IQ test
LaEna
now?" I thought, so I took an IQ test on the web. There was a time
limit for the test, approximately a minute per question. Not exactly
easiest questions but I am and have always been good at sequences,
numbers, pattern recognition so I had no problem with the test and
scored 93% better than most people. In other words I was in the top 7%
of population in this IQ test.
Quite encouraged I was eager to take another test to see if my result
will be consistent. Another time limit about 1 minute per question; I
was sweating away as if I was taking an important test in school! I
scored in the top 3% of people, 97% of people did worse. I knew that
the percentages that I lost were lost on English questions. They give
you an English word and then 5 choices what it means. Of course, the
choices were so close that I could not possibly decide on the answer so
I guessed.
'I scored so high, am I really so good?' was my question and deep down
there was a little nagging feeling. I run to the Chapters and bought a
book of IQ tests (brain teasers). I could not wait to start a test. 30
questions, 2 minutes per each. I could not solve the first question and
time was running so I left it (will do it later!) and started on the 2nd
and I could not do it in 2 minutes either, the same with the 3rd one so
I decided not to rush and just take my time. It's going to take me more
than 2 minutes so what can I do. However, I could not, in the world, do
it even if I thought about it for one hour.
So, dear friends, I'll give you two sample questions from the test:
What letter should replace the question mark? (The letters were in a
clockwise circle; I'll put them on the line)
A B D O P Q ?
Another question:
What number should replace the question mark?
74 65 61 37 58 ?
Awaiting your answers.
LaEna
Oldrich Kyn
Tyto problemy nemaji ovsem jednoznacne reseni. Jedno z pomerne
nejjednodusich reseni druheho problemu je 279 (zalozeno na predpokladu
konstantni ctvrte diference). Mnohem slozitejsi je predpoklad
polynomialu casu 4 stupne, coz dava priblizne 67100000 ale lze tezko
vyresit za dve minuty bez pocitace.
O.K.
Vladimir Bor
Konstantni ctvrta diference?
(Constant fourth difference?)
A co tohle znamena?:
Polynomial casu 4 stupne?
(Polynomial of time 4 degree?)
(Polynomial of time of fourth degree?)
(Fourth degree polynomial of time?)
(Fourth degree time polynomial?)
Frank
neznamena nic ale zni to vysoce vedecky.
Frank
"Vladimir Bor" <vladimír(dot)borNOSPAM(at)volný(dot)cz> wrote in message
news:3fccd...@news.bluewin.ch...
FEEB
>What letter should replace the question mark? (The letters were in a
>clockwise circle; I'll put them on the line)
>
>A B D O P Q ?
Nema to byt nahodou
A B D O P R ?
Frank Bures, <fe...@chem.utoronto.ca>
Vladimir Bor
Pavel Dvorak
V tom pripade by asi dalsi pismena byla D E L E (to A B na zacatku by bylo
trochu prebytecne :-))
Pavel
Lukas Hosek
> So, dear friends, I'll give you two sample questions from the test:
>
> What letter should replace the question mark? (The letters were in a
> clockwise circle; I'll put them on the line)
>
> A B D O P Q ?
>
Y.
> Another question:
>
>
> What number should replace the question mark?
>
> 74 65 61 37 58 ?
>
> Awaiting your answers.
68.
To je, co? :)
Lukas Hosek
Frank
"Lukas Hosek" <non...@your.business> wrote in message
news:bql6q7$21b2$1...@news.vol.cz...
Frank
yuko...@centrum.cz
>>
>> 68.
>>
>> To je, co? :)
>>
>> Lukas Hosek
>>
>69 is better
>Frank
>
Prosim te Frankie, jses katolik. Podle tveho
nabozenstvi sex je jenom pro "precreation". Vsechno
jine je hrichem. Tak co tu blbnes? Mas sny, ze
Beonce ti sedi na ksichte, ci co? y
gp66(remove_all_zzz).dialzzz.pipexzzz.comzzz
Výborné!!! Ha, ha, ha.
LaEna
>
>> What letter should replace the question mark? (The letters were in a
>> clockwise circle; I'll put them on the line)
>>
>> A B D O P Q ?
? = R
A B D O P Q R
Start at A and work clockwise including only letters which have enclosed
areas when printed.
>> What number should replace the question mark?
>>
>> 74 65 61 37 58 ?
>>
? = 89
74 65 61 37 58 89
Starting on the left, square each seperate digit from 2 digit numbers
and add the squares together to give the next number along, ie. 5**2 +
8**2 = 25 + 64 = 89
Of course, I gave you the answers from the 'Solutions' that are given at
the back of the IQ tests book. I would have never come up with the
answers myself; it's too advanced!
LaEna
Vladimir Bor
Ale co je "Konstantni ctvrta diference"?
A co "Polynomial casu 4 stupne" ?
> ? = R
>
>
> ? = 89
>
Paul JK
> Hi, here are the correct answers:
>
> >
> >> What letter should replace the question mark? (The letters were in a
> >> clockwise circle; I'll put them on the line)
> >>
> >> A B D O P Q ?
