@Estallar12
Yaar,Quant ka koi phobia ni hai. Bas best material se practice karna hai.
Maths mein aaj tak koi phobia ni ra. Maths to hamesha se confidence waali chiz rahi hai!
@abhinavrules said:@TONYMBAYaar,maths mein to 1st class se interest hai. Yahi to ek subject jismein shuro se top kiya hai.But every1 says,rather than doing 2-3 books partially,better to pick 1 book and do it properly.Coaching to sahi hai[Byju Sir is great],but material ka panga hai.THinking of going with Arun Sharma only.
dont worry about books then.....koi bhi book pakdo chalega....enjoy doing maths during leisure time....jab mocks doge toh apne aap strenth aur weaknesses pata chal jaega...and quants consists of 21 or 20 questions out of 60...baki sections pe bhi dhyan de do
@abhinavrules said:@TONYMBAYaar,maths mein to 1st class se interest hai. Yahi to ek subject jismein shuro se top kiya hai.But every1 says,rather than doing 2-3 books partially,better to pick 1 book and do it properly.Coaching to sahi hai[Byju Sir is great],but material ka panga hai.THinking of going with Arun Sharma only.
Move over to ShoutBox or CAT'13 Aspirants Thread Bhai.
Lets keep Quant Questions ONLY flowing here. :splat:
@Estallar12 said:--------------------------------------------Which of the following is divisible by 1897 for all n > 11) 2903^n – 803^n – 464^n + 261^n 2) 1805^n – 813^n – 179^n + 272^n 3) 2901^n – 1617^n – 794^n + 834^n 4) None of the above
approach bata...
How many int value of x and y are there such that 4X+7Y = 3
mod X
@TONYMBA said:provide the approach also...
See, exploit the question...n>1 is given, though if it true for all n expect 1 then it should be true for n=1 as well (at which the expression reduces to 1897)...this condition is given just to deter us...but this can be done with the following (longer) approach as well:
1897 = 7*271
2903 = 2710 + 193 =271*10 + 193 = 193%271
803 = 542+261 = 271*2+261 = 261%271
464 = 271+193 = 193%271
261 = 261%271
=> (2903^n €“ 803^n €“ 464^n + 261^n)%271 = (193^n - 261^n - 193^n + 261^n)%271 = 0%271 = 0
hence divisible...
@Logrhythm said:Is it option 1??
@TONYMBA said:provide the approach also...
@TONYMBA said:nai aata ...bata de with approach only
2903=271*10 +193=193(mod271)
803=271*2 + 261 = 261mod271
464= 271*1 + 193 = 193 mod 271
261 = 261 mod271
thus,
2903^n - 803^n - 464^n + 261^n = mod271(193^n - 261^n - 193^n +261^n) = 0* mod271
which means that the above equation is divisible by 271
similarly it can be proved to be divisible by 7
also 271*7=1897
thus the above equation is divisible by 1897.
803=271*2 + 261 = 261mod271
464= 271*1 + 193 = 193 mod 271
261 = 261 mod271
thus,
2903^n - 803^n - 464^n + 261^n = mod271(193^n - 261^n - 193^n +261^n) = 0* mod271
which means that the above equation is divisible by 271
similarly it can be proved to be divisible by 7
also 271*7=1897
thus the above equation is divisible by 1897.
@Estallar12
Shld be Option (1)..I simply substituted n=1..Though the soln will involve rearranging of the powers..n can be proven that 1897=7*271 is divisible.. :)
@Estallar12 said:Move over to ShoutBox or CAT'13 Aspirants Thread Bhai. Lets keep Quant Questions ONLY flowing here.
TF ka material hai kya . Their Rank Booster questions are good. For quant we should go through previous year questions and practice accordingly. What do you say ?
@Estallar12 ...I actually want to know ..about testfunda cat courseware....is it really good for CAT preparation...??
@vbhvgupta said:How many int value of x and y are there such that 4X+7Y = 3mod X
Values of x and y are forming an AP with CD = +/- 7 and +/- 4 respectively.
First Value : x = -1 and y = 1.
x here will be the Limiting Variable as the CD is more.
Max value x can take = 496
Min value x can take = - 498
Terms = (498+496)/7 + 1 = 143 Terms.
@shattereddream said:TF ka material hai kya . Their Rank Booster questions are good. For quant we should go through previous year questions and practice accordingly. What do you say ?
@smsk said:@Estallar12 ...I actually want to know ..about testfunda cat courseware....is it really good for CAT preparation...??
Lets move over to CAT'13 Aspirants Thread or may be Shoutbox keeping this Thread solely for Quant Preparation. :)
@vbhvgupta said:How many int value of x and y are there such that 4X+7Y = 3mod X
general form of solution
X=7k-1
Y=1-4k
satisfies for k=-71........71
total
143 values
Let f be a polynomial of degree 98, such that f(k)=1/k. [k=1,2,3,4...99]. Find the value of f(100)
X=7k-1
Y=1-4k
satisfies for k=-71........71
total
143 values
Let f be a polynomial of degree 98, such that f(k)=1/k. [k=1,2,3,4...99]. Find the value of f(100)
@vbhvgupta
Case(i): When y=+ve Integer only..Thus, Corresponding values of x n y are(-1,1), (-8,5)...
