@albiesriram said:
D)
for, 123456789 diff b/w alternate sum is 5.
now from 101112 to 9898... it is 450-450 (simply add all alternate no. i.e 1,2,3,4,5,6,7,8,9 for 10 times and 1111111111 similarly for all no. from 1 to 9 for 10 times).
now when we get 100 we have additional 1. so 5+1 =6.
how many roots (both real and complex) does (x^n - a)^2 = 0 have?
the square of the sum of the roots of a quadratic equation E is 8 times the product of its roots. Find the value of the square of the sum of the rootsdivided by the product of the roots of the equation whose roots are reciprocals of those of E.
@hesse said:@sur17the series is 12345678910111213 and so on till 9899100.. now for divisibility of 11 we know that the difference between the sum of alternative numbers is to be divisible by 11.for, 123456789 diff b/w alternate sum is 5.now from 101112 to 9898... it is 450-450 (simply add all alternate no. i.e 1,2,3,4,5,6,7,8,9 for 10 times and 1111111111 similarly for all no. from 1 to 9 for 10 times).now when we get 100 we have additional 1. so 5+1 =6.
here u'v made a mistake i think.....
(simply add all alternate no. i.e 1,2,3,4,5,6,7,8,9 for 10 times)- here it should be 9 times and not 10........because 1, 2 ,3...9 occur in the 10s then 20s, 30s,... 90s which is 9 times in all and not 10
(simply add all alternate no. i.e 1,2,3,4,5,6,7,8,9 for 10 times)- here it should be 9 times and not 10........because 1, 2 ,3...9 occur in the 10s then 20s, 30s,... 90s which is 9 times in all and not 10
The lowest possible degree of an equation with real coefficient two of whose roots are sqrt3 and 3+2i is.
@sur17 yupp you are right....it should be 9 times for 1,2,3,4,5,6,7,8,9 and 9 times for the first digit...:)
@vbhvgupta said:The lowest possible degree of an equation with real coefficient two of whose roots are sqrt3 and 3+2i is.
3?
(X-root3)(x-3-2i)(x-3+2i)=0satisfies.
@vbhvgupta said:the square of the sum of the roots of a quadratic equation E is 8 times the product of its roots. Find the value of the square of the sum of the rootsdivided by the product of the roots of the equation whose roots are reciprocals of those of E.
6 ?
a^2 + b^2 = 8ab
(1/a + 1/b)^2/(1/a*1/b) = (a + b)^2/(ab) = a^2 + b^2 - 2ab/ab = 8ab - 2ab/ab = 6ab/ab = 6 ?
@vbhvgupta said:the square of the sum of the roots of a quadratic equation E is 8 times the product of its roots. Find the value of the square of the sum of the rootsdivided by the product of the roots of the equation whose roots are reciprocals of those of E.
Is it 6?
@mailtoankit said:6 ?a^2 + b^2 = 8ab(1/a + 1/b)^2/(1/a*1/b) = (a + b)^2/(ab) = a^2 + b^2 - 2ab/ab = 8ab - 2ab/ab = 6ab/ab = 6 ?
@catahead said:Is it 6?
8
The sum of absolute values of all real numbers x, such that both of the fractions (x^2+4x−17)/(x^2−6x−5) and (1−x)/(1+x) are integers, can be written as a/b, where a and b are coprime positive integers. What is the value of a+b ?
@albiesriram said:
N = 123456789101112...9899100
= 1234567890*10^92 + 101112...9899*1000 + 100
For 10111213....9899 remainder by 11 will be 10 + 11 + ... + 98 + 99 or 45*109
remainder when N is divided by 11 will be:-
6*1 + 45*109*1000 + 1
or 6 + 1*(-1)*(-1) + 1
or 8
So, remainder will be 8
@albiesriram said:
12345678910.........99100
adding alternate terms and subtracting
(2 + 4 + 6 + 8 + 10*45 + 1) - (1 + 3 + 5 + 7 + 9 + 9*45) mod 11
471 - 430 mod 11
41 mod 11 = 8