@mohitjain said:there is a triangle ABC..dere is a point P inside the triangle such dat its distance from d three sides are 6,8 and 11..find the perimeter of the triangle??
i am getting 43.xx ....
@Aizen said:Is it 10?x^10 - 1 = (x - 1)(x - a)...(x - i) as they are the factorsSo (x - a)...(x - i) = (x^10 - 1)/(x - 1)Now replace x by 1.RHS will be limit x -> 1, (x^10 - 1)/(x - 1) , a 0/0 formSo apply L'Hospitals Rule: 10x^9 = 10*1 =10
right... ! can be done without L'Hospital rule as well, updating solution here in a short while..
@pratskool said:i am getting 43.xx ....
approach plz..i dont knw d OA .can u plz share ur approach.
@iLoveTorres said:
no.... it cannot be answered using 2nd statement alone,,, as p(2i) + p(2j) is already composite even if it is not mentioned... that goes without saying
p(2i) + p(2j) = 4i^2 + 4j^2 + 80 = 4( i^2 + j^2 + 20)
@pratskool said:right... ! can be done without L'Hospital rule as well, updating solution here in a short while..
You can solve it using 10th root of unity also.
@mohitjain said:approach plz..i dont knw d OA .can u plz share ur approach.
since you have mentioned nothing about the triangle... i have assumed a general case, where each of the angle of the chord (side ) is 120degree and solved for each side using cosine formula... can be wrong....
@pratskool said:Each question is followed by two statements, X and Y. Answer each question using the following instructions:Mark (A) If the question can be answered by using the statement X alone but not by using the statement Y alone.Mark (B) If the question can be answered by using the statement Y alone but not by using the statement X alone.Mark (C) If the question can be answered by using either of the statements alone.Mark (D) If the question can be answered by using both the statements together but not by either of the statements alone.Mark (E) If the question cannot be answered on the basis of the two statements.undefinedLet p(x) = x^2 + 40. Then for any two positive integers i and j where i > j, is p(i) + p(j) a composite number?(X) p(i) €“ p(j) is not a composite number(Y) p(2i) + p(2j) is a composite number
I guess it can be answered by only the 1st statement and not using 2nd statement.
As: From 1: p(i) - p(j) = i^2 - j^2 = (i + j)*(i - j)
From 2: p(2i) + p(2j) = 2^2 + 2j^2 + 40
@pratskool said:i am getting 43.xx ....
i am also getting 43.5 by taking cos = 120...par yeh jab aa raha hai..when 6,8,11 are taken from vertices....par question mein yeh distance to sides se diya hua hai na....so isn't P will be 6,8,11 away from the three sides ??
@Aizen said:I guess it can be answered by only the 1st statement and not using 2nd statement.As: From 1: p(i) - p(j) = i^2 - j^2 = (i + j)*(i - j)From 2: p(2i) + p(2j) = 2^2 + 2j^2 + 40
yeah right.... but u can elaborate the 1st 1 more so that other users can understand,,,
p(i) - p(j) = i^2 - j^2 is not a composite number only when i-j = 1 and i + j hence a prime number...
now p(i) + p(j) = i^2 + j^2 + 80 = (i-j)^2 + 2*i*j + 80 = 2*i*j + 81
now since both i and j are consecutive numbers, for i = 2, j = 1 , the expression is 85 , composite
for the values which follow like i = 3,j=2, i = 4, j=3... i*j is divisible by 3 hence it is composite...
@Nihilist1002 said:@pratskoolif the roots of the equation x^10 - 1 = 0 are 1,a,b,c.... i, then what is the value of(1-a)(1-b)... (1-i) ?is dat 2?
@iLoveTorres said:the other obvious root is -1 so the value may be 2.? else 0?
we can rewrite x^10 - 1 as (x-1)(x-a)(x-b)... (x-i)
x^10-1/ x-1 = (x-a)(x-b).. (x-i)
x^10 - 1 = (x^5 - 1)(x^5 + 1) = (x^4(x-1) + x^3(x-1) + x^2(x-1) + x(x-1) +1) (x^5 + 1)
dividing this by x-1 we ge (x^4 + x^3 + x^2 + x + 1)(x^5 + 1)
substituting x = 1 we get the value as 10....
@mailtoankit said:i am also getting 43.5 by taking cos = 120...par yeh jab aa raha hai..when 6,8,11 are taken from vertices....par question mein yeh distance to sides se diya hua hai na....so isn't P will be 6,8,11 away from the three sides ??
yeah right,,,, i would recheck my approach and would soon update my OA here...
find the remainder when 84^79 is divided by 100
a)24
b)44
c)64
d)84
@sachisurbhi said:find the remainder when 84^79 is divided by 100a)24b)44c)64d)84
64 ?
84^79 mod 100 = last two digits of 84^79
84^79 = (4*21)^79 = 2^158*21^79 = 2^8*(2^10)^15*(81) = 56*24*81 = 64
@sachisurbhi said:right ..tell me the method used ..
last two digits of (4*21)^79
2^158*21^79
(1024)^15*256 =24*56
24 raised to any odd power gives last two digit as 24
24 raised to even power gives last two digit as 76
and for 21^79=last two digit=81 multiply 24,56 and 81 and find last two digit
2^158*21^79
(1024)^15*256 =24*56
24 raised to any odd power gives last two digit as 24
24 raised to even power gives last two digit as 76
and for 21^79=last two digit=81 multiply 24,56 and 81 and find last two digit