@vbhvgupta
21578
@bs0409 said:In how many ways can 12 balls be divided among 3 people such each can get a max of 6 balls . Let me know how to take care of this max condition ?
Let a,b and c be the no. of balls wid the 3 persons.
thus, a+b+c=12;
Also max value of a,b and c can be 6.
therefore we write a=7-x; b=7-y and c=7-z (where x,y and z are +ve values)
hence, (7-x) + (7-y) + (7-z) = 12;
or x+y+z=9
which gives 8C2 as my answer??
(3^1024-1)/2^n
then find the highest value of n which will divide this..
then find the highest value of n which will divide this..
@TootaHuaDil said:(3^1024-1)/2^nthen find the highest value of n which will divide this..
11..??
[(2+1)^1024 - 1 ] / 2^n
2^1024 + ... + 1 - 1 / 2^n
2^1024 + ... + 1024C1. 2 / 2^n
1024.2 = 2^11. Hence, n = 11.
Pl correct me if I'm wrong.! :)
10000! = (100!)^K β P, where P and K are integers. What can be the maximum value of K?
a) 103
b) 104
c) 102
d) 105
a) 103
b) 104
c) 102
d) 105
@ravi.theja said:10000! = (100!)K β P, where P and K are integers. What can be the maximum value of K?a) 103b) 104c) 102d) 105
Is k getting multiplied to 100! or 100! is raised to power k?
@Tiws said:@TootaHuaDiln= 12
Ans. is 11..You can do this way easily ...found it on PG only
Use Euler for this :-
E( 2^x ) =2^x ( 1-1/2) = 2^(x-1)
Since 3 & 2 are coprime ..using euler we have :
3^ ( 2^(x-1)) mod 2^x =1
in this case 2^(x-1) = 1024 = 2^10
so x = 11
That is max value will be 11 .. & that divisor will be 2^11
Use Euler for this :-
E( 2^x ) =2^x ( 1-1/2) = 2^(x-1)
Since 3 & 2 are coprime ..using euler we have :
3^ ( 2^(x-1)) mod 2^x =1
in this case 2^(x-1) = 1024 = 2^10
so x = 11
That is max value will be 11 .. & that divisor will be 2^11
@Subhashdec2 said:Is k getting multiplied to 100! or 100! is raised to power k?
power k......edited now
The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
@ravi.theja said:The HCF of three natural nos x,y,z is 13. if the sum of x,y,z is 117 then how many ordered pairs of (x,y,z) exist?
ha+hb+hc=117
a+b+c=9
solutions->(5,3,1)-6 cases
(7,1,1)-3 cases
9 cases?
@Subhashdec2 said:ha+hb+hc=117a+b+c=9solutions->(5,3,1)-6 cases(7,1,1)-3 cases9 cases?
dnt have OA bhai :)
@ravi.theja said:10000! = (100!)^K β P, where P and K are integers. What can be the maximum value of K?a) 103b) 104c) 102d) 105
check for power of 97
100! has 1 power of 97
10000! has 103+1=104
hence 104
@TootaHuaDil
(3^1024-1)
= (3^512 -1)(3^512+1)
=(3^256-1) (3^256+1) (3^512+1)
=(3^128-1) (3^128+1) (3^256+1) (3^512+1)
=(3^64-1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^32-1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^16-1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^8-1) (3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=[(3^4-1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
=[(3^2-1) (3^2+1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
=[(3-1) (3+1)(3^2+1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
This can be written as β¬ΔΉΕ‘.
(2)(4)(2)(2)(2)(2)(2)(2)(2)(2)(2)= 2^12
(3^1024-1)
= (3^512 -1)(3^512+1)
=(3^256-1) (3^256+1) (3^512+1)
=(3^128-1) (3^128+1) (3^256+1) (3^512+1)
=(3^64-1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^32-1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^16-1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=(3^8-1) (3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)
=[(3^4-1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
=[(3^2-1) (3^2+1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
=[(3-1) (3+1)(3^2+1) (3^4+1)(3^8+1) (3^16+1) (3^32+1) (3^64+1) (3^128+1) (3^256+1) (3^512+1)]
This can be written as β¬ΔΉΕ‘.
(2)(4)(2)(2)(2)(2)(2)(2)(2)(2)(2)= 2^12
remainder when 128^1000 is divided by 153 ??
@Subhashdec2 said:ha+hb+hc=117a+b+c=9solutions->(5,3,1)-6 cases(7,1,1)-3 cases9 cases?
bhai it has to be 27 π 8c2-1 case...the case of (3,3,3) will make 39 as HCF..so its 27