@techgeek2050 said:what about x=2 , y= 44 ?
how ?
(2 - 1)(1 - 44) = 41
1*(-43) = 41
-43 = 41 ?
@gs4890 said:Find the last non-zero from right hand side of 120!
use formula 5a+b
then 2^a *a!*b!
120=5*24
Where a=5 and b=0
2^24 * 24! =6*6 =6
then 2^a *a!*b!
120=5*24
Where a=5 and b=0
2^24 * 24! =6*6 =6
@gs4890 said:Find the last non-zero from right hand side of 120!
No of 5s = 24+4 = 28
120! mod 10^29 = 6 ?
If f (x) is a quadratic polynomial, such that f (5) = 35 and f (-5) = 15, and f (p) = f (q) = 0, then find p + q.
@mailtoankit said:how ?(2 - 1)(1 - 44) = 411*(-43) = 41-43 = 41 ?
product = 44x2 = 88
sum = 44 + 2 = 46
difference = 88 - 46 = 42
@gs4890 said:Find the last non-zero from right hand side of 120!
6 ?
last non-zero digit = 2^a(R(a!)*R(b!)) ..... R = 5a + b
2^24*(2^4(R(4!)*R(4!)))
6*6*(6*6) = 6
@Subhashdec2 said:bhai ye mod itni jaldi kaise nikala...
Actually, last digit of 2*3*4*6*7*8*9 nikala 
Pen use ni karra abi to jo dimag me ata hai likh deta hun.. 😛
A is a single digit prime no and b is any natural no .
how many equation of the form X^2 - 4 sqrtA x + 38 = 0 have both real roots.
how many equation of the form X^2 - 4 sqrtA x + 38 = 0 have both real roots.
a>0 b>0 the no of roots of the equation 2X^7 - 9X^5 - 3X^4 - bX2 + 7 = 0
a>0 b>0 the no of roots of the equation X^7 - 9X^4 + X^3 - 8 = 0
find the roots of the equation 6X^4 - 6X^3 - 24X^2 - 6X +6 = 0
@Faruq said:If f (x) is a quadratic polynomial, such that f (5) = 35 and f (-5) = 15, and f (p) = f (q) = 0, then find p + q.
Faruq f(x) = ax^2 + bx + c
f (5) = 35 => 25a + 5b + c = 35
f (-5) = 15 => 25a - 5b + c = 15
solving both equations
b=2
25a+c=25
take a set of values for (a,c) which satisfy the eq. : a=1,c=0
so f(x) = x^2 + 2x
f(x) can be 0 when x=0,-2
so P+q=-2
...
@gs4890 said:Find the last non-zero from right hand side of 120!
Funda 1: Rightmost non-zero digit of n! or R(n!)
0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6! = 720; 7! = 5040.
R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]
where n = 5a + b
Example: What is the rightmost non-zero digit of 37! ?
R(24)= Last Digit of [ 2^4 x R (4!) x R (4!) ]=6
? R (120!) = Last Digit of [ 2^24 x R (24!) x R (0!) ]
=last digit of(6 x 6 x 1)=6
0! = 1; 1! = 1; 2! = 2; 3! = 6; 4! = 24; 5! = 120; 6! = 720; 7! = 5040.
R(n!) = Last Digit of [ 2^a x R(a!) x R(b!) ]
where n = 5a + b
Example: What is the rightmost non-zero digit of 37! ?
R(24)= Last Digit of [ 2^4 x R (4!) x R (4!) ]=6
? R (120!) = Last Digit of [ 2^24 x R (24!) x R (0!) ]
=last digit of(6 x 6 x 1)=6
@Sufi0469 said:Faruq f(x) = ax^2 + bx + c f (5) = 35 => 25a + 5b + c = 35 f (-5) = 15 => 25a - 5b + c = 15solving both equationsb=225a+c=25take a set of values for (a,c) which satisfy the eq. : a=1,c=0so f(x) = x^2 + 2xf(x) can be 0 when x=0,-2so P+q=-2...
a can also be equal to 2 and b equal to -25
So ans of this question is CBD
So ans of this question is CBD
If f(x ˆ' 1) + f(x + 1) = f(x) and f(2) = 6, f(0) = 1, then what is the value of f(50)?
1) ˆ'7
2) 6
3) 1
4) 7
1) ˆ'7
2) 6
3) 1
4) 7
@ScareCrow28 said:Actually, last digit of 2*3*4*6*7*8*9 nikala Pen use ni karra abi to jo dimag me ata hai likh deta hun..
bhai notepad use karlo
Engg h yaar tu...akal Lagao...:)
@vbhvgupta said:A is a single digit prime no and b is any natural no .how many equation of the form X^2 - 4 sqrtA x + 38 = 0 have both real roots.
A = 2, 3, 5 or 7
x^2 - 4_/A x + 38 = 0
D= 16A - 4*38 >= 0
=> A >= 8.5
So, 0 possible?