have had a bit of "finger" wobble putting the numbers into your calculator.

The correct answer is just over 8.1 miles, but it is a very precise fraction that makes it  "just over" 8.1

If you post a few more details of your approach, I'll see if I can spot any obvious goofs.

Later today, I will outline the three formulae, that I used in order to derive a very simple, and somewhat satisfying formula for R. I rather suspect that some people might well have written this simple formula down in their school days, along with the FULL Sine Rule and a few other formulae. I didn't, so had to derive it when I first saw this teaser !

Your answer is certainly within a gnat's whisker !!

I’m struggling as usual to upload a sketch, necessary for the explanation below to make sense. This is my last attempt - I’m providing the link as well in case it still doesn’t show.

 

https://imgur.com/a/WyfFD


Angles A,A are identical because sides X,X identical
Similarly B,B and C,C
Perpendicular from centre to any side bisects it for same reason.

Cosine rule:
Cos (A+C)   =  (15^2 + 13^2 - 14^2 ) /  2*15*13     = 0.50769
(A+C) = 59.49°

Total internal angles of triangle = 180° = 2A+2B+2C
So A+B+C=90°
So the other angle in one of the rt angle triangles with angle B and side length 7 = A+C (the angle found above)

Sin 59.49   =  7/x
∴ X =  7/sin 59.49      = 8.097 miles

  

Don, do you leave your bike at the station and return the same way, or, more sensibly, take it on the train with you to allow you to use it at your destination, with the flexibility when returning to choose the route home according to wind, either for maximum exercise or minimum effort depending how the day went

The closeness of two figures calculated different ways makes me wonder about the accuracy of our respective calculators with sines and cosines. When calculating I retained the angle returned for A+C in the calculator rather than noting the rounded value 59.49 as above However, I have now calculated twice, once with iPhone calculator, the second MS Windows calculator accessory, and both the same result.

Maybe the width of your drive comes into it...

 

Innocent Bystander posted:

The closeness of two figures calculated different ways makes me wonder about the accuracy of our respective calculators with sines and cosines. When calculating I retained the angle returned for A+C in the calculator rather than noting the rounded value 59.49 as above However, I have now calculated twice, once with iPhone calculator, the second MS Windows calculator accessory, and both the same result.

Maybe the width of your drive comes into it...

 

Ah !

Now, with my approach, although I have used the Sine Rule and the Side/Angle/Side (SAS) Rule (amongst others) , I only used these to derive a formula that doesn't actually have any Sines/Cosines/Tangets etc left in it. Just pure, wholesome numbers..........which to a large extent, cancel out. Giving a fairly neat, satisfying answer.

In other words, it doesn't depend on the accuracy of holding 16 decimal places of Logs/Sines/Cosines etc in a calculator.

But i'm sure from past experience, that a PC version of Excel, would hit the nail on the head with pin-point precision, even in a decimalised format dependent on Logs or Trig Functions !!

Anyway, I have seen your linked diagram and will look at your calcs later on to see if I can detect any obscure error...................and I'll measure the width of my driveway as well

Innocent Bystander posted:

Using Excel with my approach gives a rounder number, 8.125.  

with spacing for clarity

7/   sin (  acos (       ( 13^2 + 15^2 - 14^2 ) / (2 * 13 * 15)     )   &nbsp

So if that is what you calculate, then it is indeed the calculators at fault

 

It is indeed the calculator that is at fault, ie the machine, not you !

The correct answer is indeed 8.125 miles.

Well done !

For what it's worth, here is my solution !  Basically R = (axbxc)÷(4xArea of the triangle) 

 

Heron’s formula  Area = √[s(s-a)(s-b)(s-c)]             Where  s = semi-perimeter

Hence for triangle with sides 13,14 and 15…….

s = 21

Area = 84

 

SAS Rule  Area of Triangle = ½(abSinC)

Hence SinC = (2*Area)/ab        Eq(1)

 

Sine Rule  a/SinA = b/Sinb = c/Sinc = 2R    where R = Radius of the Circumcentre

Hence SinC = c/(2R)          Eq(2)

 

Combining Eq(1) and (2)

(2*Area)/ab = c/(2R)

Hence R = abc/4*(Area)            Eq(3)

 

Substituting the Area from Heron’s formula in Eq(3)

R = (13x14x15) ÷ (4x84) = 65/8 = 8.125 miles

Innocent Bystander posted:

Hmm, I’ve never come across the first of those. Second I’m not sure - vaguely familiar so probably have seen before.

Heron's Formula ? I have to admit. until a couple of years ago, I didn't know it was called Heron's Formula, but we had it drilled into us at school, nontheless !

Likewise the Side/Angle/Side Formula was drilled into us and the Full Sine Rule including the "2R" element at the end.

As I indicated above, some of these formulae appeared in the Farmer Watt's Field teaser back on Page 8

Anyway, you solved it - brilliant !

And you now know your calculator is useless !!!!!!!!! 

Don Atkinson posted:

And you now know your calculator is useless !!!!!!!!! 

And I always thought my pocket wonder-computer that happens to be usable as a phone was a great device because it combined so many useful things including a scientific calculator... 

Have to bin it and lug a laptop to have Excel as a calculator...

