TEST UR MATH!!!

First topic message reminder :

mga ka cg pips....patunayan ntin na ang mga artist ay magaling din sa math!!!!
common mind buggling problems lng po...para maiba nman.. post ntin lahat dito... umpisahan ko na...hehhe
kumbaga, mind exercise lng para magkaron din tau ng kakaibang exercises....weeeeeee


ok....game nah...hehehe

Problem no. 1


There is a bus with 7 girls .

Each girl has 7 bags .

In each bag, there are 7 big cats

Each big cat has 7 little cats.

Each cat has 4 legs .

Question: How many legs are present in the bus?

nway nice thread po mam.... nakaka-calibrate ng utak ung mga problem .. hehehe!!

ako isip din ako ng itatanong..

alwin wrote:share ko po!
kunyari:
pareho tayo nagtinda nang balut, tapos nagkasalubong tayo,
sabi mo sa akin ilan nalang ang natira sa balut mo?
sabi ko, kung bigyan mo ako nang isa kalahati lang yung sayo kesa sa akin,
pero kung bigyan kita nang isa pareho tayo karami.
ang tanong: ilan yung sa kanya at ilan rin yung sa akin?

ganda ng thread na to a
sagot ko ay ung sa akin = 7, ung sa kanya = 5 tama kaya?

[quote="kietsmark"]

alwin wrote:share ko po!
kunyari:
pareho tayo nagtinda nang balut, tapos nagkasalubong tayo,
sabi mo sa akin ilan nalang ang natira sa balut mo?
sabi ko, kung bigyan mo ako nang isa kalahati lang yung sayo kesa sa akin,
pero kung bigyan kita nang isa pareho tayo karami.
ang tanong: ilan yung sa kanya at ilan rin yung sa akin?[/quote]

ganda ng thread na to a
sagot ko ay ung sa akin = 7, ung sa kanya = 5 tama kaya?

hmmmm...i agree po sa answer ni sir kietsmark...un din answer ku ei...7 and 5...tma po ba?weeeeeee

bokkins wrote:haha, nagets ko na. 6:59pm! galing nito ah! thanks rica.

ang laking jar siguro nito.

ok sir!!!galing..kuha nyu agad!!!exciting noh?
liit lng ng jar na yan...parang ung sa bagoong lng ng "barrio fiesta"...waheheheh
hahahha...ok sir bok..nakakagulo ng utak noh?wahehehe...eto ang hobby ko pg wla akong mgawa ei..wahehehe

corpsegrinder wrote:yup 6:59!!! half of jar + 1 min(half again/double)= 7:00(full)

astig!!!kuha nyo din sir!!!congrats po...hehehehe
share din po kau ng mga tinatago nyu dyan!!!!

nauubusan ako ng dugo dito ah....

Hay naku mam rica..
Problem 1 palang ako nahihilo na ako hinay hinay lang ha..

nabobo nakng tuluyan sa math.. nyahhahah...

Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers? (Computer assistance allowed.)

[quote="rica"]

kietsmark wrote:

alwin wrote:share ko po!
kunyari:
pareho tayo nagtinda nang balut, tapos nagkasalubong tayo,
sabi mo sa akin ilan nalang ang natira sa balut mo?
sabi ko, kung bigyan mo ako nang isa kalahati lang yung sayo kesa sa akin,
pero kung bigyan kita nang isa pareho tayo karami.
ang tanong: ilan yung sa kanya at ilan rin yung sa akin?[/quote]

ganda ng thread na to a
sagot ko ay ung sa akin = 7, ung sa kanya = 5 tama kaya?

hmmmm...i agree po sa answer ni sir kietsmark...un din answer ku ei...7 and 5...tma po ba?weeeeeee

yap tama!!!! gets nyo weeeeeeeeeeeeee

ARCHITHEKTHURA wrote:Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers? (Computer assistance allowed.)

mahirap to, may mga great than... hahaha

Dugo ilong ko dito... hilo hilo pa.. waaaaaa

nakupo to na problema ko. its time for my leftside of the brain to work. puro kasi right side nalang.

ARCHITHEKTHURA wrote:Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers? (Computer assistance allowed.)

ask ko lang sir kung may absolute value ba ito o derivative din ang sagot?

ARCHITHEKTHURA wrote:Two perfect logicians, S and P, are told that integers x and y have been chosen such that 1 < x < y and x+y < 100. S is given the value x+y and P is given the value xy. They then have the following conversation.

P: I cannot determine the two numbers.
S: I knew that.
P: Now I can determine them.
S: So can I.

Given that the above statements are true, what are the two numbers? (Computer assistance allowed.)

hello po sir, hmmmm...mkhang mhirap tung problem nyu ah..waheheh
ask ko lng po...la po bang range ung difference ng two numbers?marami kcing possible answer ei..

e2 try ko lng...wahehehe...khit mali ok din...hehee
x=4 and y=13
S=17 and P=52...hahahaa..wild guess lang pu...try kpa ulit mg-icip...

