@ishu1991 said:how many zeroes will be there at the end of 1003*1001*999*....123?
odd series...so zero ?
remainder when 7^87 mod 97= ?
@ishu1991 said:@Logrhythmbhai thda detail me btao i dont havbe OA
7^87%97 = 7^3*7^84%97 = 52*52^28%97 = 52*85^14%97 = 52*(-12)^14%97 = 52*47^7%97 = 52*47*75*75^3%97 = 52*47*75*96%97 = -(52*47*75)%97 = -67 or 30
ye toh bohot hi bekar method hai...i mean arduous calculation wala...i guess self made question hoga??
but if agar humare pass options hote toh hum isko ase kar sakte the..
e(97) is 96..
hence 7^96%97=1
=> 7^9*7^87%97=1
let the remainder with 7^87 be x
=> 7^9*x%97=1
55*x%97=1
now 30 wld be in the options...try that
55*30%97 gives 1..
so using this method wld be far better... 
@DespicableA said:Find sum of digit of the least number n, such that 2n is square and 3n is a cube.a, 9b. 10c. 11d. 12
ans is 9 (72)
3n is cube
2n in square
n=3^2 x 2^3
@Deb.csen said:Ram, Shyam and Hari went out for a 100 km journey. Ram and Hari started the journey in Ram's car at the rate of 25 kmph, while shyam walked at 5 kmph. After sometime, Hari got off and started walking at the rate of 5kmph and Ram went back to pick up shyam. All three reached the destination simultaneously. The number of hours required for the trip was:A. 8B. 7C. 6D. 5E. 4Solve this with explanation...
8..
...... .................... ..................................
5x 20x 5y
25x+5y=100
2*20x/30 + 5y/25 = y
Q1
@DespicableA said:Here is one question from my side..How Many Two Digit Numbers Have Exactly Four Factorsa. 29b. 30c. 31d. 32
31
@vbhvgupta said:explain
root(2) a =32
a = 32/root(2) first inner square side
similarly
b = 32/root(2)root(2) .... second inner square side
so sum of perimeter = 4x32(1+1/root(2) + 1/2 + 1/2*root(2)......
4x32(root(2)/(root(2)-1))
128(root(2)(root(2)+1))
@DespicableA said:Here is one question from my side..How Many Two Digit Numbers Have Exactly Four Factorsa. 29b. 30c. 31d. 32
a^3 or a*b
a^3 => a=2 ,3 [ 2 values]
a*b = 2* b ===> all primes values of b less than 50 and greater than or equal to 5 ==> total no.'s = 13
a*b = 3*b ==> all prime values of b less than 33 and greater than 3. ==> 5 ,7.....31 ==> total = 9
a*b = 5*b ==> all primes values of b less than 20 and greater than 5 ==> 7,11,13,17,19==> 5 values
a*b = 7*b ==> all prime values of b greater than 7 and less than 14...11,13 ==> 2 values
total = 2+13+9+5+2 = 31.
#Ravi Theja
a^3 => a=2 ,3 [ 2 values]
a*b = 2* b ===> all primes values of b less than 50 and greater than or equal to 5 ==> total no.'s = 13
a*b = 3*b ==> all prime values of b less than 33 and greater than 3. ==> 5 ,7.....31 ==> total = 9
a*b = 5*b ==> all primes values of b less than 20 and greater than 5 ==> 7,11,13,17,19==> 5 values
a*b = 7*b ==> all prime values of b greater than 7 and less than 14...11,13 ==> 2 values
total = 2+13+9+5+2 = 31.
#Ravi Theja
@DespicableA said:Find sum of digit of the least number n, such that 2n is square and 3n is a cube.a, 9b. 10c. 11d. 12
n = a^3 *b ^2 Here a=2 b=3 { assumption as per question}
2n = 2^4*3^2 ==> perfect square .
3n = 2^3*3^3 ==> perfect cube .
NUmber is 2^3* 3^2 = 8*9 =72 ===> digital sum = 9
2n = 2^4*3^2 ==> perfect square .
3n = 2^3*3^3 ==> perfect cube .
NUmber is 2^3* 3^2 = 8*9 =72 ===> digital sum = 9
@bs0409 said:V anand and G kasparov play a series of 5 chess games. the prob that anand wins a game is 2/5 and kasparov is 3/5. there is no prob of draw . the series will be won by a person who wins 3 matches . find prob that Anand wins the series( The series ends the moment w/n any of the two wins 3 matches)
WWW 8/125
WLWW 24*3/625
LLWWW 6*72/3125
LWLWW
LWWLW
WLLWW
WLWLW
WWLLW
992/3125....is this right?