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General Forum: Love | | PUZZLES PUZZLES AND PUZZLES | |
| simple question fast gaa cheppali ans...
In order to write counting upto 100,how many times the key of the keyboard will have to be pressed?????
Posted by: Mrs. Aruna At: 15, Oct 2003 12:57:20 PM IST
yeah....u know it...but cannot bring it out....
vikram.....now urs turn to give some goo dquestion......i think u gave very less questions.....
Posted by: Raj Sekhar At: 14, Oct 2003 8:12:38 PM IST
chinnaga chala baaga cheppavu pallantla.
thelisi cheppalekapovatam lone ila naala problems vuntaayi. :o)
Posted by: MONARCH 007 At: 14, Oct 2003 8:06:49 PM IST
naaku nee bhaadhanthaa ardham ayyindi....chaalaa baagaa try chesaavu.....
aa equation ee problem ki key point....kaani aa equation ni baagaa explain cheyyledu.....
/*****
for 14n+3 and 21n+4 to be divisible by a factor ...lets say f, f has to divide the left hand side of equation : 3(14n+3)-2(21n+4).
And as the above equation states,
this should also divide the right hand side of the equation 1.
As 1(RHS) is only divisible by 1, LHS is also divisible only by 1.
As LHS is divisible by 1, 14n+3 and 21n+4 are divisible only by 1.****/
alaa cheppkoodadu....adi reasonable kaadu.....
left side divide chesthe right side divide cheyyaalani rule ledu kadaa....u better think on this point again....
3(14n+3) - 2(21n+4) = 1
the above equation can be best explained like this.....
it means.....the numbers 3(14n+3) and 2(21n+4) are consecutive numbers and hence they r "relatively prime".......
"any two consecutive numbers r relatively prime"
then obviously.....14n+3 and 21n+4 r relatively prime....
if p*q and r*s do not have any factor in common,then (p,r),(p,s),(q,r),(q,s) will be relatively prime....
anyhow well done vikram.....great work......
Posted by: Raj Sekhar At: 14, Oct 2003 7:59:45 PM IST
okay.. alagaithe argument wrong, answer correct avuthundi.
supporting the answer, i can state this:
consider the equation
3(14n+3) - 2(21n+4) = 1 (it is true)
for 14n+3 and 21n+4 to be divisible by a factor ...lets say f, f has to divide the left hand side of equation : 3(14n+3)-2(21n+4).
And as the above equation states,
this should also divide the right hand side of the equation 1.
As 1(RHS) is only divisible by 1, LHS is also divisible only by 1.
As LHS is divisible by 1, 14n+3 and 21n+4 are divisible only by 1.
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indulo neeku enthavaraku ardamavuthundanedi naaku doubte. I am sure I couldn't explain it better.
lets see if this satisfies your idea.
Posted by: MONARCH 007 At: 14, Oct 2003 7:37:18 PM IST
oohooo alaa antaavaa???
aithe ikkada koodaa oka chikku undi...
what happens if "n" is a multiple of "6"
then 21n+4 will be even and hence will not be prime
and 14n+3 will be a multiple of "3"and hence not prime....
then????
Posted by: Raj Sekhar At: 14, Oct 2003 6:43:16 PM IST
because "one of the numbers" will always become a prime for any 'n'
I shall stress on the "one of the numbers" part....
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consider for n=3.
then the numbers are
21n+4 = 63+4 = 67 (not even?)
14n+3 = 42+3 = 45.
I guess 67 is a prime.
Posted by: MONARCH 007 At: 14, Oct 2003 5:57:47 PM IST
no not at all...
21n + 4 will be even for odd "n" and so cant be prime
and 14n+3 =45 for n = 3.....not a prime
and more over if "n" is a multiple of "3" then......14n+3 is a multiple of "3".....
Posted by: Raj Sekhar At: 14, Oct 2003 5:46:05 PM IST
because one of the numbers will always become a prime for any 'n'
Posted by: MONARCH 007 At: 14, Oct 2003 5:38:09 PM IST
yes...but how??
Posted by: Raj Sekhar At: 14, Oct 2003 5:20:32 PM IST
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