A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???

jain bhai - aap mat solve karna , you know this one dearly :)

We know that => 30 = 1*2*3*5

Also, 32 = 2^5

Let the number = 2^a * 3^b * 5^c

(a+1)*(b+1)*(c+1)= 32 = 2*4*4 or 2*2*8

Possibilities are 3!/2 + 3!/2 = 3! = 6 numbers

If the sum of 101 distinct terms in arithmetic progression is zero, in how many ways can three if these terms be selected such that their sum is zero ?

N = 2^m

We need to calculate:
R = (2^m)! MOD

E(2^2^m) = 2^(2^m - 1)

2^m! contains the maximum power of 2 as = (2^m)/2 + 2^m/4 + 2^m/8 + 2^m/16 + .....2^m/2^m

=> Total = (2^2^m - 1)

=> 2^m! will leave a remainder of 2^(2^m - 1) with 2^2^m

Therefore remainder = 2^(2^m - 1)

varun tyagi bhai,,
i have a doubt,,
in dis if u put m=2,, u get remainder as 8

but actually on putting m=2 N=2^2=4
so N!/2^N = 4*3*2/16,, remainder is 1???

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

Answer: 6

A no. divisible by 56 has factors of 7*8.Thus,it is divisible by 8 & 7.N for 8,last 3 digits has to be divided by 8.So,last digit has to be 2.
Checking for the options,x=4 divides the no. completely.
x=4
y=2
x+y=6

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

Denominator always has an extra power of 2. So the number will be divisible as long till 2^(N-1)...since its not further divisible , it must be odd and hence its remainder is 1.
now we have multiply it with 2^(N-1) to get the actual remainder

Thanks Varun Sir .......:)

p.s - With every new quant thread, tension badte jaati hain ; CAT's approaching

p.p.s - Ready to burn midnight oil today :)

Question :

Q) 16 people are to be arranged around an octagonal table with alternate sides of length 6m and 8m. People either sit at the corner or in the middle of the side .In how many ways can these people can be arranged ??

First person will have 4 choices with him : The two corners and two middle points.
=> 4C1 = 4 ways.
Rest 15 persons can be arranged in 15! ways.

Thus, Total = 4*15! ways.

A no. has exactly 32 factors out of which 4 are not composite.Product of these 4 factors (which are not composite) is 30.How many such no.s are possible???

jain bhai - aap mat solve karna , you know this one dearly

Question :

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

N = 2^m

We need to calculate:
R = (2^m)! MOD

E(2^2^m) = 2^(2^m - 1)

2^m! contains the maximum power of 2 as = (2^m)/2 + 2^m/4 + 2^m/8 + 2^m/16 + .....2^m/2^m

=> Total = (2^2^m - 1)

=> 2^m! will leave a remainder of 2^(2^m - 1) with 2^2^m

Therefore remainder = 2^(2^m - 1)

N=2^m
let m=3
N=8
8!%2^8
max power of 2 in 8!=(4+2+1)=7
so ans. 1 since it is not 0

fedbite bhai,,put m=1,,u will get 2 as remainder,,
for any value of m other than 1,,, we get remainder as 1,,
bas ab isko generalize karna hai,,
uske lie options k bharose,,,

A,B,C,D each had some money.D doubled the amounts with the others.C then doubled the amounts with the others.B then doubled the amount with the others.A then doubled the amount with the others.At this stage,each of them had Rs 80.Find the initial amount with C.

A) 45
B) 65
C) 95
D) 85

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

N=2^m
let m=3
N=8
8!%2^8
max power of 2 in 8!=(4+2+1)=7
so ans. 1 since it is not 0

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

15x67y is divisible by 56 = 7*8

Divisibility by 8 = Last three digits should be divisble by 8
=> 67y mod 8 = 0 and thus, y = 2

Divisibility by 7 = Double the units digit and subtract it from the remaining number and keep on doing this.
=> 15x672 => 15x63 => 15x => (15 - 2x) mod 7 = 0
=> x = 4

Thus, Number = 154672 ; x = 4 and y = 2.

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

x=4 and y=2,
so x+y=6,,

Approach,, 56=8*7
for 8 last 3 digits mst be divisible by 8,,hence y=2
ab isse x bhi nikal jaaega dat will be 4
so 4+2=6???

Varun bhai - is it 7^6/10^6


allani bhai - is it 82 ??

___//\\___ chevy bhai, catgirl, faruq bhai
Question :

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

N! mod 2^n = ?

if N is a power of 2, then N! will have (N-1) 2s

now make both numerator and denominator coprime
it means when we make N! and 2^N (denominator) co-prime, only 2 will left as denominator and no 2 will left with numerator

and as we have taken out all 2s from numerator (N!) so now numerator is odd that will give 1 reminder when divided by 2

now again multiply 1 (reminder) by common factor

so ans will be 2^(N-1)

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

it means 15x67y is divisible by 8 as well as 7
for div by 8 67y should be divisible by 8 so y=2
and for div by 7 x=4
so oa is 6
good night puys ...

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

y+670+x*1000+150000
56=7*8
(y-2)%8=0
((y-2)+6x+4)mod7=0
(y+6x+2)mod7=0
y=2
(4+6x)%7
x=4
so ans. 6

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

Should be 6



No. will be 154672

:)

varun.tyagi Says

q.) ten tickets are numbered 1,2,3.........10. Six tickets are selected at random one at a time with replacement. Prob that largest no. Appearing on the selected ticket is 7 is ??

