@Exodia said:Find the remainder (30^72)^87 is divided by 11.
5 Aayega eska
n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has how many factors?
1) 35
2) 32
3) 28
4) None of these
n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has how many factors?
1) 35
2) 32
3) 28
4) None of these
@scrabbler said:...rt2:1 hai kya? Painful calculation regardsscrabbler
yes its rt2:1..can u plz xplain!!!
@Exodia said:n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has how many factors?1) 352) 323) 284) None of these
35?
@Exodia said:Another problem on remainders:Find the remainder when 10^10 + 10^100 + ......... 10^10000000000 is divided by 7:1) 02) 13) 24) 5
OA-5
@Exodia
Shld be 35..
Here, 2n = 2^(k+1)*3^l...n, 3n = 2^k*3^(l + 1)
Thus, (k+2)(l+1) = 28...n, (k+1)(l + 2)= 35..
On Solving, k=5 n l=3..
Thus, 6n will be 2^6*3^4..Hence, Number of factors will be 35..
@Exodia said:n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has how many factors?1) 352) 323) 284) None of these
35 factors..
n=2^5*3^3..by simple observation..
n=2^5*3^3..by simple observation..
If x^3-ax^2+bx-a=0 has 3 real roots, then which is true ?
a=1
a not equal to 1
b=1
b not equal to 1
@mohitjain said:yes its rt2:1..can u plz xplain!!!
See attached....it was much easier to draw on paper :(
regards
scrabbler
regards
scrabbler
@amresh_maverick said:If x^3-ax^2+bx-a=0 has 3 real roots, then which is true ?a=1a not equal to 1b=1b not equal to 1
@scrabbler said:See attached....it was much easier to draw on paper regardsscrabbler
thanx a lot mate..:)
@amresh_maverick said:If x^3-ax^2+bx-a=0 has 3 real roots, then which is true ?a=1a not equal to 1b=1b not equal to 1
@scrabbler approach batana
Put b
@amresh_maverick said:@scrabbler approach batana
Put b equal to1 and the equation will have complex roots...so b not equal to 1
@Exodia said:n is a number such that 2n has 28 factors and 3n has 30 factors. 6n has how many factors?1) 352) 323) 284) None of these
35 ?
n = 2^5*3^3
2n = 2^6*3^3
3n = 2^5*3^4
6n = 2^6*3^4
@raj37 said:Put b Put b equal to1 and the equation will have complex roots...so b not equal to 1
Nice... 
regards
scrabbler
regards
scrabbler
@scrabbler said:Nice... regardsscrabbler
bro thoda detail mein samjha do how to solve this types of questions
@iLoveTorres said:bro thoda detail mein samjha do how to solve this types of questions
x^3-ax^2+bx-a=0
b=1
x^3-ax^2+x-a=0
x^2 (x-a)+1 (x-a)=0
(x^2+1)(x-a)=0
x^2+1 will give complex roots
hence b cannot be 1
b=1
x^3-ax^2+x-a=0
x^2 (x-a)+1 (x-a)=0
(x^2+1)(x-a)=0
x^2+1 will give complex roots
hence b cannot be 1