Innocent Bystander posted:
Don Atkinson posted:

And you now know your calculator is useless !!!!!!!!! 

And I always thought my pocket wonder-computer that happens to be usable as a phone was a great device because it combined so many useful things including a scientific calculator... 

Have to bin it and lug a laptop to have Excel as a calculator...

Well, at least this particular Brain Teaser thread has proved useful at last

Don Atkinson posted:
Innocent Bystander posted:

2 7 3                         3 2 7                      2 1 9

5 4 6                          6 5 4                      4 3 8

8 1 9                          9 8 1                       6 5 7

 

On the same idea, there is just one arrangement where each row is double the one above. What is it? (This may be a quicker one)

 

I think i'll give others a crack at this one before I respond

(you will obviously need to read my intro to this, a couple of posts above)

Hi IB

"Others" have had loads of time to think about this one. Nobody has come up with a solution.

Now, you've used the term "On the same idea". I have presumed this to mean "using the same rules" ie three numbers, each with three digits, spanning each and all of the digits 1 to 9.

The target is the only difference to my initial puzzle. ie each row to be double the one above. So that :-

123

246

492

would conform to the target, but would NOT conform to the rules (it doesn't use all the digits).

So would be invalid.

Innocent Bystander posted:
Innocent Bystander posted:

Hi Don, 

Yes, same rules

 

But I may have cocked it up! I can’t find my original scribble, and now I try I can’t find a solution that works...  sorry!

It’s ok IB, we all screw up occasionally

To find the four solutions to my initial puzzle, I wrote a simple Excel program to generate a column of the "top line" numbers from 123 to 334 ie 211 rows. I then generated two corresponding columns based on 2x and 3x the "top line" numbers. I did a manual search of the three columns to find the four rows that contained each of the digits 1 to 9. It was easy to do.

A few days ago I created a similar table to search for the one arrangement where each row is double the one above. This only ran from 123 to 250 for the column with the "top line" numbers ie 127 rows. At that time, using a manual search , I was unable to identify a row in which all the digits 1 to 9 appeared.

I wasn’t sure if you were somehow working to a different set of rules.

No spreadsheet in my method, just recognising combinations that won’t work, reducing, identifying first the limited possibilities for left colum, then start with each option for right colum, and in a large proportion of cases the middle column options are immediately evident as causing number duplication.

Don Atkinson posted:
notnaim man posted:

Total frustration, I have a pile of envelopes with numbers scribbled on. I got 347, then 375. Then, I have been told there is a similar puzzle about getting to see a mistress without being seen en route every day of the year, so without counting again, 365?

Tantalising, Frustrating, annoying............That's what some of these Brain Teasers are all about Also a bit of fun andgreat satisfaction when you get there !

Take a careful look at the "13" Node. ie there are 13 different routes to that specific Node.

They are quite easy to follow. and en-route you will have passed through Nodes that could have been reached by 5, 3 and another 5 routes respectively. (You might have noticed, there is only 1 route to the NW corner and there is only 1 route to the SE corner).

There is a "pattern" developing, with a Pub at the end.

(but sorry, the number of routes isn't 345, 347, 365 or 375)

More Frustration ! or Great satisfaction ?

Decided to have one more go: 321

So, for the person with a lover, they’d have been seen more than once 44 or 45 times.

Glad you didnt use a grid of 6x6, as the answer would have been a rather large 1683

Innocent Bystander posted:

I’m struggling as usual to upload a sketch, necessary for the explanation below to make sense. This is my last attempt - I’m providing the link as well in case it still doesn’t show.

 

https://imgur.com/a/WyfFD


Angles A,A are identical because sides X,X identical
Similarly B,B and C,C
Perpendicular from centre to any side bisects it for same reason.

Cosine rule:
Cos (A+C)   =  (15^2 + 13^2 - 14^2 ) /  2*15*13     = 0.50769
(A+C) = 59.49°

Total internal angles of triangle = 180° = 2A+2B+2C
So A+B+C=90°
So the other angle in one of the rt angle triangles with angle B and side length 7 = A+C (the angle found above)

Sin 59.49   =  7/x
∴ X =  7/sin 59.49      = 8.097 miles

  

Don, do you leave your bike at the station and return the same way, or, more sensibly, take it on the train with you to allow you to use it at your destination, with the flexibility when returning to choose the route home according to wind, either for maximum exercise or minimum effort depending how the day went

IB,

Your approach to this particular puzzle has inspired a new puzzle concerning circles and polygons (eg a triangle).

I'll post it later today - the training airfield is closed due to snow !

3,4,5 Triangles et al

Now I guess we have all heard of the “3,4,5” triangle ?

It’s a right angled triangle. The sides are all integers.

3^2 = 9;    4^2 = 16;  9+16 = 25, the root of which is 5

All nice and tidy.

The next such triangle has sides 5,12 and 13. But I guess many of us knew that as well ?

So, the question is…..

What’s the next one ?

……….And the next ?

…………… And the next ?

………………..In fact, how many such triangles are there ?

 

PS Note: I skipped the 6,8,10 triangle and others of a similar nature, because that’s just the same as the 3,4,5 triangle using a different scale !

PPS: this not the "inspirational" puzzle I referred to above ! But I hope it's not too dull and boring !!

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