1-(.6)^6...?

got it ..cal mistake..

(6c1*6^5+6c2*6^4+....6c6*1*6^0)/(10^6)

varun.tyagi Says

Q.) Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??

since irs replacement,,
total ways of selecting tickets = 10c1*10c1...(6 times) = 10^6

largest shud be 7,,so consider only first 7 tickets,,
Favourable cases = total cases - cases wid no 7 =
total cases for first 7 = 7c1*7c1...(6 times)=7^6
similarly fr nt 7 = 6^6

(7^6 - 6^6)/10^6 hai kya??

naye thread pe pehla answer mera(galat ho ya shi ho)!!!

catgirl2012 Says

7c6/10c6 ?

Faruq Says

But chevy Bhai not waale cases lene ki kya zaroorat hai. Dont u think the ans should be -> 7^6/10^6

Varun bhai - is it 7^6/10^6


allani bhai - is it 82 ??

___//\\___ chevy bhai, catgirl, faruq bhai
Question :

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

chevy Says

isliye faruq bhai,,coz its replacement,,so aisa bhi toh ho sakta hai he picks out f dose 7 tickets but he always picks d same ticket,,replacement hai na

O.A. is (7^6 - 6^6)/10^6


The number of ways of selecting 1,2,3,4,5,6,7 = (7^6)
The number of ways in which any one of 1,2,3,4,5,6 is the highest number = (6^6)

Required ways = ( 7^6 - 6^6 )

Total ways = 10^6

Probability = (7^6 - 6^6)/10^6

Varun bhai - is it 7^6/10^6


allani bhai - is it 82 ??

___//\\___ chevy bhai, catgirl, faruq bhai
Question :

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

fr m=1,, N=2
2!/2^2 = 2/4,,, remainder is 2

fr m=2 ,,N=4

4!/2^4 = 4*3*2/16 remainder will be 1,,


we get remainder as 1 for all values of m except for m=1,,
so we have to generalize accordingly

Faruq Says

But chevy Bhai not waale cases lene ki kya zaroorat hai. Dont u think the ans should be -> 7^6/10^6

isliye faruq bhai,,coz its replacement,,so aisa bhi toh ho sakta hai he picks out f dose 7 tickets but he always picks d same ticket,,replacement hai na

varun.tyagi Says

Q.) Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??

If F(x) = x^4 -360x^2 +400, then for x E Integer, if F(x) is a prime number, then what is the sum of all possible F(x)??

Varun bhai - is it 7^6/10^6


allani bhai - is it 82 ??

___//\\___ chevy bhai, catgirl, faruq bhai
Question :

If m is a natural no. and N = 2^m, what is the remainder when N! is divided by 2^N????

if 15x67y is divisible by 56, find x+y
options : 8,6,4,2

If F(x) = x^4 -360x^2 +400, then for x E Integer, if F(x) is a prime number, then what is the sum of all possible F(x)??

f(x) = x^4 - 360x^2 + 400 = (x^2 + 20)^2 - 400x^2

since irs replacement,,
total ways of selecting tickets = 10c1*10c1...(6 times) = 10^6

largest shud be 7,,so consider only first 7 tickets,,
Favourable cases = total cases - cases wid no 7 =
total cases for first 7 = 7c1*7c1...(6 times)=7^6
similarly fr nt 7 = 6^6

(7^6 - 6^6)/10^6 hai kya??

naye thread pe pehla answer mera(galat ho ya shi ho)!!!

But chevy Bhai not waale cases lene ki kya zaroorat hai. Dont u think the ans should be -> 7^6/10^6

varun.tyagi Says

Q.) Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??

7c6/10c6 ?

catgirl2012 Says

Not fair varun.tyagi :'(...iske upar ek Groan banta hai :P

I understand your emotions

The problem is the PG server. It becomes really a burden on a server when any thread nears 1000 pages. It is better to remove some burden off the server. I hope you understand

P.S. Groan to moh maaya hai, Jo de uska bhi bhala jo na de uska bhi bhala

varun.tyagi Says

Q.) Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??

since irs replacement,,
total ways of selecting tickets = 10c1*10c1...(6 times) = 10^6

largest shud be 7,,so consider only first 7 tickets,,
Favourable cases = total cases - cases wid no 7 =
total cases for first 7 = 7c1*7c1...(6 times)=7^6
similarly fr nt 7 = 6^6

(7^6 - 6^6)/10^6 hai kya??

naye thread pe pehla answer mera(galat ho ya shi ho)!!!

Not fair varun.tyagi :'(...iske upar ek Groan banta hai

Thanks Varun Sir .......:)

p.s - With every new quant thread, tension badte jaati hain ; CAT's approaching

p.p.s - Ready to burn midnight oil today :)

Question :

Q) 16 people are to be arranged around an octagonal table with alternate sides of length 6m and 8m. People either sit at the corner or in the middle of the side .In how many ways can these people can be arranged ??

If F(x) = x^4 -360x^2 +400, then for x E Integer, if F(x) is a prime number, then what is the sum of all possible F(x)??