>
>
> ? = R
>
> A B D O P Q R
>
> Start at A and work clockwise including only letters which have enclosed
> areas when printed.
>
>
>
> >> What number should replace the question mark?
> >>
> >> 74 65 61 37 58 ?
> >>
>
> ? = 89
>
> 74 65 61 37 58 89
>
> Starting on the left, square each seperate digit from 2 digit numbers
> and add the squares together to give the next number along, ie. 5**2 +
> 8**2 = 25 + 64 = 89
Well, what happens next? After 89 it goes into 3 digits.
I wonder if it is divergent or if it always drops back
below 100. If it ever hits 86 then it collapses to 1 and
gets stuck there forever.
But I am too lazy to do the calculations by hand, perhaps
a little Pascal program is called for....
Paul JK
Vladimir Bor
of 1 to 3 digit numbers or converging to 1, as can be easily shown.
In fact, this particular sequences starts to repeat very soon:
<<74 65 61 37 58 89 145 42 18 >> <<65 61 37 58 89 145 42 18 >> etc.
An Excel-Macro would be simpler than a Pascal program.
"Paul JK" <pa...@nzx.com> wrote in message
news:84e8f4c7.03120...@posting.google.com...
Vladimir Bor
74 65 61 37 <<58 89 145 42 20 4 16 37 >> <<58 89 145 42 20 4 16 37>> etc.
"Vladimir Bor" <vladimír(dot)borNOSPAM(at)volný(dot)cz> wrote in message
news:3fd4736f$1...@news.bluewin.ch...
Zdislav V. Kovarik
On 7 Dec 2003, Paul JK wrote:
[...]
> > >> What number should replace the question mark?
> > >>
> > >> 74 65 61 37 58 ?
> > >>
> >
> > ? = 89
> >
> > 74 65 61 37 58 89
> >
> > Starting on the left, square each seperate digit from 2 digit numbers
> > and add the squares together to give the next number along, ie. 5**2 +
> > 8**2 = 25 + 64 = 89
>
> Well, what happens next? After 89 it goes into 3 digits.
> I wonder if it is divergent or if it always drops back
> below 100. If it ever hits 86 then it collapses to 1 and
> gets stuck there forever.
>
> But I am too lazy to do the calculations by hand, perhaps
> a little Pascal program is called for....
>
> Paul JK
This is a toy problem form a book by Hugo Steinhaus (I could get
the data if I were in my office), and in his form, more than a
Pascal program is needed:
Start with any positive number in its decimal representation
(the more digits, the merrier), sum up the squares of its
digits, apply the same procedure to the result, thus creating a
sequence.
Prove: After a finite number of steps, you hit either 1
(forever) or 89 (starting a loop).
Cheers, Slavek(ZVK)
Zdislav V. Kovarik
Reference:
"One Hundred Problems in Elementary Mathematics"
by Hugo Steinhaus
Dover Publications 1979
ISBN 0-486-23875-X
It is Problem #2.
Cheers, Slavek(ZVK)
Paul JK
> On 7 Dec 2003, Paul JK wrote:
> [...]
> > > >> What number should replace the question mark?
> > > >>
> > > >> 74 65 61 37 58 ?
> > > >>
> > >
> > > ? = 89
> > >
> > > 74 65 61 37 58 89
> > >
> > > Starting on the left, square each seperate digit from 2 digit numbers
> > > and add the squares together to give the next number along, ie. 5**2 +
> > > 8**2 = 25 + 64 = 89
> >
> > Well, what happens next? After 89 it goes into 3 digits.
> > I wonder if it is divergent or if it always drops back
> > below 100. If it ever hits 86 then it collapses to 1 and
> > gets stuck there forever.
> >
> > But I am too lazy to do the calculations by hand, perhaps
> > a little Pascal program is called for....
> >
> > Paul JK
Tak jste mi to lidi zkazili, chystal jsem si napsat programek
ale uz to nema cenu, kdyz vim jak to dopadne (udelal to butler).
> This is a toy problem form a book by Hugo Steinhaus (I could get
> the data if I were in my office), and in his form, more than a
> Pascal program is needed:
Proc by Pascal programek nestacil, to prece neni nijak
slozity ukol?
Programek by zacal serii pro kazde cislo od 1 do rekneme 10^n.
V kazde serii by daval cleny serie do pole a sledoval, jestli
se nekdy zacne opakovat. Jakmile se zacne opakovat, nebo preleze
zvolene maximum rekneme 10^2n (jina moznost neni) tak skonci serii,
vytiskne vysledek a zacne dalsi serii.
To prece neni nijak slozity ukol pro pritele packala?
Paul JK
Zdislav V. Kovarik
Kdybyste mel cely Turinguv stroj (vcetne nekonecne pasky), a
nekonecne casu k dispozici, tak snad by Pascal pomohl. (*)
Tak jak to Steinhaus formuloval, je to totiz otazka tykajici se
vsech prirozenych cisel.
