"Result the same"

sophia

when some numbers are subtracted or divided the result will be the same . can you give the numbers ? (Fractional numbers can be used)

the soln is given as (4.5 , 3), (5 1/3, 4), (6 1/4, 5)

NB:- 5 1/3, should be read as 5 and 1 by 3
6 1/4,should be read as 6 and 1 by 4


richard heathfield

I found 37 solutions in the range (-10, -10) to (+10, +10), of which 20
were negative.

Qurqirish dragon

From the given examples, I hope you can see that there is an infinite family of solutions of the form:
x*(1 + 1/(x-1)), x for all x other than 1.

Thus, you can extend the sequence of pairs with (using integers only):
(7 1/5, 6) (8 1/6, 7) (9 1/7, 8) and so on.

Your task is now to prove the apparent family I stated does indeed work.


Phillipe 92

Infinitely many :
for any integers p, q
a = p/q and b = p^2/(q*(p-q))
b - a = b/a = p/(p-q)

for instance p=16, q=2 result into a = 8, b = 64/7

64/7 - 8 = 8/7
(64/7)/8 = 8/7

> NB:- 5 1/3, should be read as 5 and 1 by 3

5 1/3 = 16/3 without any ambiguity
64/7 = 9 1/7 if you prefer this ambiguous writing, and needing a 'NB:'to explain...


Bill

"b - a = b/a" can be solved to give b = (a^2) / (a-1)
FYI: b + a = b*a = a = b / (b = 1)


Phillipe 92

Bill wrote :


> "b - a = b/a" can be solved to give b = (a^2) / (a-1)

As it seemed to be restricted to _rational_ numbers, I wrote 'a' as any rational number : p/q, and we get then the same formula...(p/q)^2/(p/q - 1 ) = p^2/(p*q - q^2) = p^2/(q*(p-q))

BTW this gives *all* rational solutions.

Restricting even more to integers results into much less solutions...
(the only solution is 4 - 2 = 4/2)

There is no interest in allowing a,b to be any _real_ number.

"year without change"

sulekha

the question is as follows, mention a past year that has no change when put upside down ?
the answer is given as 1961, is there any other solution ?


mensanator

sulekha wrote:
> the question is as follows, mention a past year that has no change
> when put upside down ?
> the answer is given as 1961, is there any other solution ?

You haven't defined what the "upside down" transformation actually is.

Do the glyphs pivot about the units, such that the thousands glyph becomes the units and hundreds become tens:
__________
| ___ |
| | | |
1961 abcd

so that 'a' is '1' uside down, 'b' is '6' upside down, etc.

Or are the glyphs inverted in place? (Probably not intended as that would be 1691, hardly unchanged.)

Either way the answer is wrong because the glyph '1' is not symetrical about the horizontal axis.

Ignoring that (assuming they ARE symmetrical about the horizontal axis), what glyphs are legit when inverted?

1 <--> 1
2 <--> 2
5 <--> 5
6 <--> 9
8 <--> 8
9 <--> 6
0 <--> 0

So, is 1961 the only year you turn upside down?

rick lones

sulekha wrote:
> the question is as follows, mention a past year that has no change
> when put upside down ?
> the answer is given as 1961, is there any other solution ?

Does anyone else remember the January 1961 special issue of Mad magazine, the one which was readable from front to back or vice versa (if you could even decide which was which)?

"1961 The first upside down year since 1881"
"1961 The last upside down year until 6009"

Of course, there are plenty of others.


eric sosman

1881, 1691, 1001, 986, 916, 906, 818, 808, 689, 619, 609,
96, 88, 69, 11, 8, 1, and perhaps 0 by courtesy.


CBFalconer

How about 1001, 101, 96, 69, 88, 888, 1881, 181?
If you don't insist on past years 6699 and 9966, 8888.


mark brader

Eric Sosman writes:
> 1881, 1691, 1001, 986, 916, 906, 818, 808, 689, 619, 609,
> 96, 88, 69, 11, 8, 1, and perhaps 0 by courtesy.

You missed one there.

If Wikipedia correctly describes the modern Assyrian calendar and we allow its use, then 6009, 6119, and 6699 are also years in the past.


jonnie103

Eric Sosman wrote:
> sulekha wrote:
> > the question is as follows, mention a past year that has no change
> > when put upside down ?
> > the answer is given as 1961, is there any other solution ?
> 1881, 1691, 1001, 986, 916, 906, 818, 808, 689, 619, 609,
> 96, 88, 69, 11, 8, 1, and perhaps 0 by courtesy.

