16. If a = tan6 tan 42 and B = cot66 cot 78. Then,
(a) A = 2B
(b) A = 3B
(c) A = B
(d) 3A = 2B

a=tan6 tan42
b=cot66 cot78 can be written as tan24 tan12 bec cot(90-t)=tant;
tanA tanB = tanC tanD if A+B=C+D; or A+B =C-D or A-B=C+D or A-B=C-D;
as 42-6=24+12;
Hence A= B

17. What is the remainder when 128^1000 is divided by 153?

128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153
= [(2*3*17+1)^125]mod153
At this point,observe that 153=17*(3^2);
Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.
Actually,the above line can be written since only except the last two
terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,
[153*68] as one of its factors.
Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;
[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) +
1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);
which means [125*2*3*17 + 1] = 83*153 + 52;
which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 =
52mod153 ;
So, remainder is 52.

18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6?

log(abc)=6
abc=10^6 because of b^2=ac=100,
b^3=10^6,
b=10^2=100
and the combinations are
(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).

19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48times^2 is divided by 9?

First, let us consider a general case :
(222222222222.....{2 is repeated n times})^2
=[2(111111111111........{1 is repeated n times})]^2
=4(111111111111........{1 is repeated n times})^2
=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]
Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);
Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
Now, looking at the [(eqn1)] ,
we can find that here k =(n-1), (n-2), (n-3),....,1.
Now,for 10^(n-1) = 9*(a1) + 1;
10^(n-2) = 9*(a2) + 1;
10^(n-3) = 9*(a3) + 1;
10^(n-4) = 9*(a4) + 1;
.
.
.
.
10^1 =9 +1;
1 = 1
So,now in [(eqn1)] , we can write
(222222222222.....{2 is repeated n times})^2
=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2
=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.
=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]
Now this a general expression for (222222222222.....{2 is repeated n times})^2.
For the given problem,we can find that n=1,2,3,4,5,6,....48.
In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
to consider the last term,i.e, 4*(n^2);
Summing up [4*(n^2)] for n=1,2,...48. we get
4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}
=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
=4*8*49*97
=4(9- 1)(9*5 +4)(9*11 -2)
=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
=9*B + 32
=9*B + 9*3 + 5;
So, clearly the remainder will be 5.

20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 gives remainder 2?

for 5
7 12 17.........197....39 terms sum=3978
for 7
9 16 23 30....... 198...28 terms sum =2848
but some terms are counted more than 1 ..ie which are common in 5 and 7
eg 37 72 107 142 177..sum=535
so 6876-535=6341

in Q no: 18. i think there is one more possible combination... (1,100,10000)...correct me if im wrong

ReplyDeletein question 20, shouldn't we add 2 to the sum...?? dividing 2 by 5 or 7 gives 0 as quotient and 2 as remainder...

ReplyDelete