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 ?

50 hain kya



????

wovfactorAPS Says

50 it is...

Sum of AP = (n/2)* = 0
=> a + = 0

=> Every same number term from the beginning and the last have their sum equal to zero.
Or, 1st + 101th term = 0
2nd + 100th term = 0
=> 50th + 52th term = 0

Now, we are left with 51st term which should be zero to make the sum Zero.

Now selecting 3 terms => 50C1 ways.

If the sum of odd numbers is zero and the series is an A.P., we can safely say that the middle number is zero.

=> Middle term = 51st term = 0

The series can be -50d, -49d, -48d,......0, d, 2d,3d,.....49d, 50d

We need to find that sum of three numbers selected must be zero

Case 1: When zero is one of the three numbers

We need to pick one positive and one negative complimentary
so 50 ways
Case 2: When zero is not picked

We need to pick numbers such as 25d + 20d - 45d....so that the sum becomes zero i.e. select two positive and one negative OR two negatoves and one positive.

Select two positive numbers in C(50,2) ways and the third one is an obvious choice

OR

Select two negative numbers in C(50,2) ways and the third one is an obvious choice

Total ways = 2*C(50,2)

Grand Total = 50 + 2*C(50,2) = 50 + 2450 = 2500

mba4mheart Says

Is it 1250????

-50 -49......-1 0 1 2.....50

ways of selecting three terms out f dem such dat sum is 0,,

50???

It is 1250.
Consider -50 to 50 as the numbers in AP

one can take 2 +ve -1 -ve , or 2-ve -1+ve or obvious 1+ve -0 and 1 -ve

first number values of second number | number of values for second no.

1 2,3,.....47,48,49 48
2 3,......,48 46
.
.
..
.
24 24,25 2
25 | 0

so summation of it is 600 ways for 2 +ve -1 -ve
similarly for 2-ve -1+ve
1+ve -0 and -ve in 50 ways
600*2+50 =1250 ways

But Varun Bhai Agar hum second case mein 50 mein se 2 choose karte hai.
-50,-49,-48...........0.............1,2,3,,,,,49,50

If we will choose 2 numbers 49+50=99 then we will have no any option in negative no. as -99

You are right bhai!! I realized it later. Actually C(50,2) mein se 25 cases aise subtract hongey, jinme ye flaw hai....Uske baad vo humein jo pichle 50 the usme add karne hongey

Total = 50 + C(50,2) - 25 = 50 + 1225 - 25 = 1250

guys ,
i have a doubt with the following geometry question , kindly help.
c and d are points on the circle with diametre AB such that AQB = 2* COD , where o is the centre of the circle. Q is the point of intersection of ac and db inside the circle. the tangents at c and meet at p. the circle has radius 1 . the distance of p from its centre is ?
Following are the options
2/3 ,2/3 ,3/2,1

If the sum of odd numbers is zero and the series is an A.P., we can safely say that the middle number is zero.

=> Middle term = 51st term = 0

The series can be -50d, -49d, -48d,......0, d, 2d,3d,.....49d, 50d

We need to find that sum of three numbers selected must be zero


Case 1: When zero is one of the three numbers

We need to pick one positive and one negative complimentary
so 50 ways
Case 2: When zero is not picked

We need to pick numbers such as 25d + 20d - 45d....so that the sum becomes zero i.e. select two positive and one negative OR two negatoves and one positive.

Select two positive numbers in C(50,2) ways and the third one is an obvious choice

OR

Select two negative numbers in C(50,2) ways and the third one is an obvious choice

Total ways = 2*C(50,2)

Grand Total = 50 + 2*C(50,2) = 50 + 2450 = 2500


EDIT: :oops:
2500 are ordered pairs and we are supposed to find unorderd pairs. So, In this case just divide by 2.

Total ways = 1250

But Varun Bhai Agar hum second case mein 50 mein se 2 choose karte hai.
-50,-49,-48...........0.............1,2,3,,,,,49,50

If we will choose 2 numbers 49+50=99 then we will have no any option in negative no. as -99

catgirl2012 Says

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 ?

let series be -50,-49,.....,0,........,49,50
ways of pattern (-50,0,50),(-49,0,49)=50
-50,1,49
a+b=50
-3- 1case
-4- 1case
-5- 2case
-6 -2case
-7 -3case
-8 -3case

so for 48 numbers we have 24*25=600 ways
similarly for positive
so ans. 1250

catgirl2012 Says

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 ?