I've always understood there was no Year 0, but that dates went straight from 1BC to 1AD.

Mensanator

Depends on who you ask.

> but that dates went straight from 1BC to 1AD.

"AD" was invented long after the transition, so that never happened.

And since going from 1BC to 1AD is mathematically stupid,those who actually make calculations (astronomers trying to backdate previous comet appearances, for example) use a Year 0.

Think of it this way, there WAS NO such transition, we just switched calendars:

BC: 4 3 2 1 0 -1 -2 -3 -4
AD: -3 -2 -1 0 1 2 3 4 5

It's just that the 0's don't coincide.


jim waters

mensanator wrote:
> jonnie303 wrote:
>> Eric Sosmanwrote:
>> > sulekhas wrote:
>> > > the question is as follows, mention a past year that has no change
>> > > when put upside down ?
>> > > the answer is given as 1961, is there any other solution ?
>> > 1881, 1691, 1001, 986, 916, 906, 818, 808, 689, 619, 609,
>> > 96, 88, 69, 11, 8, 1, and perhaps 0 by courtesy.
>> > Eric Sosman

I noticed in one of your earlier posts you had included the digits 2 and 5 as also inverting unto themselves, which is accurate if you use calculator displays, so how come no one yet has brought up the recent year 2002 ?


eric sosman

ObPuzzle: What about 910, 862, 798, 782, ...?

mensanator

What number is 7 supposed to be when upside down? That doesn't even work on a calculator.


mensanator

jim waters wrote
> I noticed in one of your earlier posts you had included the digits
> 2 and 5 as also inverting unto themselves, which is accurate if you
> use calculator displays, so how come no one yet has brought up the
> recent year 2002 ?

Because 5005 isn't the same year?


"King Of Pain"

"King Of Pain" wrote
"Mark Brader" wrote
>> Eric Sosman writes:
>>> 1881, 1691, 1001, 986, 916, 906, 818, 808, 689, 619, 609,
>>> 96, 88, 69, 11, 8, 1, and perhaps 0 by courtesy.
>> You missed one there.
> Which one of the three he missed are you talking about?

Scratch that - he missed several :)

101,111,181, 888, 1111


ashish garg

Phil Carmody wrote:
> Eric Sosman writes:
> > Mensanator wrote:
> >> Eric Sosman
> >>>ObPuzzle: What about 910, 862, 798, 782, ...?

Since we are talking about ambigrams http://en.wikipedia.org/wiki/Ambigram
If we are including calculator display...we should also include all form of modern calligraphic and some artist are so smart that they can write any number such that they are same up side down....have a look at some ambigrams

http://ambigram.net/


Rich Grise

This guy does them too:
http://www.scottkim.com/inversions/index.html

alphabets

sulekha

this is the question which was asked to me by my friend: -

suppose we are writing all the digits from one to hundred in words,
which all alphabets in english language are not used ?

the answer she gave me was A,B,C,J,K,L,M,P,Q,Z

How correct is this answer, assuming all the words are spelled
correctly ?

richard heathfield

The straight answer is: not very. Consider eLeven, which uses L.

The twisty answer, the "gry" answer if you like, is:

A

which is the only letter in "ENGLISH LANGUAGE" that is not used in the first hundred counting numbers as normally rendered in English. The letters are: A E G H I L N S U; EIGHT eliminates all but ALNSU, ELEVEN eliminates L and N, SIX eliminates S, and FOUR eliminates U, leaving A.

In English, A is eliminated in the very next number, ONE-HUNDRED-AND-ONE.In Usanian, we have to wait until ONE-THOUSAND.


dgates

It's funny. At first glance, you'd think that over the course of 100 or 1,000 numbers, you'd pass through a lot of letters. But there really aren't that many times while counting that new letters are used.

"one" through "ten," okay. (not that each one necessarily uses a new letter, but each word is made of a couple parts/sounds/syllables that weren't already used in the previous ones.)

"eleven," "twelve," sure. (again, at least there are new parts/sounds/syllables being introduced)

"thirteen" at least is something other than the combination of a previously used number plus "teen."

then no fresh parts/sounds/syllables until "twenty" and (maybe) "thirty"

and then (depending on what you think of "forty") nothing new till "hundred," and nothing new after that until "thousand."

By my count, that's only about 17 different words/sounds that get you
all the way up to 1,000 (or, actually, up to 999,999).

one
two
three
four
five

six
seven
eight
nine
ten

eleven
twelve
thirteen
twenty
forty

hundred
thousand

Esra Sdrawkcab

"and"
It's easy for the speaking-clock stylee programs.

