Brain Teaser No 1

2.4m. I was going to ask the width of the alley - but then I twigged.

quote:

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I sent some polish to my Polish friend.

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Don,

Thanks for polishing that one off!

cheers,

Dan

After getting the answer to the two ladders, I have spent the last hour trying to figure why it crunches down to obey the rule: (X x Y) / (X + Y).
Can anyone put me out of my misery?

Rasher,

This isn't the why unless you can place blind faith in algabra.

Generate following equations to track lines (origin at base of wall height A and other wall height B, gap is M)

y = -A/M * x + A

and

y = B/M * x (or x = M/B * y)

elimnate x gives

y = -A/M * M/B * y + A

which simplifies to y = AB/(A+B).

Matthew

I saw ladder lengths rather than touching the wall heights.

I'll go back under my stone now.

Paul

Paul,

You still managed to get the answer right! I thought that was the question but when there was no indication of wall spacing I read the question again.

Matthew

Matthew
I understand the re-arrangements of the formula, but still can't see where the equation originates. I am still looking for some simple geometric pattern in the triangles, but with the 2 variables it is just making the cooling fans in my head come on. I wish I was smarter sometimes. I found the equation by muliplying the two sides first to keep the number crunching nicely rounded up, and then spotted what was happening when trying different values. So I've found the answer without understanding how!

Rasher,

You seem smart enough to me ! answers = answers

I have added a few letters to my original diagram.

You should now be able to see two triangles, ABC and BCD

They share a common side BC and also a common dimension EF (which is parallel to the walls)

To make the geometry/algebra a bit easier (I hope) I have called

AB = a; DC = b; BC = x; BF = x'; EF = y

Triangles ABC and EFC are similar (equal angles)

Hence EF/AB = FC/BF (hope this rings a few bells?)

Substituting gives

y/a = (x-x')/x (1) (take your time, it does work)

Triangles BCD and BFE are similar

Hence EF/DC = BF/BC

Substituting gives

y/b = x'/x (2)

Hope you're with me so far ?

(1) gives xy = a(x-x') (3)

(2) gives x' = xy/b (4)

Substituting for x' in (3) gives

xy = ax - axy/b Note, each term contains an "x"

y = a - ay/b (ie the width of the ally is irrelevant)

by = ba - ay

(a + b)y = ba

y = ab/(a + b) QED

Hope this helps, but if not, just ask!

Cheers

Don

Cheers Don
It was the EFC triangle that I needed, and I just couldn't see it.
Thanks.

Take a bath!

It takes one tap 12 minutes to fill the bath by itself. It takes the other tap 6 minutes to fill the bath by itself. The filled bath will empty in eight minutes following plug removal.

If both taps are on and the plug is left out, how long will take the empty bath to fill, if indeed it will fill at all.

You can assume that the depth of water in the bath will not affect its rate of filling or emptying.

Cheers

Don

Going round in circles ?

In the circle the numbers represent the size of the sectors. By combining adjacent sectors it is possible to find sectors in this circle of sizes 1, 2, 3.....all the way up to 13, (13 being the whole circle)

In a similar fashion, given a circle divided up into 5 sectors in a certain way, it is possible to combine adjacent sectors to give sizes 1, 2, 3.....up to the biggest possible in the circumstances.

What are the sizes of the five sectors (you can produce the drawing if that helps)

Cheers

Don

Palindromic Dates,

Many dates are palindromic when written in the form "day/month/year" without any breaks and without any leading zeros.

Eg September 27th 1972 is written 2791972 and is palindromic.

How many palindromic dates are there in this millenium ? (ie the one that started on 1st January 2000)

Cheers

Don

Your views count

Of course, I would welcome your views on the last two teasers, eg do they look so easy they aren't worth bothering with; perfectly straight forward - I just need to find 5 minutes to tidy up my thinking; difficult; nigh-on impossible.. so I won't even start......etc

Only printable comments please........

Cheers

Don

Don,

31 (hopes it is right, knee jerk answer, probably very easy)

and

that looks like a complete and utter nightmare!

thanks

Matthew

The segments of a circle teaser probably touches on profound group and set theory, or trial and error... Is there a mathematician in the house?

Datier matters,

1-1-2211...9-1-2219
1-2-2221...9-2-2229
...
1-9-2292...9-9-2299

10-1-2101...31-1-2113
10-2-2201...29-2-2292
...
10-9-2901...31-9-2913

1-10-2011...9-10-2019
1-11-2111...9-11-2119
1-12-2211...9-12-2219
10-10...31-10 - no palindrome...
10-11...30-11 - no palindrome...
10-12-2101...31-12-2113

Of the 366 possible dates 10 Oct to 31 Oct and 10 Nov to 30 Nov can't be palindromic in this millenium. So the answer is possibly 323.

Or not.

Paul

I think the circle can have an area no bigger than 21.

Paul

Of the 366 possible dates 10 Oct to 31 Oct and 10 Nov to 30 Nov can't be palindromic in this millenium. So the answer is possibly 323.

