Three logicians — A, B and C are wearing hats, which they know are either black or white but not
all white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is asked
in turn if they know the colour of their own hats. The answers are 'A : No, B : No, C : Yes'. What is
the colour of C's hat?
a. Black b. White
c. Black if A is wearing a white hat d. Cannot be determined
Find the number of triples of integers a
@joyjitpal said:Three logicians — A, B and C are wearing hats, which they know are either black or white but notall white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is askedin turn if they know the colour of their own hats. The answers are 'A : No, B : No, C : Yes'. What isthe colour of C's hat?a. Black b. Whitec. Black if A is wearing a white hat d. Cannot be determined
Black...
See, A must have seen a black hat...otherwise if he saw both B and C wearing white hats he would have been sure that he is wearing a black one... (but he couldn't)
Now B must also have seen the same, and the black hat B must have seen wld have been on C otherwise B would have known that she is wearing a black hat as A wld have seen at least 1 black hat...Hence, C can be sure that he is wearing a black hat..
@joyjitpal said:Three logicians — A, B and C are wearing hats, which they know are either black or white but notall white. A can see the hats of B and C; B can see the hats of A and C; C is blind. Each is askedin turn if they know the colour of their own hats. The answers are 'A : No, B : No, C : Yes'. What isthe colour of C's hat?a. Black b. Whitec. Black if A is wearing a white hat d. Cannot be determined
when A see B and C must be having a Black Hat and white Hat(that is why he cant say color of his hat)
When B see A and C Hats (B is unable to say because he might have seen Black and white hats)
There fore C who is blind(can hear the answers of A and B) thinks all this and says his hat is Black :)
When B see A and C Hats (B is unable to say because he might have seen Black and white hats)
There fore C who is blind(can hear the answers of A and B) thinks all this and says his hat is Black :)
Find the largest positive integer n such that for all real numbers a1, a2, …,a(n+1), the equation a(n+1)x^2 − 2x*root[a1^2 + a2^2 +⋯+ a(n+1)^2] + (a1+a2+⋯+an) = 0 has real roots.
The expression 1/2^2 + 2/2^3 + 3/2^4 + €Ś+ 99/2^100 can be expressed as 1 ˆ' (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
@jain4444 said:Find the number of triples of integers a
The expression is cyclic i.e if we replace a with b ..b with c and c with a we get the same expression, hence all the factors must be symmetric in nature.
if we put a=b then the value of expression = 0, =>(a-b) is a factor
similarly (b-c) and (c-a) should be the factor of the above expression.
Since in the above expression all the terms are of degree 4 hence there must be a fourth factor of degree 1.
a(b−c)^3 + b(c−a)^3 + c(a−b)^3 = k(a-b)(b-c)(c-a)(a+b+c)
now we need to find the value of k by replacing a=0,b=1,c=2
0(1-2)^3 + 1*(2-0)^3 + 2*(0-1)^3 = k(0-1)(1-2)(2-0)(0+1+2)
=>0+8-2 = k(2*3)
=>k=1
hence the above expression can be written as (a-b)(b-c)(c-a)(a+b+c) = 0
since a (a-b) or (b-c) or (c-a) cannot be 0,
hence (a+b+c) = 0 => all a,b,c cannot be +ve
for a = -10 we get b+c = 10 =>b can take values from -9 onwards upto 4 hence 14 solutions
for a = -9 we get b+c = 9 =>b can take values from -8 upto 4 hence 13 solutions
for a = -8 we get b+c = 8 =>b can take values from -7 upto 3 hence 11 solutions
for a = -7 we get b+c = 7 =>b can take values from -6 upto 3 hence 10 solutions
for a = -6 we get b+c = 6 =>b can take values from -5 upto 2 hence 8 solutions
for a = -5 we get b+c = 5 =>b can take values from -4 upto 2 hence 7 solutions
for a = -4 we get b+c = 4 =>b can take values from -3 upto 1 hence 5 solutions
for a = -3 we get b+c = 3 =>b can take values from -2 upto 1 hence 4 solutions
for a = -2 we get b+c = 2 =>b can take values from -1 upto 0 hence 2 solutions
for a = -1 we get b+c = 1 =>b can take values 0 hence 1 solutions
hence total of 14+13+11+10+8+7+5+4+2+1 = 75.
for a = -5 we get b+c = 5 =>b can take values from -4 upto 2 hence 7 solutions
for a = -4 we get b+c = 4 =>b can take values from -3 upto 1 hence 5 solutions
for a = -3 we get b+c = 3 =>b can take values from -2 upto 1 hence 4 solutions
for a = -2 we get b+c = 2 =>b can take values from -1 upto 0 hence 2 solutions
for a = -1 we get b+c = 1 =>b can take values 0 hence 1 solutions
hence total of 14+13+11+10+8+7+5+4+2+1 = 75.
