@vbhvgupta said:explain kar do.
Find the sum upto 21 terms (-) sum upto 20 terms
N is the product of first 100 multiples of 5 starting from 5, then which is the rightmost nonzero digit of N ?
a 6
b 4
c 2
d NOT
@vbhvgupta said:expalin bhai...
i simply took an example and solved it.
1, (1/3), (1/9), (1/27)...
toh option B was giving the right ans.
otherwise, you can always go with the traditional method of solving though equations.
@Tiws said:N is the product of first 100 multiples of 5 starting from 5, then which is the rightmost nonzero digit of N ?a 6 b 4 c 2 d NOT
5 ni hoga ??
Q. Number of 5 digit numbers with only 3 distinct digits. I do not have the options ppl
@coonal2006 said:Q. Number of 5 digit numbers with only 3 distinct digits. I do not have the options ppl
p&c; 
@coonal2006 said:Q. Number of 5 digit numbers with only 3 distinct digits. I do not have the options ppl
PnC kamzor hai ...
agr 3 distinct me 0 include krte hain to 2*3*3*3*3 hona chaiye .. nhi include krte 0 to 3*3*3*3*3 ..
answer aur method dono postkrte tym tag kr dena :)
agr 3 distinct me 0 include krte hain to 2*3*3*3*3 hona chaiye .. nhi include krte 0 to 3*3*3*3*3 ..
answer aur method dono postkrte tym tag kr dena :)
@Tiws said:if f(x)= x^4 - 9x^3 + 26x^2 - 24x + 5 then one root of f(x) lies btwna)-1 &0b)0&2c)2&3d)4&5@raopradeepf(0) = 5 ; F(1) = -1 ;hence thr is a root between x=0 and x=1 thus option B is correct
there is the doubt sign change happens between 0&2 and
4&5
@Tiws said:find remainder when S is devided by 13.S =[ 5+ 5^2 +......+ 5^ 55]a.0b.1c.7d.12
5^2mod13 =-1
5^3mod13=-5
5^4mod13=1
5^5mod13=5
a pair of 4 will cancel each other
so remaining ones are
5+ (-1) + (-5)
so remainder -1 or 12
OA d
@Tiws said:N is the product of first 100 multiples of 5 starting from 5, then which is the rightmost nonzero digit of N ?a 6 b 4 c 2 d NOT
i am getting 5 what is the answer?
How many three-digit numbers are composed of three
distinct digits such that one digit is the average of the
other two?
(A) 96 (B) 104 (C) 112 (D) 120 (E) 256
distinct digits such that one digit is the average of the
other two?
(A) 96 (B) 104 (C) 112 (D) 120 (E) 256
@Tiws said:N is the product of first 100 multiples of 5 starting from 5, then which is the rightmost nonzero digit of N ?a 6 b 4 c 2 d NOT
5?
@ishu1991 said:How many three-digit numbers are composed of threedistinct digits such that one digit is the average of theother two?(A) 96 (B) 104 (C) 112 (D) 120 (E) 256
112 ways??
@Tiws said:N is the product of first 100 multiples of 5 starting from 5, then which is the rightmost nonzero digit of N ?a 6 b 4 c 2 d NOT
5*10*15...TILL 100 multiples.
5^100 (1*2*3......... *20)
= 5^104 * 2^18 * 3^8 * 7*2 * 11 * 13 * 17 * 19
this product would have 18 zeros.
so just find the unit digit of 5 ^ 86 * 3^8 * 7*2 * 11 * 13 * 17 * 19
= 5 * 1 * 9 * 1 * 3 * 7 * 9
Unit digit = 5
@ishu1991 said:How many three-digit numbers are composed of threedistinct digits such that one digit is the average of theother two?(A) 96 (B) 104 (C) 112 (D) 120 (E) 256
From 0 to 9, select 2 odd digits or 2 even digits. That will give all possible AP's between 0 to 9.
So 2*5C2=20
For odd it will be 5C2*3!=60
For even, 0 will occur in 4 cases out of 10, so combinations will be 6*3!+4*(3!-2!)=52
ANS=112
So 2*5C2=20
For odd it will be 5C2*3!=60
For even, 0 will occur in 4 cases out of 10, so combinations will be 6*3!+4*(3!-2!)=52
ANS=112