Galing rica!If you dont mind pwde mo i post yung solution mo..Thanks!

sagot is x = 4, y=13

eto solution
First of all, trivially, xy cannot be prime. It also cannot be the square of a prime, for that would imply x = y.

We now deduce as much as possible from each of the logicians' statements. We have only public information: the problem statement, the logicians' statements, and the knowledge that the logicians, being perfect, will always make correct and complete deductions. Each logician has, in addition, one piece of private information: sum or product.

P: I cannot determine the two numbers.

P's statement implies that xy cannot have exactly two distinct proper factors whose sum is less than 100. Call such a pair of factors eligible.

For example, xy cannot be the product of two distinct primes, for then P could deduce the numbers. Likewise, xy cannot be the cube of a prime, such as 33 = 27, for then 3×9 would be a unique factorization; or the fourth power of a prime.

Other combinations are ruled out by the fact that the sum of the two factors must be less than 100. For example, xy cannot be 242 = 2×112, since 11×22 is the unique eligible factorization; 2×121 being ineligible. Similarly for xy = 318 = 2×3×53.

S: I knew that.

If S was sure that P could not deduce the numbers, then none of the possible summands of x+y can be such that their product has exactly one pair of eligible factors. For example, x+y could not be 51, since summands 17 and 34 produce xy = 578, which would permit P to deduce the numbers.

We can generate a list of values of x+y that are never the sum of precisely two eligible factors. The following list is generated by JavaScript; the function may be inspected by viewing JavaScript: function genSum (plain text.)

Eligible sums: 11, 17, 23, 27, 29, 35, 37, 41, 47, 53.

(We can use Goldbach's Conjecture, which states that every even integer greater than 2 can be expressed as the sum of two primes, to deduce that the above list can contain only odd numbers. Although the conjecture remains unproven, it has been verified empirically up to 1018.)

P: Now I can determine them.

P now knows that x+y is one of the values listed above. If this enables P to deduce x and y, then, of the eligible factorizations of xy, there must be precisely one for which the sum of the factors is in the list. The table below, generated by JavaScript (view plain text JavaScript: function genProd), shows all such xy, together with the corresponding x, y, and x+y. The table is sorted by sum and then product.

Note that a product may be absent from the table for one of two reasons. Either none of its eligible factorizations appears in the above list of eligible sums (example: 12 = 2×6 and 3×4; sums 8 and 7), or more than one such factorization appears (example: 30 = 2×15 and 5×6; sums 17 and 11.)

S: So can I.

If S can deduce the numbers from the table below, there must be a sum that appears exactly once in the table. Checking the table, we find just one such sum: 17.

Therefore, we are able to deduce that the numbers are x = 4 and y = 13.

DUGO ILONG DITO HEHEHE

astig! hahahah ang talino mo sir mitch.. parang nakita ko to sa wikianswers ah parang lang naman.. hehehe

christine wrote:astig! hahahah ang talino mo sir mitch.. parang nakita ko to sa wikianswers ah parang lang naman..

ahahaha sinunod ko lang ung instruction ni parekoy... computer assistance is allowed wehehehe read instruction first kumbaga

HAHAHAHAHHAA.Tumpak parekoy!parehas ata tayo ng pinagkuhaan a!hehehe..

ARCHITHEKTHURA wrote:HAHAHAHAHHAA.Tumpak parekoy!parehas ata tayo ng pinagkuhaan a!hehehe..

hehehe eto ba ung 1-160 ung sample puzzles? hindi naman cheating ung ginawa ko ano parekoy... nasunod ko ung instruction mo e hehehe

ang gagaling nyo naman guys,ambunan nyo naman kami ng konting numero nyo

hindi po naituro sa akin yan nung college eh.(IBA INAATUPAG KO)

ang tanong .. naintindihan mo ba un isinagot mo bago ni-copy-paste??? ???

KettleRenderer wrote:ang tanong .. naintindihan mo ba un isinagot mo bago ni-copy-paste??? ???

hindi... saka wala akong pakialam hehehe tapos na ko sa school e

kietsmark wrote:

KettleRenderer wrote:ang tanong .. naintindihan mo ba un isinagot mo bago ni-copy-paste??? ???

hindi... saka wala akong pakialam hehehe tapos na ko sa school e

qui gon wrote:ang gagaling nyo naman guys,ambunan nyo naman kami ng konting numero nyo

hindi po naituro sa akin yan nung college eh.(IBA INAATUPAG KO)

dre si mr. google ang magaling hehehe

oops.. sorry iba na pala yung tanong.. :p