Number System

[vaibhav2cool]

[quote=naresh007;1045904]take eg: as 2^6 × 3^5 × 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
where AXB are number formed by multipliying each element of A for each element of B, and taking into account
that aixbi = bixai
for instance n(AXB)=30
apply the same funda here...
so finally total no: of divisiors for:
2^66*3^33 + 2^23*17^23 + 2^47
| |
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

i hope u got now...?

but there is another formula for calculating the number of divisors...
i.e if N= a^p x b^q....
then the total number of divisors are...(p+1)(q+1)

for ex-N= 2^6 X 3^5..
THEN no of divisors is ....(6+1)(5+1)=42

but going by ur method...the no of divisors cm out to 6+5+6x5=41...
also here...
2^66*3^33 + 2^23*17^23 + 2^47
| |
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

can u explain the funda of getting the total number of divisors by adding total number of divisors for each term...

[naresh007]

@vaibhav,
i proved the result with 2 methods...plz chek below
eg: as 2^6 × 3^5 × 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U!

[quote=vaibhav2cool;1046060]

Originally Posted by naresh007

take eg: as 2^6 × 3^5 × 5^2
A={2,4,8..} n(A)=6
B={3,9,27,.} n(B)=5
C={5,25} n(C)=2
the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
where AXB are number formed by multipliying each element of A for each element of B, and taking into account
that aixbi = bixai
for instance n(AXB)=30
apply the same funda here...
so finally total no: of divisiors for:
2^66*3^33 + 2^23*17^23 + 2^47
| |
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

i hope u got now...?

but there is another formula for calculating the number of divisors...
i.e if N= a^p x b^q....
then the total number of divisors are...(p+1)(q+1)

for ex-N= 2^6 X 3^5..
THEN no of divisors is ....(6+1)(5+1)=42

but going by ur method...the no of divisors cm out to 6+5+6x5=41...
also here...
2^66*3^33 + 2^23*17^23 + 2^47
| |
= {66+33+ 66*33} + {23+23+23*23} + 47
= 2178+99 + 46+ 529+47 = 2899

can u explain the funda of getting the total number of divisors by adding total number of divisors for each term...

[sandeep jha]

[quote=naresh007;1046426]@vaibhav,

the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U!

hey plz check the bold part ...u r writing no. of divisors and finding sum of divisors and also in the second bold part the total no of divisors formula seems incorrect to me....(why it shld be (p+1)..(r+1)-1)
plz explain how u got the formula!!

[vaibhav2cool]

[quote=sandeep jha;1047139]

Originally Posted by naresh007

@vaibhav,

the total number of divisors are
n(A)+n(B)+n(C)+ n(AXB)+ n(AxC)+ n(BXC)+ n(AxBxC)
TOTAL SUM OF DIVISORS...= 6 + 5+ 2 + 30+12+10+60 = 125

NOW USING ANOTHER METHOD:
I.E. =Total no: of divisors in an expansion a^p * b^q * c^r is (p+1)(q+1)(r+1) - 1.
so, we can apply to this..
i.e. = (6+1)*(5+1)*(2+1) - 1 = 126-1 = 125
matching..right!

we can check for our problem too..i.e. 2^66*3^33 + 2^23*17^23 + 2^47
{67*34 - 1 }+ {24*24 - 1} + {48 - 1}
= 2277+575+47 = 2899

Hope now its clear for U!

hey plz check the bold part ...u r writing no. of divisors and finding sum of divisors and also in the second bold part the total no of divisors formula seems incorrect to me....(why it shld be (p+1)..(r+1)-1)
plz explain how u got the formula!!

exactly.....this is the confusion causing part....
also...
let a= 2^66*3^33 , b= 2^23*17^23, c= 2^47


can u plz explain y the sum of divisors of a, b and c will give u the total number of divisors for the entire term....

[naresh007]

yea, accepted that is not the formulae for the Sum of divisors but its for the unique divisors ...OMG! what a kind of mess.....
Thanks Vaibhav & Sandeep....for makin me aware of my mistake..

Finally, plz find the formulae i'm attached below for the sum of the divisors:
i hope you might find clear explanation now:
First check one Eg:
σ0(12) is the number of the divisors of 12:
σ0(12) = 10 + 20 + 30 + 40 + 60 + 120
= 1 + 1 + 1 + 1 + 1 + 1 = 6.
while σ1(12) is the sum of all the divisors:
σ1(12) = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 + 12^1
= 1 + 2 + 3 + 4 + 6 + 12 = 28.
sum of its divisors equals :
2^2 * 3 = {(2^3 - 1)/2-1 } * { (3^2 -1)/3-1} = 7*4 = 28