ATDH.
@jain4444 said:The expression 1/2^2 + 2/2^3 + 3/2^4 + €Ś+ 99/2^100 can be expressed as 1 ˆ' (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
1/2^2 + 2/2^3 = 4/2^3 = 2^2/2^3
1/2^2+2/2^3+3/2^4 = 2^2+4+3/2^4 = 11/2^4
11/2^4 + 4/2^5 = 22+4/2^5 = 26/2^5
26/2^5 + 5/2^6 = 57/2^5
57/2^5 + 6/2^7 = 120/2^7
4,11,26,57,120
7,15,31,63
8,16,32....now this wld go as twice of previous term and we can backtrack the value of a..
1/2^2+2/2^3+3/2^4 = 2^2+4+3/2^4 = 11/2^4
11/2^4 + 4/2^5 = 22+4/2^5 = 26/2^5
26/2^5 + 5/2^6 = 57/2^5
57/2^5 + 6/2^7 = 120/2^7
4,11,26,57,120
7,15,31,63
8,16,32....now this wld go as twice of previous term and we can backtrack the value of a..
and b would be 100...
fir a+b find karlo
@jain4444 said:The expression 1/2^2 + 2/2^3 + 3/2^4 + €Ś+ 99/2^100 can be expressed as 1 ˆ' (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
ans 101
@jain4444 said:The expression 1/2^2 + 2/2^3 + 3/2^4 +…+ 99/2^100 can be expressed as 1− (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
Let S=1/2^2 + 2/2^3 + 3/2^4 +…+ 99/2^100
S/2= 1/2^3+2/2^4+........................+99/2^101
S/2=1/2^2+1/2^3+1/2^4+..................+1/2^100 -99/2^101
S/2=(2/4)(1-(1/2)^99)-99/2^101
S=1- 1/2^99 - 99/2^100
=1 - (2+99) /2^100= 1 - 101/2^100
a+b=201
what is the remainder when 50^56^52 is divided by 11?
a-7
b-5
c-9
d-10
@jain4444 said:The expression 1/2^2 + 2/2^3 + 3/2^4 + €Ś+ 99/2^100 can be expressed as 1 ˆ' (a/2^b), where a is an odd positive integer, and b is a non-negative integer. What is the value of a+b ?
1/2^1(1-1/2^98) + 1/2^2(1-1/2^97) + 1/2^3(1-1/2^96) + .... + 1/2^99(1-1/2^0)
= 1/2^1+1/2^2+...+1/2^99 - (99/2^99)
= (1-1/2^99) - 99/2^99
= 1 - 100/2^99
= 1/2^1+1/2^2+...+1/2^99 - (99/2^99)
= (1-1/2^99) - 99/2^99
= 1 - 100/2^99
a+b=199
@bs0409
Shld be 5..
E(11) = 10..
Thus, Rem[50^10/11] = 1..Nw, Rem [56^52/10] = 6..Thus, 56^52 = 10k+6
Now, Rem[50^(10k + 6)/11] = Rem[50^6/11] = Rem[6^6/11] = Rem[3^3/11] = 5..
@amresh_maverick said:OA :1> a >(11/9)2> 2 solns
bro can you share your approach for the first question i.e ("Values of a forwhich roots of eq : x^2 -6ax +2 -2a +9a^2 =0 are greater than 3")?
Q36 and 37
Q36 and 37
@bs0409 said:what is the remainder when 50^56^52 is divided by 11?a-7b-5c-9d-10
[EDIT] 5?? 
Let, 56^52 = a
Mod(50^a/11) = Mod(6^a/11) ------(1)
Now,
Mod(6^2/11) = 3
Mod(6^4/11) = 9
till..
Mod(6^10/11) = 1, Hence the Period cycle of 10. ----(2)
from(2)
Now, for a,
Mod(56^52/10) = Mod(6^52/10) = 6 = a ----(3)
From (2) ans (3)-
Mod(6^6/11) = 5 = OA
Let, 56^52 = a
Mod(50^a/11) = Mod(6^a/11) ------(1)
Now,
Mod(6^2/11) = 3
Mod(6^4/11) = 9
till..
Mod(6^10/11) = 1, Hence the Period cycle of 10. ----(2)
from(2)
Now, for a,
Mod(56^52/10) = Mod(6^52/10) = 6 = a ----(3)
From (2) ans (3)-
Mod(6^6/11) = 5 = OA