-50d, -49d, -48d, . . .-d, 0, d,. . . 48d, 49d, 50d

if out of 3 selected terms 1 is 0, then rest 2 should be -nd and +nd => 50 ways

if 1 number is negative and 2 positive =>
if 1 number is -50d, then sum of other 2 positive numbers should be 50d => a+b = 50d => 24 solutions (unordered, and both number different)

1 number is -49d, sum of other 2 positive numbers is 49d => 24 solutions
1 number is -48d sum of other 2 positive numbers is 48d => 23 solutions
1 number is -47d , sum of 2 positive numbers should be 47d => 23 solutions
.
.
.
so on (also no solution for -2d and -1 d in this case - 1 -ve 2 positive)

2(24+23+22+...+1) = 600 solutions

same will be for when 1 positive and 2 negative

so it becomes 50+600+600 = 1250 ways

wrong post,,,,,,,,

catgirl2012 Says

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 ?

Is it 1250????

catgirl2012 Says

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 ?

If the sum of odd numbers is zero and the series is an A.P., we can safely say that the middle number is zero.

=> Middle term = 51st term = 0

The series can be -50d, -49d, -48d,......0, d, 2d,3d,.....49d, 50d

We need to find that sum of three numbers selected must be zero


Case 1: When zero is one of the three numbers

We need to pick one positive and one negative complimentary
so 50 ways
Case 2: When zero is not picked

We need to pick numbers such as 25d + 20d - 45d....so that the sum becomes zero i.e. select two positive and one negative OR two negatoves and one positive.

Select two positive numbers in C(50,2) ways and the third one is an obvious choice

OR

Select two negative numbers in C(50,2) ways and the third one is an obvious choice

But we have to subtract case such as 50 + 49 - 99, Since we do not have 99 here. There will be 25 such cases

Total ways = 2*(C(50,2) - 25)

For un-ordered pairs, Total ways = C(50,2) - 25 = 1200

Grand Total = 50 + 1200 = 1250

catgirl2012 Says

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 ?

Sum of AP = (n/2)* = 0
=> a + = 0

=> Every same number term from the beginning and the last have their sum equal to zero.
Or, 1st + 101th term = 0
2nd + 100th term = 0
=> 50th + 52th term = 0

Now, we are left with 51st term which should be zero to make the sum Zero.

Now selecting 3 terms => 50C1 ways.

varun.tyagi Says

24 = 8 MOD 16......Remainder 1 kyu hoga bhai?? 8 he to hai remainder :)

oooo,,, i did a blunder,,mera dimaag kharaab ho gya hai,,,

chuck it, got that ! Thank You! :D

invincible2910 Says

how are you people checking for the divisibility of 7...??

catgirl2012 Says

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 ?

50 it is...

catgirl2012 Says

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 ?

50 hain kya



????

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 :)

Product of 4 NOT composite factor is 30
1*2*3*5=30
2^a *3^b*5^c Where (a+1)(b+1)(c+1) is 32 (No. of factors)
now a b c can take 3,3,1 & 7,1,1 value .=3!/2!=3*2=6 values possible

how are you people checking for the divisibility of 7...??

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 ... :-)

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

Should be 6



No. will be 154672

:)

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 :)

ans. 6??

we have to arrange 4,4,2 =3 ways
2,8,2=3ways

Originally Posted by gs4890
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

Same as varun bhai's method
Ans 6

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

In such questions, starting from the last is a good approach.

-- A ---- B ---- C ---- D
- 25 --- 45 --- 85 --- 165

- 50 --- 90 --- 170 --- 10
-100 ---180 --- 20 --- 20
-200 --- 40 --- 40 ---- 40
- 80 ---- 80 --- 80 ---- 80

Thus, C has Rs.85 as the initial amount.

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???

24 = 8 MOD 16......Remainder 1 kyu hoga bhai?? 8 he to hai remainder

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

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)??

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 ??