I pronounce fifty differently to five-ty, and thirty to three-ty, but on the plus side forty sounds like four-ty. (same with the 3 and 5 teens)


CBFalconer

Richard Heathfield wrote:


> The twisty answer, the "gry" answer if you like, is:

> A

> which is the only letter in "ENGLISH LANGUAGE" that is not used
> in the first hundred counting numbers as normally rendered in
> English. The letters are: A E G H I L N S U; EIGHT eliminates all
> but ALNSU, ELEVEN eliminates L and N, SIX eliminates S, and FOUR
> eliminates U, leaving A.

> In English, A is eliminated in the very next number,
> ONE-HUNDRED-AND-ONE. In Usanian, we have to wait until ONE-THOUSAND.

How about "four and twenty". :-)

crescent

sophia

This is the question which i saw in a book

can you divide a crescent into 6 parts by drawing just two straight
lines ?

The solution is as given here:-



can any one give/suggest an alternative solution ?


Richard Heathfield

You could simply switch geometries. In hyperbolic geometry, a "straight line" is defined as one that intersects the edge of the (circular) universe at right angles, which means we should be able to use a couple of what we would normally call circles to cut the crescent into six chunks. I
scratched out a solution on paper which seems to work, but converting this to ASCII art proved to be beyond my meagre artistic skills.

even numbers & prime numbers

sophia

this is the question which I saw in a book :- can any even number greater than 2 be expressed as the product of primes ?

the answer is given as : people just don't know


Jose Carlos Santos

sophia wrote

> this is the question which I saw in a book :- can any even number
> greater than 2 be expressed as the product of primes ?

> the answer is given as : people just don't know



The answer is: *any* number (even or odd, greater than 2 or not) can be expressed as a product of primes.This was proved a long, long time ago.

What people do not know about even numbers greater than 2 is whether or not they can always be expressed as the _sum_ of _two_ prime numbers. The assertion "yes, it is aways possible" is known as "Goldbach conjecture".


sophia

yeah you are correct, i quote the question wrongly

amzoti

http://mathworld.wolfram.com/PrimeFactorization.html


Stephen Montgomery-Smith

sophia wrote:
> this is the question which I saw in a book :- can any even number
> greater than 2 be expressed as the product of primes ?
> the answer is given as : people just don't know

I think you are referring to the famous Goldbach conjecture, which asks whether any even number greater than 2 can always be expressed as the SUM of two primes. People just don't know.

Goldbach conjecture: can every even number greater than two always be expressed as the sum of two primes?

Silverbach conjecture: can every sufficiently large odd number always be expressed as the sum of three primes?

Bronzebach conjecture: can every even number always be expressed as the sum of two odd numbers?

Woodenbach conjecture: can every odd number always be expressed as the sum of two even numbers?


bill

Leadbach conjecture: All primes are the sum of two smaller primes

factorial

sophia

There are many factorials that can be expressed as the product of two or more other factorials.

ex:- 7! x 6! = 10!
5! x 3! = 6!
3! x 5! x 7! = 10!
2! x 47! x 4! = 48!
2! x 287! x 4! x 3! = 288!

how many others can you find ?

Is there any formula by which we can express a factorial as a product 2 or more factorials (excluding 0! and 1!) ?


Eric sosman

There is an infinite supply matching the pattern of your second example:

3! * 5! = 6!
4! * 23! = 24!
5! * 119! = 120!
6! * 719! = 720!
7! * 5039! = 5040!
...
n! * (n! - 1)! = (n!)!


CBFaloner

sophia wrote:

> There are many factorials that can be expressed as the product of
> two or more other factorials.

> ex:- 7! x 6! = 10!
> 5! x 3! = 6!
> 3! x 5! x 7! = 10!
> 2! x 47! x 4! = 48!
> 2! x 287! x 4! x 3! = 288!

> how many others can you find ?

> Is there any formula by which we can express a factorial as a
> product 2 or more factorials (excluding 0! and 1!) ?

I think the key is express each factorial as a product of prime factors. The only numbers needed are the powers of those primes. For example,

5 3 2 (<- the primes)
3! = 3 * 2 = 1,1.
5! = 5*4*3*2 = 5*2*2*3*2 = 3,1,1
6! = 6*5*4*3*2 = 3*2*5*2*2*3*2 = 4,2,1

Tripod

sophia

the following is the question which i saw in a book
why is that a tripod stands firmly, even when its three legs are of different length ?

the answer is given as follows,The three legs of a tripod always rest on the floor because through any three points of space there can pass one plane and only one. the reason is purely geometrical and not physical. can any one give an alternative answer ?


barabara bailey

No, that's pretty much why a tripod or three-legged stool always has all three feet in firm contact with the surface it's standing on. Whether it's "standing firmly is a function of the angle that the upper surface is at an the position of the center of gravity to the center of mass. A tall tripod with a heavy camera mounted on it, and standing at an angle that is extremely off-vertical will still tip over.

mensanator

And the plane determined by the 3 points mitght not be the the surface; such as when you take thr tripod outside.