That is in fact correct. Here's the complete list:
1102011,2102012,3102013,4102014,5102015,6102016,
7102017,8102018,9102019,1012101,10122101,2012102,
20122102,3012103,30122103,1112111,1112111,
11122111,2112112,2112112,21122112,3112113,3112113
,31122113,4112114,5112115,6112116,7112117,8112118
,9112119,1212121,12122121,2212122,22122122,13121
31,13122131,2312132,23122132,1412141,14122141,2412142,
24122142,1512151,15122151,2512152,25122152,1612161,
16122161,2612162,26122162,1712171,17122171,2712172,
27122172,1812181,18122181,2812182,28122182,1912191,
19122191,2912192,29122192,1022201,2022202,112211,
1122211,1122211,212212,2122212,2122212,312213,3122213,
412214,4122214,512215,5122215,612216,6122216,712217,
7122217,812218,8122218,912219,9122219,122221,1222221,
222222,2222222,322223,422224,522225,622226,722227,
822228,922229,1322231,132231,2322232,232232,332233,
432234,532235,632236,732237,832238,932239,1422241,
142241,2422242,242242,342243,442244,542245,642246,
742247,842248,942249,1522251,152251,2522252,252252,
352253,452254,552255,652256,752257,852258,952259,
1622261,162261,2622262,262262,362263,462264,562265,
662266,762267,862268,962269,1722271,172271,2722272,
272272,372273,472274,572275,672276,772277,872278,
972279,1822281,182281,2822282,282282,382283,482284,
582285,682286,782287,882288,982289,1922291,192291,
2922292,292292,392293,492294,592295,692296,792297,
892298,992299,1032301,2032302,3032303,1132311,2132312,
3132313,1232321,2232322,1332331,2332332,1432341,
2432342,1532351,2532352,1632361,2632362,1732371,2732372,
1832381,2832382,1932391,2932392,1042401,2042402,
3042403,1142411,2142412,1242421,2242422,1342431,
2342432,1442441,2442442,1542451,2542452,1642461,
2642462,1742471,2742472,1842481,2842482,1942491,
2942492,1052501,2052502,3052503,1152511,2152512,
3152513,1252521,2252522,1352531,2352532,1452541,
2452542,1552551,2552552,1652561,2652562,1752571,
2752572,1852581,2852582,1952591,2952592,1062601,
2062602,3062603,1162611,2162612,1262621,2262622,
1362631,2362632,1462641,2462642,1562651,2562652,
1662661,2662662,1762671,2762672,1862681,2862682,
1962691,2962692,1072701,2072702,3072703,1172711,
2172712,3172713,1272721,2272722,1372731,2372732,
1472741,2472742,1572751,2572752,1672761,2672762,
1772771,2772772,1872781,2872782,1972791,2972792,
1082801,2082802,3082803,1182811,2182812,3182813,
1282821,2282822,1382831,2382832,1482841,2482842,
1582851,2582852,1682861,2682862,1782871,2782872,
1882881,2882882,1982891,2982892,1092901,2092902,
3092903,1192911,2192912,1292921,2292922,1392931,
2392932,1492941,2492942,1592951,2592952,1692961,
2692962,1792971,2792972,1892981,2892982,1992991,
2992992

Here's what I used to generate it:
public static void palindromicDates()
{
Date date = Utilities.makeDateTime( "2000-01-01" );
Date end = Utilities.makeDateTime( "3000-01-01" );
ArrayList pals = new ArrayList();
DateFormat format = new Utilities.GMT_DateFormat( "dMyyyy" );
for( int i=0; date.before( end ); ++i )
{
String p = Utilities.dateToString( date, format );
// System.out.println( p );
char[] ch = p.toCharArray();
boolean isPalindrome = true;
int n = ch.length/2;
for( int j=0; j<n && isPalindrome; ++j )
{
if( !(ch[j]==ch[ch.length-(j+1)]) ) isPalindrome = false;
}
if( isPalindrome ) pals.add( p );
date = Utilities.addDays( date, 1 );
}
System.out.println( pals.size() );
for( Iterator itr=pals.iterator(); itr.hasNext(); )
{
System.out.print( itr.next() + "," );
}
}

John

What are the sizes of the five sectors (you can produce the drawing if that helps)

I think the answer is 1, 2, 6, 5, 4 giving an answer of 18. The logic is thus:
In the 4 sector example the adjacent pairs add up to 3, 5, 8, 10. You can thus insert one of these numbers as the fifth element and still maintain the original situation. The biggest of these that works is 5.
John

By a simple progression you would expect the answer to be as following:
Sectors Numbers Total Diff
1 1 1
2 1,2 3 2
3 1,2,4 7 4
4 1,2,4,6 13 6
5 1,2,4,6,8 21 8

But I am reasonably sure this is not the case because with an 8 in place you cannot make 9 other than by putting it next to the 1 which obviously doesn't work. I'm pretty sure therefore my original answer is the correct one, unless someone can prove otherwise!
John

The example makes use of all possible combinations of segment without duplicating a size. So logically the 5 segment circle should add up to 21 and produce all sizes from 1 to 21.

Paul

Paul,
You misread it is possible to combine adjacent sectors ...which makes the problem a lot harder.
John

Paul,

John is correct in that you are only allowed to combine adjacent sectors. However, you can combine more than two adjacent sectors in order to generate the numbers.

Hope this clarifies

Cheers

Don

Matthew T,

Sorry Matthew, 31 isn't right.... keep hoping ...and yes, the date dilemma looks daunting, even with John's program in front of you..

Cheers

Don

Paul,

I went for the trial and error technique.

But I would like to see the profund thesis on groups and sets....

Cheers

Don

Paul, John,

323 it is. Well done.

I can see how John did it. Paul ?

My method was manual, based on examining 6-digit; 7-digit and 8-digit dates. but I never actually needed to write all of the dates down.

Cheers

Don