(Definice totiz tvrdi, ze pokud nemate zaruku (dukaz), ze
pocitani skonci po konecnem poctu kroku, tak nemate algoritmus.)
Nastesti rozumova cast dukazu (na rozdil od strojove) je
snadna (a Vy jste ji mozna naznacil, ale ne dost jasne pro me):
Pokud ma cislo rekneme n cifer, nasledujici cislo je nanejvys
81*n. Tedy (snadna indukce): pokud pocet cifer je 4 nebo vice,
cisla v te posloupnosti (ostre) klesaji.
Zaver: Pascal potom pomuze, nebot je pouze zapotrebi proverit
vsechna trojciferna cisla do 324 vcetne (4*9*9).
(*) Nebo, kdyby kvantova mechanika neplatila, mohl by jeden
doufat v pocitac, ktery proveri prvni cislo za pul sekundy, a
kazde dalsi za polovinu predchazejiciho casu, takze po jedne
sekunde by to bylo vyreseno.
Cheers, Slavek(ZVK)
-------------------------
Na divadle v La Scala,
nevim, proc jsi mlaskala. (V+S)
FEEB
>Na divadle v La Scala,
>nevim, proc jsi mlaskala. (V+S)
>
Do zidli jsi triskala,
Madonna mia.
Misto abys tleskala,
orisky sis louskala,
Porco bestia.
Frank Bures, <fe...@chem.utoronto.ca>
Paul JK
> On 8 Dec 2003, Paul JK wrote:
>
> > "Zdislav V. Kovarik" <kov...@mcmail.cis.mcmaster.ca> wrote in message news:<Pine.SOL.4.33.031208...@mcmail.cis.mcmaster.ca>...
> > > On 7 Dec 2003, Paul JK wrote:
> > > [...]
[...]
> > Programek by zacal serii pro kazde cislo od 1 do rekneme 10^n.
> > V kazde serii by daval cleny serie do pole a sledoval, jestli
> > se nekdy zacne opakovat. Jakmile se zacne opakovat, nebo preleze
> > zvolene maximum rekneme 10^2n (jina moznost neni) tak skonci serii,
> > vytiskne vysledek a zacne dalsi serii.
> >
> > To prece neni nijak slozity ukol pro pritele packala?
> >
> > Paul JK
> >
> > > Start with any positive number in its decimal representation
> > > (the more digits, the merrier), sum up the squares of its
> > > digits, apply the same procedure to the result, thus creating a
> > > sequence.
> > > Prove: After a finite number of steps, you hit either 1
> > > (forever) or 89 (starting a loop).
> > >
> > > Cheers, Slavek(ZVK)
> >
>
> Kdybyste mel cely Turinguv stroj (vcetne nekonecne pasky), a
> nekonecne casu k dispozici, tak snad by Pascal pomohl. (*)
To chcete rict, ze vy ho jeste nemate? Ja pouzivam kapesni versi
pre-beta-released Microsoft Turinguv stroj implemented v Excelu
s nekonecnou columnkou :-)
Prozatim to bohuzel dodavaji bez nekonecneho casu.
Ten se musi koupit od third party suppliers.
> Tak jak to Steinhaus formuloval, je to totiz otazka tykajici se
> vsech prirozenych cisel.
> (Definice totiz tvrdi, ze pokud nemate zaruku (dukaz), ze
> pocitani skonci po konecnem poctu kroku, tak nemate algoritmus.)
No jo, vec je ta, ze ja jsem vzdelanim el.inzenyr.
Ja zacnu premyslet o dukazech, limitach, atd. atd.
az kdyz mne k tomu fyzikalni svet dokope.
Kdyz zplichtim programek, ktery po par minutach stale jeste
bezi, nebo pretece, tak teprve potom zacnu skrabat vzorecky
na papire a premyslet o konvergencich a nekonecnu.
> Nastesti rozumova cast dukazu (na rozdil od strojove) je
> snadna (a Vy jste ji mozna naznacil, ale ne dost jasne pro me):
(tak to nevim, jestli jsem to naznacil, kdyz to ale rikate,
tak, mozna....)
> Pokud ma cislo rekneme n cifer, nasledujici cislo je nanejvys
> 81*n. Tedy (snadna indukce): pokud pocet cifer je 4 nebo vice,
> cisla v te posloupnosti (ostre) klesaji.
> Zaver: Pascal potom pomuze, nebot je pouze zapotrebi proverit
> vsechna trojciferna cisla do 324 vcetne (4*9*9).
>
> (*) Nebo, kdyby kvantova mechanika neplatila, mohl by jeden
> doufat v pocitac, ktery proveri prvni cislo za pul sekundy, a
> kazde dalsi za polovinu predchazejiciho casu, takze po jedne
> sekunde by to bylo vyreseno.
>
> Cheers, Slavek(ZVK)
> -------------------------
> Na divadle v La Scala,
> nevim, proc jsi mlaskala. (V+S)
Paul JK
Na divadle v La Scala,
koho's to tam laskala? (PJK)