BRAIN STRAIN !!!
Q1) How many three digit no's in base 11 are possible, when expressd in base 9, has its digit reversed?
Q2) Find no. of integral solutions of eqn a*b*c*d = 4 ?
Q3) Find no. of ways in which three no's in A.P. can be selected from first 51 natural no's?
Q4) Find the remainder of (19^92)/92 ?
ques 3 :
for three naturl numbers to be AP , the sum of the first and last nos. must be even
out of first 51 natural nos 26 are odd and 25 are even.
hence the total possible ways 26C2 + 25C2 = 325 + 300 = 625
i THINK THIS IS A lot easier solution.
ques : 92 = 23 * 4
19 = 23-4
(23-4)92 = 2392 + 492 + k*92
2392 = x mod(23*4) 492 = y mod(23 *4 )
2391 = x/23 mod(4) 491 = y/4 mod(23)
cyclicity = 2 2182 = y/4 mod(23)
thus x/23 = 3 cyclicity = 22
x = 69 thus y/4 = 18
y = 72
thus remainder = (72+69) mod 92 = 49
hello rajorshi,
i am very weak in maths..
plz help me out with the chinese remainder theorem
i hv gone thru d proof but not able to understand anything..
i would really appreciate if u kindly explain the theorem.
Thanks,
sonal
Qn no.1 is absolutely fine..i',ve checked it..m giving u the options also..& solutn to qn no.2 is wrong.
Q1) How many three digit no's in base 11 are possible, when expressd in base 9, has its digit reversed?
a)1 b)2 c)3 d)4
Q2) Find no. of integral solutions of eqn a*b*c*d = 4 ?
a)80 b)76 c)56 d)72
Try again !!
anita you missed out negative values
Adding to the two diff cases you have mentioned
For case 1 No of Possibilities = 4!/(2! 2!) =6For case 2 No of Possibilities = 4!/(3! ) =4so total number of solutions 6 + 4 = 10
4 = -1 x -1 x 1 x 4 Possibilities = 4!/2! = 12
= 1 x 1 x -1 x -4 4!/2! = 12
= -1 x -1 x -1 x -4 4!/3! = 4
= 1 x 1 x -2 x -2 4!/(2! x 2!) = 6
= -1 x -1 x 2 x 2 6
= -1 x 1 x -2 x 2 24
= -1 x -1 x -2 x -2 6
Total = 80
n/a
In the solution of Qn no.4, i coudnt understand how did u find the cyclicity as 2 & 22 & the solution beyond it..could you please explain Rajorshi???????
Times of India se uthye hue h yeh questions........haina...coming mondy ansrs aa jayenge...check it frm thr.....
cheers.....
ansr of questn 1(base wala) is 1
hint
121a+11b+c=81c+9b+a
this will give
60a+b=40c
check fr values
only one value will satisfy
hii all
For solving questn no 4, go it by EULER METHOD.
It will be lot easierr....
Rt nw in office...il post it 2nite with complete theory of euler and solution to this also.
till thn tke dis mantra:
we can solve remainder probs by:
1. by binomial
2. by cyclicity
3.by euler method
4. by manipulations by making the remainder of number 1 or -1
so try solving the questions by above techniques....it vil be helpfull...
neone of you cn suggest me a buk from where i can find problems that anita has solved a*b*c*d, some probs like 1/a+1/b+1/c=1/a+b+c....finding no of solutions .....
thankssss
and after calculations I get the answer as 49.
Answer Q2. Find no. of integral solutions of eqn a*b*c*d = 4 ?
Q3) Find no. of ways in which three no's in A.P. can be selected from first 51 natural no's?
Could you please Check Question number 1 ? ? ?is there any TYPO???