We see that the value of x has a CD of -7 n a=-1[We dn't look for y, as CD =4..]
Therefore, -1 + (n-1)*(-7)
Case(ii): When y =-ve Integer only..Thus, Corresponding values of x n y are (6,-3), (13,-7)...
We see that the value of x has a CD of 7 n a=6[We dn't luk for y, as CD = -4]
Therefore, 6 + (n-1)*7
Thus, a total of 143 Solns..
Given, 4x + 7y = 3...(i), or, x = (3-7y)/4
Case(i): When y=+ve Integer only..Thus, Corresponding values of x n y are(-1,1), (-8,5)...
We see that the value of x has a CD of -7 n a=-1[We dn't look for y, as CD =4..]
Therefore, -1 + (n-1)*(-7)
Case(ii): When y =-ve Integer only..Thus, Corresponding values of x n y are (6,-3), (13,-7)...
We see that the value of x has a CD of 7 n a=6[We dn't luk for y, as CD = -4]
Therefore, 6 + (n-1)*7
Thus, a total of 143 Solns..
@bs0409
f(k) = 1/k
=> k*f(k) - 1 = 0
Let h(k) = k*f(k) - 1
f(k) is a polynomial of degree 98 => h(k) = k*(f(k) - 1 => h(k) is a polynomial of degree 99
Given that k = 1,2,3,4,.....99
=> h(k) = a*(k-1)*(k-2)*......*(k-99)
k*f(k) - 1 = a*(k-1)*......*(k-99)
k = 0
=> -1 = a*(-1)*(-2)*......*(-99)
=> a = 1/99!
=> k*f(k) - 1 = (1/99!) * (k-1)*(k-2)*.......*(k-99)
Put k = 100
100*f(100) - 1 = (1/99!)*(99*98*......*1
=> 100*f(100) - 1 = 99!/99!
=> f(100) = 2/100 = 1/50
=> k*f(k) - 1 = 0
Let h(k) = k*f(k) - 1
f(k) is a polynomial of degree 98 => h(k) = k*(f(k) - 1 => h(k) is a polynomial of degree 99
Given that k = 1,2,3,4,.....99
=> h(k) = a*(k-1)*(k-2)*......*(k-99)
k*f(k) - 1 = a*(k-1)*......*(k-99)
k = 0
=> -1 = a*(-1)*(-2)*......*(-99)
=> a = 1/99!
=> k*f(k) - 1 = (1/99!) * (k-1)*(k-2)*.......*(k-99)
Put k = 100
100*f(100) - 1 = (1/99!)*(99*98*......*1
=> 100*f(100) - 1 = 99!/99!
=> f(100) = 2/100 = 1/50
Courtesy:@@varuntyagi sir... :)
PS: Sir ne sikhaya tha..
@bs0409 said:
Let f be a polynomial of degree 98, such that f(k)=1/k. [k=1,2,3,4...99]. Find the value of f(100)
f(k) = 1/k
=> k*f(k) = 1
=> k*f(k) - 1 = 0 = h(k) [Suppose]
f(k) = 98 Degree => h(k) = 99 degree polynomial.
So, h(k) can be written as = A*(k-1)(k-2) ... (k-99)
=> k*f(k) - 1 = A*(k-1)(k-2) ... (k-99)
For k = 0, A = 1/99!
=> k*f(k) - 1 = [(k-1)(k-2) ... (k-99)] / 99!
For k = 100,
=> 100*f(100) - 1 = 1
Thus, f(100) = 1/50.
@pyashraj said:@bs0409f(k) = 1/k=> k*f(k) - 1 = 0Let h(k) = k*f(k) - 1f(k) is a polynomial of degree 98 => h(k) = k*(f(k) - 1 => h(k) is a polynomial of degree 99Given that k = 1,2,3,4,.....99=> h(k) = a*(k-1)*(k-2)*......*(k-99)k*f(k) - 1 = a*(k-1)*......*(k-99)k = 0=> -1 = a*(-1)*(-2)*......*(-99)=> a = 1/99!=> k*f(k) - 1 = (1/99!) * (k-1)*(k-2)*.......*(k-99)Put k = 100100*f(100) - 1 = (1/99!)*(99*98*......*1=> 100*f(100) - 1 = 99!/99!=> f(100) = 2/100 = 1/50Courtesy:@@varuntyagi sir... PS: Sir ne sikhaya tha..
Bhai, pls tell ,For any other polynomial what is the approach ?
is the approach to make the f(x) into polynomial which is equal to zero ??