pratiksahoo33

@pratiksahoo33

procrastinator17
above and beyond @procrastinator17
Consider the set S = {1, 2, 3, …., 1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and with 1000 and have at least 3 elements? (1) 3 (2) 4 (3) 6 (4) 7 (5) 8 Please explain the method as well. Thanks.
pratiksahoo33
Pratik Sahoo @pratiksahoo33

4.7 1000 = 1+(n-1)d =>999/d +1 = n Given n>=3 So 999/d +1>=3=>999/d >=2 d will be a factor of 999 so as to be within the S so no of factors of 999 = 3^3 * 37^1  8 factors out of these 8 factors, 999 will be not possible since it will giving 999/999=1 So 8-1 = 7 values of d

chuck_gopal
Deepak Gopalakrishnan @chuck_gopal
[image] [image] *PS:* Columbia Business School asked for a 140-character essay as part of its admissions application for the MBA admissions of 2012. _Concept, illustrations and graphics by *Deepak Gopalakrishnan* aka chuck_gopal, a Mallu-turned Mumbaiker who blogs here and t...
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 01 Aug '11
> If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by: a>50 b>60 c>70 d>80 Please justify ............. Is it (b)60? take the min pythagorean triplets 3*4*5 = 60
If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:

a>50
b>60
c>70
d>80

Please justify .............



Is it (b)60?

take the min pythagorean triplets 3*4*5 = 60
  • 1 Like  
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 01 Aug '11
> If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by: a>50 b>60 c>70 d>80 Please justify ............. Is it (b)60? take the min pythagorean triplets 3*4*5 = 60
If a,b,c are the lengths of a right angled triangle, then a*b*c is divisible by:

a>50
b>60
c>70
d>80

Please justify .............



Is it (b)60?

take the min pythagorean triplets 3*4*5 = 60
  • 1 Like  
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 16 Jul '11
> Can someone elaborate the approach as to how to solve such questions : Q-Find the remainder when 2^89 is divided by 89 ? A)2 B)1 C)87 D)88 OA-A Use Eulers theorm Euler no of 89 = 88 (2^8 :: *2/89 = 2/89 So remainder = 2
Can someone elaborate the approach as to how to solve such questions :

Q-Find the remainder when 2^89 is divided by 89 ?

A)2 B)1 C)87 D)88

OA-A


Use Eulers theorm Euler no of 89 = 88
(2^8*2/89 = 2/89

So remainder = 2
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 16 Jul '11
> MM is rite!! ;) *Two ants Amy and Pamy are running on a graph paper. Amy is at a distance 2 cm from the X-axis and 3 cm right of Y-axis. Pamy is directly 2 cm below the X-axis and 4 cm to the right side of Y-axis. They spot a food particle dropped on the first quadrant of the paper. The fo...
MM is rite!! ;)

Two ants Amy and Pamy are running on a graph paper. Amy is at a distance 2 cm from the X-axis and 3 cm right of Y-axis. Pamy is directly 2 cm below the X-axis and 4 cm to the right side of Y-axis. They spot a food particle dropped on the first quadrant of the paper. The food particle's shortest distance from the X-axis is 6 cm and 8 cm from the Y-axis. If Amy can run at a speed of 1 cm/sec and Pamy at speed of 1.5 cm/sec and, Pamy is further away from the food particle than Amy, then in approximately how many seconds would the food particle be grabbed by either of the ants?

5.2 seconds
7 seconds
6.4 seconds
5.9 seconds


Suja


Is it (d ) 5.9 sec?
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 13 Jul '11
> Bhai log, you have ignored the bold part. by going with that condition. After 4th round, a-b,2b-c,2c-d,2d-a+b Now, a-b=48, 2b-c=48, 2c-d=48, 2d-a+b=48 2d-a+b+a-b = 96, but here we'll get d = 48. Please correct me if/where I am wrong? If d gives a-b coins to a, then a wi...
Bhai log, you have ignored the bold part.




by going with that condition. After 4th round, a-b,2b-c,2c-d,2d-a+b
Now, a-b=48, 2b-c=48, 2c-d=48, 2d-a+b=48
2d-a+b+a-b = 96, but here we'll get d = 48. Please correct me if/where I am wrong?


If d gives a-b coins to a, then a will have 2a-2b coins
  • 1 Like  
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 13 Jul '11
> plz see the ques below: find the sum of the first 20 terms of the series: (1*2^2)+(2*3^2)+(3*4^2)+........ i am getting answer as 50050 but the sol says 47180. where am i going wrong ? :: Gen Term = nv=> n(n^2 + 2n + 1)=>n^3 +2*n^2 + n S = ^2 + n(n+1)/2 + 2*n*(n+1)*(2n...
plz see the ques below:

find the sum of the first 20 terms of the series:
(1*2^2)+(2*3^2)+(3*4^2)+........

i am getting answer as 50050 but the sol says 47180. where am i going wrong ?


Gen Term = nv=> n(n^2 + 2n + 1)=>n^3 +2*n^2 + n

S = ^2 + n(n+1)/2 + 2*n*(n+1)*(2n+1)/6

Put n = 20

S = 50050
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 13 Jul '11
> Father gives a certain amount of one rupee Coins to his four sons a,b,c,d. now, a gives to b as many coins as b has,b gives to c as many coins as c has,c gives to d as many coins as d has,and d gives to a as many coins a has. now, all of them has got 48 coins. what is the maximum amount of coin...
Father gives a certain amount of one rupee Coins to his four sons a,b,c,d. now, a gives to b as many coins as b has,b gives to c as many coins as c has,c gives to d as many coins as d has,and d gives to a as many coins a has. now, all of them has got 48 coins. what is the maximum amount of coins that anyone has at any point of time?

1.60
2.54
c.90
d.72


Let initially a,b,c,d

After 1st rnd, a-b,2b,c,d
After 2nd rnd,a-b,2b-c,2c,d
After 3rd rnd, a-b,2b-c,2c-d,2d

2d=48=>d=24
2c-d=48=>c=36
2b-c=48=>b=42
a-b=48=>a=90

So at any time we see that a has highest amount

So (c) 90
Whats the OA?
  • 3 Likes  
pratiksahoo33
Pratik Sahoo @pratiksahoo33 77
Posted 13 Jul '11
> akkitheaviator Says > > find the ratio of the 17th terms of two AP if the ratio of the sum upto n terms of two APs is (3n+2):(4n-13) Is it 101:119?? :=(3n+2):(4n-13) =>: =(3n+2):(4n-13) =>* = (3n+2):(4n-13) Put n = 33 and get the ans !!*
akkitheaviator Says
find the ratio of the 17th terms of two AP if the ratio of the sum upto n terms of two APs is (3n+2):(4n-13)


Is it 101:119??

:=(3n+2):(4n-13)

=>: =(3n+2):(4n-13)
=> = (3n+2):(4n-13)

Put n = 33 and get the ans !!