what is the remainder when
1. 128^1000 is divided by 153
2.50^56^52 is divided by 11
33^34^35 is divided by 7
is the ans for 50^56^52 by 11= 3
is the ans for 50^56^52 by 11= 3
2a + 5b + 10c = 2010
a should be a multiple of 5, say, a = 5k, then
2k + b + 2c = 402
b should be even
=> b = 2n
=> k + n + c = 201
=> C(201 + 3 - 1, 3 - 1) = C(203, 2) = 20503 ways
Possible rectangles = (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12)
Sum of perimeters = 2*Sum of all factors = 2*(1 + 2 + 4 + 8 )(1 + 3)(1 + 5) = 2*15*4*6 = 720
128^1000(mod 153)
153 = 9*17
9 = 2^3 + 1
=> 128^1000 = 2^7000 = 2*2^6999 = 7(mod 9)
17 = 2^4 + 1
=> 128^1000 = 2^7000 = 1 (mod 17)
=> N = 9k + 7 = 17n + 1
=> k = 2n - (n + 6)/9, k will be integer for n = 3
=> N = 153k + 52
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11)
33^34^35 is divided by 7
33^34^35 = 2^34^35 (mod 7)
We know that 2^3 = 1 (mod 7)
34^35 = 1 (mod 3)
=> 2^34^35 = 2^(3k + 1) (mod 7) = 2 (mod 7)
chillfactor plz where to learn fermat theorem from...cannot find it in any quant books ..sry for the spam
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11
I dont understand how 56^ 52=6 mod10.......pls explain??
i dont know eulers theorem ,fermat theorem...can u pls explain it details and how i apply this theorem to find out remainder
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11
I dont understand how 56^ 52=6 mod10.......pls explain??
i dont know eulers theorem ,fermat theorem...can u pls explain it details and how i apply this theorem to find out remainder
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prats92 Sayschillfactor plz where to learn fermat theorem from...cannot find it in any quant books ..sry for the spam
jaldi download maar lena..
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Loving the enthusiasm in this forum. Here's another little chestnut.
A box contains 5 red and 5 blue balls . A ball is picked from the box and is replaced by a ball of the other colour. For instance , if a blue ball is picked then it is replaced by a red ball and vice- versa. The process is repeated ten times and then a ball is picked from the box. What is the probability that this ball is red?
a.)1- {1/2 + (1/2)^2 +(1/2)^3 + ......+(1/2)^10}
b.)1-{1/2-(1/2)^2+(1/2)^3-(1/4)^4+.........-(1/2)^10}
c.)1/2-(1/2)^3+(1/2)^3-(1/2)^4+.........-(1/2)^10
d.)None of these
:shocked:
sorry deleting this as posted it wrongly
1)Computer takes an input of a number N and X where X is a factor of the number N.N is 83p796161q and X is equal to 11 where o
Q.Computer takes an input of a number N and X where X is a factor of the number N.N is 83p796161q and X is equal to 11 where o
As X is a factor of N, therefore N is divisible by X.
And as X = 11, the given number N is divisible by 11.
Now use the rule :
when a number is divisible by 11 then the difference between the Sum of digits at odd places and the sum of its digits at even places must be 0 or divisible by 11.
so sum of digits at odd places = 8+p+9+1+1 = 19 + p.
so sum of digits at even places = 3+7+6+6+q = 22 +q.
so either ( 22 +q ) - (19 + p) = 0 or {( 22 +q ) - (19 + p) } divisible by 11.
simplify the two eqns we get
1) 3 + q -p = 0 or p = q + 3.
2)3 + q - p = 11k
now as it is given that otherefor the eqn (1) cannot be true as it makes p > q.
so taking the 2nd eqn .
k = 0 is same as eqn 1.
Case 1.
for k = 1.
3 + q - p = 11
or q = p + 8.
so possible values of p = 0 and 1.
if p >1 then q becomes > 9 i.e 10 , 11, ... so on.
which is not possible .
but P > 0 as it is given that o
17961619
R1 = R {N /(p + q)} = 9
R2 = R{N/p} = 0
so R1 + R2 = 9.
Case 2.
for k = 2.
3 + q - p = 22
or q = p + 19.
this is not possible as q is a digit so it can vary from 0 to 9.
So answer is 9.