now similary, following the above principle for a= 2^66*3^33 , b= 2^23*17^23, c= 2^47
to find the total sum of divisors of a + b + c,
1st we calculate individually a, b, c and add them..
for a: 2^66*3^33
= {(2^67 - 1)/2-1 } * { (3^34 -1)/3-1} = {2^67 - 1}*{3^34-1} /2
for b: 2^23*17^23,
= {(2^24 - 1)/2-1 } * { (17^24 -1)/17-1} = {(2^24 - 1) } * { (17^24 -1)/16}
for c: 2^47
= {(2^48 - 1)/2-1 } = {(2^48 - 1)}
finally,
total sum of its divisors of a + b + c =
i.e. {2^67 - 1}*{3^34-1} /2 + {2^24 - 1} * { (17^24 -1)/16} + {2^48 - 1}
=> {{2^70 - 8}*{3^34 - 1} + {2^24 - 1} * { (17^24 -1)} + {2^52 - 16} } /16

well, i hope the Method is correct now...

[quote=vaibhav2cool;1048054]

Originally Posted by sandeep jha

exactly.....this is the confusion causing part....
also...
let a= 2^66*3^33 , b= 2^23*17^23, c= 2^47


can u plz explain y the sum of divisors of a, b and c will give u the total number of divisors for the entire term....

[vaibhav2cool]

[quote=naresh007;1048902]yea, accepted that is not the formulae for the Sum of divisors but its for the unique divisors ...OMG! what a kind of mess.....
Thanks Vaibhav & Sandeep....for makin me aware of my mistake..

Finally, plz find the formulae i'm attached below for the sum of the divisors:
i hope you might find clear explanation now:
First check one Eg:
σ0(12) is the number of the divisors of 12:
σ0(12) = 10 + 20 + 30 + 40 + 60 + 120
= 1 + 1 + 1 + 1 + 1 + 1 = 6.
while σ1(12) is the sum of all the divisors:
σ1(12) = 1^1 + 2^1 + 3^1 + 4^1 + 6^1 + 12^1
= 1 + 2 + 3 + 4 + 6 + 12 = 28.
sum of its divisors equals :
2^2 * 3 = {(2^3 - 1)/2-1 } * { (3^2 -1)/3-1} = 7*4 = 28

now similary, following the above principle for a= 2^66*3^33 , b= 2^23*17^23, c= 2^47
to find the total sum of divisors of a + b + c,
1st we calculate individually a, b, c and add them..
for a: 2^66*3^33
= {(2^67 - 1)/2-1 } * { (3^34 -1)/3-1} = {2^67 - 1}*{3^34-1} /2
for b: 2^23*17^23,
= {(2^24 - 1)/2-1 } * { (17^24 -1)/17-1} = {(2^24 - 1) } * { (17^24 -1)/16}
for c: 2^47
= {(2^48 - 1)/2-1 } = {(2^48 - 1)}
finally,
total sum of its divisors of a + b + c =
i.e. {2^67 - 1}*{3^34-1} /2 + {2^24 - 1} * { (17^24 -1)/16} + {2^48 - 1}
=> {{2^70 - 8}*{3^34 - 1} + {2^24 - 1} * { (17^24 -1)} + {2^52 - 16} } /16

well, i hope the Method is correct now...

sir ji everything is clear now except onething....that is to find the total sum of of divisors of expression a+b+c..y do v add the individual sum of each of a, b and c...

[trevis]

if (bx-ay)/b =(cy-bz)/c = (az-cx)/a and bx-ay NE 0 and cy-bz NE 0 and az-cx NE 0 ..then whcih of the following is true .??
a. a+b+c = 0
b. a+b+-c = 0
c. ab+bc+ca = 0
d. ab+bc-ca= 0
ps: NE is Not Equal to

[naresh007]

bx-ay = bk ---- 1
cy-bz = ck ---- 2
az-cx = ak ---- 3

from eq.3 we get x,
as, a(z-k)/c = x

from eq.1,

b( a(z-k)/c) - a( bz+ck)/c = bk
ab(z-k)- a(bz+ck) = bck
abz-abk-abz-ack = bck
-abk-ack = bck
cancel k both sides...
bc = -(ac+ab)
i.e. ab+bc+ca= 0
so ans: C

Originally Posted by trevis

if (bx-ay)/b =(cy-bz)/c = (az-cx)/a and bx-ay NE 0 and cy-bz NE 0 and az-cx NE 0 ..then whcih of the following is true .??
a. a+b+c = 0
b. a+b+-c = 0
c. ab+bc+ca = 0
d. ab+bc-ca= 0
ps: NE is Not Equal to

[trevis]

thanks a lot bro !!

[avinav2712]

Originally Posted by trevis

thanks a lot bro !!

Just the "Thanks" button would have done.... No need for a separate post buddy.....