we'll convert binary to decimal to find the sum- 7 digit no -> first digit will be always 1...remaining 6 places wil be filled by 3 1's and 3 0's... so 2^6 will be repeated 6!/(3!*3!) ways sum=2^6 *(6!/3!*3!)
now 6th digit- no of times 1 will be there
1 1 _ _ _ _ _
remaining blanks will be filled by 2 1's and 3 0's -> 5!(2!*3!) ways sum = 2^5 * 5!/(2!*3!)
similarly for rest of terms....
so total sum = 2^6 (6!/3!*3!) + 5!/(2!*3!) (2^5+2^4+2^3+2^2+2^1+2^0)
= 1910
my take is 1910
hi abhishek i could not get the 1 in bold (P&C; ) can u elaborate
hi abhishek i could not get the 1 in bold (P&C; ) can u elaborate
_ _ _ _ _ _ _
we'll convert binary to decimal to find the sum-
7 digit no -> first digit will be always 1...remaining 6 places wil be filled by 3 1's and 3 0's... so 2^6 will be repeated 6!/(3!*3!) ways sum=2^6 *(6!/3!*3!)
now 6th digit- no of times 1 will be there
1 1 _ _ _ _ _
remaining blanks will be filled by 2 1's and 3 0's -> 5!(2!*3!) ways sum = 2^5 * 5!/(2!*3!)
similarly for rest of terms....
so total sum = 2^6 (6!/3!*3!) + 5!/(2!*3!) (2^5+2^4+2^3+2^2+2^1+2^0)
= 1910
my take is 1910
hey kinscool...here we have
1 _ _ _ _ _ _
as the case... now these 6 places can be filled by....2^6 ways.....and can be arranged by 6! / (3! * 3!)...because of 3 1s and 3 0s...hope u get the concept.....
In an infinite G.P every term is 7 time of the sum of all terms that follow. t1+t2=9. Find the sum..
let sum of all the terms following t2 = s2 and sum of all the terms following t1 = s1 According to the Q => t2 = 7*s2 so, s2 = t2/7 --(i) similarly, t1 = 7*s1 s1 = t2 + s2 also from (i) s1 = t2 + t2/7 = (8*t2)/7 so t1 = 8*t2 --(ii) given : t1 + t2 = 9 from (ii) 8*t2 + t2 = 9 => t2 = 1 => t1 = 8 --(iiia) and s1 = 8/7 --(iiib)
sum of series = t1+ s1 from (iiia) and (iiib) we get sum of series = 8 + 8/7 = 64/7 = Ans ..any further queries are most welcomed
let sum of all the terms following t2 = s2 and sum of all the terms following t1 = s1 According to the Q => t2 = 7*s2 so, s2 = t2/7 --(i) similarly, t1 = 7*s1 s1 = t2 + s2 also from (i) s1 = t2 + t2/7 = (8*t2)/7 so t1 = 8*t2 --(ii) given : t1 + t2 = 9 from (ii) 8*t2 + t2 = 9 => t2 = 1 => t1 = 8 --(iiia) and s1 = 8/7 --(iiib)
sum of series = t1+ s1 from (iiia) and (iiib) we get sum of series = 8 + 8/7 = 64/7 = Ans ..any further queries are most welcomed
what was wrong with my approach....i've posted few blocks earlier...just have a look.....
here we have t1+t2 = 9 t3 = 7(t1+t2) t4 = 7(t1+t2+t3) = 7*8(t1+t2) t5 = 7(t1+t2+t3+t4) = 7*8^2*(t1+t2) t6 = 7(t1+t2+t3+t4+t5) = 7*8^3(t1+t2)...and so on
so series becomes=> (t1+t2) + (t3+t4+t5.....infinity) (t1+t2) + 7 proceed.......
divishth Says
what was wrong with my approach....i've posted few blocks earlier...just have a look.....
the bolden text above is not matching the requirements of the Q. have a re-look.
14)A quadratic with integral coefficients has two distinct positive integers as roots, the sum of its coefficients is prime and it takes the value -55 for some integer. The sum of the roots is (a) 32 (b) 20 (c) 24 (d) none of the these
14)A quadratic with integral coefficients has two distinct positive integers as roots, the sum of its coefficients is prime and it takes the value -55 for some integer. The sum of the roots is (a) 32 (b) 20 (c) 24 (d) none of the these
is the ans 24? My approach... for the sum of coeficients to be prime the roots shud be even:) let the quadratic eqn be (x-a)(x-b) = 0 also given that for sum integer the value is -55 = -5 * 11 from the options let the roots be 20 n 4. we can see tht if we subsitute 9 we get (9-4)(9-20) = 5*-11 = -55 hence all conditions are satisfied here... so the sum of the roots must be 20+4 = 24. PS-Did a bit of hit n trial before taking the values:)
Plz explain method of finding 10s digit for any given number..
the method of finding 10's digit is to divide the number by 100 and finding out the remainder so here rem(2^1001/100) or rem(2x((2^10)^100))/100 or (2*(24^100))/100 or (2*(76^50))/100 now 76 ^n always results in last two digits being 76 so finally it will be remainder of(2*76)/100 that is 52 got it??
hey i think the approach you took has some error coz see if we consider the option 20 and proceed in your way then also we can get a solution like (7-2)(7-11) cant we?
hey i think the approach you took has some error coz see if we consider the option 20 and proceed in your way then also we can get a solution like (7-2)(7-11) cant we?
but in this case the sum of coeficients is not a prime no. chk tht out its 10....for (x-2)(x-11)
here is the answer provided in the material consider a,b as the roots of the equation where a,b are positive integers, then clearly there is another equation with same sum of roots but product of roots is ab+55.clealry if we take option a we can find (13,7) and(18,2) so option b)
here is the answer provided in the material consider a,b as the roots of the equation where a,b are positive integers, then clearly there is another equation with same sum of roots but product of roots is ab+55.clealry if we take option a we can find (13,7) and(18,2) so option b)
didnt get it...if the roots are 13 and 7, pls chk it the sum of coeficients is 72...so how can it be.
) can u elaborate