@ abhilas
2) x and y are two positive integes,what will be the sum of the coefficients of the expansion
(x+y)^44?
Ans = 2^ 44.
it is the basic property of binomial theorem .
if expression is (x + y )^ n ,
then sum of the coefficients of the expansion is always 2^n.
you can check the detailed solution in any higher mathematics book for class XI .
2) x and y are two positive integes,what will be the sum of the coefficients of the expansion
(x+y)^44?
Ans = 2^ 44.
it is the basic property of binomial theorem .
if expression is (x + y )^ n ,
then sum of the coefficients of the expansion is always 2^n.
you can check the detailed solution in any higher mathematics book for class XI .
2a + 5b + 10c = 2010
a should be a multiple of 5, say, a = 5k, then
2k + b + 2c = 402
b should be even
=> b = 2n
=> k + n + c = 201
=> C(201 + 3 - 1, 3 - 1) = C(203, 2) = 20503 ways
Possible rectangles = (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12)
Sum of perimeters = 2*Sum of all factors = 2*(1 + 2 + 4 + 8 )(1 + 3)(1 + 5) = 2*15*4*6 = 720
128^1000(mod 153)
153 = 9*17
9 = 2^3 + 1
=> 128^1000 = 2^7000 = 2*2^6999 = 7(mod 9)
17 = 2^4 + 1
=> 128^1000 = 2^7000 = 1 (mod 17)
=> N = 9k + 7 = 17n + 1
=> k = 2n - (n + 6)/9, k will be integer for n = 3
=> N = 153k + 52
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11)
33^34^35 is divided by 7
33^34^35 = 2^34^35 (mod 7)
We know that 2^3 = 1 (mod 7)
34^35 = 1 (mod 3)
=> 2^34^35 = 2^(3k + 1) (mod 7) = 2 (mod 7)
awesome..i just solved and matched each step...
2a + 5b + 10c = 2010
a should be a multiple of 5, say, a = 5k, then
2k + b + 2c = 402
b should be even
=> b = 2n
=> k + n + c = 201
=> C(201 + 3 - 1, 3 - 1) = C(203, 2) = 20503 ways
Possible rectangles = (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12)
Sum of perimeters = 2*Sum of all factors = 2*(1 + 2 + 4 + 8 )(1 + 3)(1 + 5) = 2*15*4*6 = 720
128^1000(mod 153)
153 = 9*17
9 = 2^3 + 1
=> 128^1000 = 2^7000 = 2*2^6999 = 7(mod 9)
17 = 2^4 + 1
=> 128^1000 = 2^7000 = 1 (mod 17)
=> N = 9k + 7 = 17n + 1
=> k = 2n - (n + 6)/9, k will be integer for n = 3
=> N = 153k + 52
50^56^52 is divided by 11
Eulers number of 11 is 10
56^52 = 6 (mod 10)
=> 50^56^52 = 50^6(mod 11) = 6^6 (mod 11) = 3^3 (mod 11) = 5 (mod 11)
33^34^35 is divided by 7
33^34^35 = 2^34^35 (mod 7)
We know that 2^3 = 1 (mod 7)
34^35 = 1 (mod 3)
=> 2^34^35 = 2^(3k + 1) (mod 7) = 2 (mod 7)
can you please explain the highlighted parts?
bhaskar_goel Sayscan you please explain the highlighted parts?
2a + 5b + 10c = 2010
as the last digit in RHS is 0.
and the last digit of the 3rd term in the LHS is also 0.
so the last digit of the expression (2a + 5b ) has to be zero then only the eqn will be valid .
therefore (2a + 5b ) will give last term as 0 , when a = 5k and b = 2n
so that the expression becomes (10k + 10n )
So the eqn becomes 10k + 10n + 10c = 2010.
no dividing both sides by 10 ,
we get k + n + c = 201.
hope you got it !
Loving the enthusiasm in this forum. Here's another little chestnut.
A box contains 5 red and 5 blue balls . A ball is picked from the box and is replaced by a ball of the other colour. For instance , if a blue ball is picked then it is replaced by a red ball and vice- versa. The process is repeated ten times and then a ball is picked from the box. What is the probability that this ball is red?
a.)1- {1/2 + (1/2)^2 +(1/2)^3 + ......+(1/2)^10}
b.)1-{1/2-(1/2)^2+(1/2)^3-(1/4)^4+.........-(1/2)^10}
c.)1/2-(1/2)^3+(1/2)^3-(1/2)^4+.........-(1/2)^10
d.)None of these
:shocked:
Yar, u dont have to do any calculations for this.
