@mailtoankit said:yaar 13c1 zaara bata na kaise aaya.....4c2 to samjh gaya..
13 type k cards hain na.
so picking one type by C(13, 1) methods
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbersA 37B 31C 41D 43
31?
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbersA 37B 31C 41D 43
n=15
so 31
@nick_baba said:In how many ways can Rs. 18.75 be paid in exactly 85 coins consisting of 50-paisa, 25-paisa and 10-paisa coins?
x+y+z=85
x/2+y/4+z/10=18.75
=>10x+5y+2z=375
=>y is an odd number, z is a multiple of 5
z=55
=>x=23,y=7
from here, z decreases by 5
z=50
=>x=20,y=15
...
x=17,14,11,8,5,2
so 8 values
@deedeedudu , @rkshtsurana bhai __/\__ itni jaldi kaise kia ?
x/2+y/4+z/10=18.75
=>10x+5y+2z=375
=>y is an odd number, z is a multiple of 5
z=55
=>x=23,y=7
from here, z decreases by 5
z=50
=>x=20,y=15
...
x=17,14,11,8,5,2
so 8 values
@deedeedudu , @rkshtsurana bhai __/\__ itni jaldi kaise kia ?
@Torque024 @mailtoankit
a+ b +c = 85
50a + 25b + 10c = 1875
eliminate c
8a + 3b = 205
a = 23 b = 7
a = 20 b = 15
..
..
a = 2 b = 63
total 8 solutions
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbersA 37B 31C 41D 43
31
@rkshtsurana said:@Torque024@mailtoankita+ b +c = 8550a + 25b + 10c = 1875eliminate c8a + 3b = 205a = 23 b = 7a = 20 b = 15....a = 2 b = 63total 8 solutions
thanks bhai..
@nick_baba said:V is the natural number below 1000, which has the maximum number of divisors. Find the sum of the digits of V.
whats the OA and soln :banghead:
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbersA 37B 31C 41D 43
n/2(1+n)=120
=>n^2+n-240=0
=>(n-15)(n+16)=0
n=15
15*16*31/6
b)31
=>n^2+n-240=0
=>(n-15)(n+16)=0
n=15
15*16*31/6
b)31
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbersA 37B 31C 41D 43
n(n+1) = 240k
n =15 satisfies for smallest
sum of squares = 15*16*31/6
largest prime = 31
@rkshtsurana said:@Torque024@mailtoankita+ b +c = 8550a + 25b + 10c = 1875eliminate c8a + 3b = 205a = 23 b = 7a = 20 b = 15....a = 2 b = 63total 8 solutions
mera dimag tel lene gya hai, hit and trial se kia :splat:
@vijay_chandola said:13 type k cards hain na.so picking one type by C(13, 1) methods
thanks.....Q hi galat padh liya tha meine..
...Q hi galat padh liya tha meine..
@rubikmath @Torque024
New Question
Find the coefficient of x^30 in expasnion of
(1-x^5)^6 * (1+x^3)^10
a) -5038
b)5038
c) -5040
d) 5039
OA 31 for below
@techsurge said:If the sum of first n natural numbers is divisble by 5!. For the smallest possible value of n, find the largest prime that divides the sum of squares of first n natural numbers
A 37
B 31
C 41
D 43
@techsurge said:whats the OA and soln
see bro...we have to take max possbl combo of powers of 2,3,5,7 which can gave a value less than 1000...2^3*3*5*7..lucky for me coz i got it in first hit..
@rkshtsurana said:@Torque024@mailtoankita+ b +c = 8550a + 25b + 10c = 1875eliminate c8a + 3b = 205a = 23 b = 7a = 20 b = 15....a = 2 b = 63total 8 solutions
Yaar koi shortcut pta hai aise questions ke liye..?
A milkman has 16 lt of pure milk in a container. At first house he sells 1 lt of milk and adds water to fill the container. At every subsequent house he sells twice the volume of the solution as he did at the previous house and adds water to fill the container. If he claims to sell at the cost price . Find his overall profit, when container is empty.
@maddy2807 said:A milkman has 16 lt of pure milk in a container. At first house he sells 1 lt of milk and adds water to fill the container. At every subsequent house he sells twice the volume of the solution as he did at the previous house and adds water to fill the container. If he claims to sell at the cost price . Find his overall profit, when container is empty.
93.75 i think
@techsurge said:@rubikmath@Torque024@vijay_chandolaOA 31Find the coefficient of x^30 in expasnion of (1-x^5)^6 * (1+x^3)^10a) -5038b)5038c) -5040d) 5039
3 type ki coefficients
x^0, x^30
x^15, x^15
x^30, x^0
answer= -5038
@techsurge said:@rubikmath@Torque024@vijay_chandolaOA 31Find the coefficient of x^30 in expasnion of (1-x^5)^6 * (1+x^3)^10a) -5038b)5038c) -5040d) 5039
(1 - 6x^5 + 15x^10 - 20x^15+ 15x^20 - 6x^25 + x^30) * ( 1+ 10x^3 + 45 x^6 + 120 x^9 + 210 x^12 + 252 x^15 + 120x^18 + 120x^21 + 45x^24 + 10x^27 + x^30)
1 + (-20)(252) + 1
2 - 5040
-5038
@maddy2807 said:A milkman has 16 lt of pure milk in a container. At first house he sells 1 lt of milk and adds water to fill the container. At every subsequent house he sells twice the volume of the solution as he did at the previous house and adds water to fill the container. If he claims to sell at the cost price . Find his overall profit, when container is empty.
93.75
for totally sells 31 litres --> 16ltr milk , 15 ltr water .
profit --> 15/16 *100
for totally sells 31 litres --> 16ltr milk , 15 ltr water .
profit --> 15/16 *100