CBFalconer

Barbara Bailey wrote:
> sophia wrote:

>> the following is the question which i saw in a book

>> why is that a tripod stands firmly, even when its three legs are
>> of different length ?

>> the answer is given as follows,
>> The three legs of a tripod always rest on the floor because
>> through any three points of space there can pass one plane and
>> only one. the reason is purely geometrical and not physical.

>> can any one give an alternative answer ?

> No, that's pretty much why a tripod or three-legged stool always
> has all three feet in firm contact with the surface it's standing
> on. Whether it's "standing firmly is a function of the angle that
> the upper surface is at an the position of the center of gravity
> to the center of mass. A tall tripod with a heavy camera mounted
> on it, and standing at an angle that is extremely off-vertical
> will still tip over.

The stability depends on the location of the Center of Gravity (CG) of the system. If it lies within the triangle of the floor contact points, the system is stable.



mark brader

Sophia Agnes:

>>> why is that a tripod stands firmly, even when its three legs are
>>> of different length ?

Barbara Bailey:

>> No, that's pretty much why a tripod or three-legged stool always
>> has all three feet in firm contact with the surface it's standing
>> on. Whether it's "standing firmly is a function of the angle that
>> the upper surface is at an the position of the center of gravity
>> to the center of mass. A tall tripod with a heavy camera mounted
>> on it, and standing at an angle that is extremely off-vertical
>> will still tip over.

Chuck Falconer:

> The stability depends on the location of the Center of Gravity (CG)
> of the system. If it lies within the triangle of the floor contact
> points, the system is stable.

No; if it lies within the triangle and there is no wind or similar disturbing force, then it won't tip over immediately. But to be "stable" or "standing firmly", it also needs to be true that a small disturbing force won't move the thing enough that the CG -- or rather, the point directly below the CG -- goes outside of the triangle.


mathew .t russotto

Mark Brader wrote:

>No; if it lies within the triangle and there is no wind or similar
>disturbing force, then it won't tip over immediately. But to be
>"stable" or "standing firmly", it also needs to be true that a small
>disturbing force won't move the thing enough that the CG -- or rather,
>the point directly below the CG -- goes outside of the triangle.

I think that's true if the CG is over the interior of the triangle; the only unstable equilibria are where the CG is over an edge or vertex of the triangle.


mark brader

Mark Brader:

>> No; if it lies within the triangle and there is no wind or similar
>> disturbing force, then it won't tip over immediately. But to be
>> "stable" or "standing firmly", it also needs to be true that a small
>> disturbing force won't move the thing enough that the CG -- or rather,
>> the point directly below the CG -- goes outside of the triangle.

Matthew Russotto:

> I think that's true if the CG is over the interior of the triangle;
> the only unstable equilibria are where the CG is over an edge or
> vertex of the triangle.

If you take "stable" to be an all-or-nothing condition, that's true.


CBFalconer
Mark Brader wrote:
> Sophia Agnes:
>>>> why is that a tripod stands firmly, even when its three legs are
>>>> of different length ?

> Barbara Bailey:
>>> No, that's pretty much why a tripod or three-legged stool always
>>> has all three feet in firm contact with the surface it's standing
>>> on. Whether it's "standing firmly is a function of the angle that
>>> the upper surface is at an the position of the center of gravity
>>> to the center of mass. A tall tripod with a heavy camera mounted
>>> on it, and standing at an angle that is extremely off-vertical
>>> will still tip over.

> Chuck Falconer:
>> The stability depends on the location of the Center of Gravity (CG)
>> of the system. If it lies within the triangle of the floor contact
>> points, the system is stable.

> No; if it lies within the triangle and there is no wind or similar
> disturbing force, then it won't tip over immediately. But to be
> "stable" or "standing firmly", it also needs to be true that a small
> disturbing force won't move the thing enough that the CG -- or rather,
> the point directly below the CG -- goes outside of the triangle.

Hey, I'm speaking ideal triangles, without disturbing influences.We find the stability by dropping an ideal string, weighted with an ideal pin, from the CG.