If probability of picking Red ball at the end of the process is P, you might as well say that probability for picking the Blue ball is P. Since there isnt any distinction in the two events, and both are similar.
So P(R) = P(B)
and P(R) + P(B) = 1
So probability for each case = 1/2
Hence option D)
I Don't have a clue so please some
1.)In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.
2.) How many 3 digit no's composed of 3 distinct digits are there such that one digit is the average of other two.
a.)96
b.)104
c.)112
d.)120
e.)256
I Don't have a clue so please some
1.)In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.
2.) How many 3 digit no's composed of 3 distinct digits are there such that one digit is the average of other two.
a.)96
b.)104
c.)112
d.)120
e.)256
2.) How many 3 digit no's composed of 3 distinct digits are there such that one digit is the average of other two.
Ans : 96.
here is my approach....
Denote the three digit number as abc
then c = (a+b)/2.
therefore for c to be an integer (a+b) has to be an even number i.e 2n
now lets fix values for a.
if a = 1 ; b = 3,5,7,9 ; c = 2,3,4,5 (Note1 : b can't be 1 because then all the digits will not be distinct.)
therefore numbers formed are 132 ,153,174,195.
now the 3 digits in can be arranged in 3! ways .
therefore total count of such numbers = 3! * 4 = 24.
now a=2 ; b = 4,6,8 ; c = 3,4,5 .
numbers formed = 3.
total count = 3! * 3 = 18
a = 3 ; b = 5,7,9 , c = 4,5,6 (note2 : as the case of 1 & 3 is already discussed so we will not take 3,1 )
numbers formed = 3.
total count = 3! * 3 = 18
a = 4 ; b = 6,8 c = 5,6 (Excluding 2 & 4 : reason same as note 2 & note 1)
numbers formed = 2.
total count = 12.
a = 5 ; b = 7,9 c = 6,7 ( excluding 1,3,5 )
numbers formed = 2
total count = 12.
a = 6 ; b = 8 ; c = 7 (excluding 2,4,6)
numbers formed = 1
total count = 3! = 6 .
a = 7; b = 9 ; c= 8 (excluding 1,3,5,7)
numbers formed = 1
total count = 3! = 6 .
so total count of numbers = 24 + 18 +18+12+12+6+6 = 96.
so ans is option (a).
Please correct if I am wrong.
I Don't have a clue so please some
1.)In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.
Maximum wuld be 31%
ksagarwa SaysMaximum wuld be 31%
Please post complete solution if you are answering for any questions.....
that helps to figure out the approach..
I Don't have a clue so please some
2.) How many 3 digit no's composed of 3 distinct digits are there such that one digit is the average of other two.
a.)96
b.)104
c.)112
d.)120
e.)256
it is coming 96....here's my approach
u need to just calculate numbers from 100-200 which satisfies the given condition and the u can arrange the digits wid out repetition that will give 96 as answer. correct me if iam wrong.thanx
it is coming 96....here's my approach
u need to just calculate numbers from 100-200 which satisfies the given condition and the u can arrange the digits wid out repetition that will give 96 as answer. correct me if iam wrong.thanx
Can u plz elaborate...
1.)In an office where working in one dept is necessary 78% of the employees are in operations,69% are in finance and 87% are in HR. what are the maximum and minimum %ages of employees that could have been working in all 3 depts.
2.) How many 3 digit no's composed of 3 distinct digits are there such that one digit is the average of other two.
a.)96
b.)104
c.)112
d.)120
e.)256
I think its 112.
When two digits are odd, then we can have third one as average of them.
So, C(5, 2)*3! = 60 such numbers
When two digits are even barring 0, then we can have third one as average of them.
So, C(4, 2)*3! = 36 such numbers
When one of the digit is 0, then if second digit is x, then third will be 2x as (1, 2), (2, 4), (3, 6), (4, 8 )
So, 4*2*2 = 16 such numbers (2*2 as these numbers can be permuted in 2*2 ways)
So, total 112 such numbers