Posts and Comments

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 15 Jun '12

> My scores:
OA______45A____34C____91
Sec1____20A____16C____44
Sec2____25A____18C____47
PS: Can anyone tell me how to view the poll result, as in names of guys with >120 score ??
The guy who started the thread should have made it public while creating a poll, then its possible t...

My scores:

OA______45A____34C____91

Sec1____20A____16C____44

Sec2____25A____18C____47

PS: Can anyone tell me how to view the poll result, as in names of guys with >120 score ??

The guy who started the thread should have made it public while creating a poll, then its possible to see the names of guys who polled.

I think, now a moderator can change it.

- 1 Like

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 28 Sep '11

> ***How many 3 digit numbers are divisible by 3 or 5 but not by 7?*
*1\. 51*
*2\. 300*
*3\. 360*
*4\. 180*
*5\. 128*
We need to find numbers divisible by 3,5 and eliminate those divisible by 21,35,15 and add those divisible by 105.
divisible by 3 => 102 to 999 => 897/3 +...

How many 3 digit numbers are divisible by 3 or 5 but not by 7?

1. 512. 3003. 3604. 1805. 128

We need to find numbers divisible by 3,5 and eliminate those divisible by 21,35,15 and add those divisible by 105.

divisible by 3 => 102 to 999 => 897/3 + 1 => 300

divisible by 5 => 105 to 995 => 895/5 + 1 => 180

Divisible by 21 => 21*47 is near 999 so , 47 - 4 = 43 such numbers

Divisible by 35 => 35*28 is near 999 so, 28 - 2 = 26 such numbers

Divisible by 15 => 105 to 990 => 885/15 + 1 => 60

Divisible by 105 are 9 numbers

so, Answer is 300 + 180 - 129 + 9 => 360

- 4 Likes

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 11 Sep '11

> emanresu Says
>
> *P is a group of four numbers 1, 2, 3 and 1. In every step, 1 is added to any two numbers in group P. In how many such steps is it possible to make all the four numbers in group P equal? *
Its not possible to make all of them equal.

emanresu SaysP is a group of four numbers 1, 2, 3 and 1. In every step, 1 is added to any two numbers in group P. In how many such steps is it possible to make all the four numbers in group P equal?

Its not possible to make all of them equal.

- 2 Likes

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 09 Sep '11

> *
Find the sum of the last three digits of S=871x873x875x878x881x883?
*
129*127*125*122*119*117
Last digit is 0
now, Last two digits of
29*27*25*61*19*17 is 25
> h.kagda Says
>
> *How many ordered pairs (x,y) of integers are possible for xy/(x+y) = 196?*
1/x + 1/y ...

Find the sum of the last three digits of S=871x873x875x878x881x883?

129*127*125*122*119*117

Last digit is 0

now, Last two digits of

29*27*25*61*19*17 is 25

h.kagda SaysHow many ordered pairs (x,y) of integers are possible for xy/(x+y) = 196?

1/x + 1/y = 1/196

(x+196)(y+196) = 196^2

RHS takes all the divisors of 196^2 = 14^4 => 7^4 * 2^4 => 25

(Oops Ordered pairs i ignored)

@RC89 : Wish you a Happy Birthday

- 3 Likes

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 09 Sep '11

> yeh wala try kariye zara bhai log .. mera frnd planet of the apes dekh ke aaya aur mujhe pucha .. he said usi movie me dekha tha ...
*there are 4 wts . a,b,c,d
wts = a>b>c>d*
*there are 3 pans . the wts are kept in the first pan such that the heaviest is at the bottom.
u need to re-...

yeh wala try kariye zara bhai log .. mera frnd planet of the apes dekh ke aaya aur mujhe pucha .. he said usi movie me dekha tha ...there are 4 wts . a,b,c,d

wts = a>b>c>dthere are 3 pans . the wts are kept in the first pan such that the heaviest is at the bottom.

u need to re-shuffle them and place them back again using all the 3 pans such that the same formation is obtained ( each wt. needs to be moved atleast once ).

rule = at no point can u keep the heavier wt on a lighter one

d

c

b

a

_ _ _

1 2 3question is , minimum how many moves are needed ?

Its 2^n - 1, n is number of weights

(For more info search for towers of Hanoi)

- 1 Like

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 05 Sep '11

> not sure if am right .. jus approximating the calculation
number of 5 will be 2*67 = 134
number of 5^2 will be 1*13 = 13
number of 5^3 will be 1*6 = 6
number of 5^4 will be 1*3 = 1
so i think its around 154
It will be 168

not sure if am right .. jus approximating the calculation

number of 5 will be 2*67 = 134

number of 5^2 will be 1*13 = 13

number of 5^3 will be 1*6 = 6

number of 5^4 will be 1*3 = 1

so i think its around 154

It will be 168

- 2 Likes

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 05 Sep '11

> Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is???
7^6 is the number of cases in which the 6 tickets chosen are numbered between 1 to 7 i.e., wher...

Ten tickets are numbered 1,2,3,...,10. Six tickets are selected at random one at a time with replacement. The probability that the largest number appearing on the selected ticket is 7 is???

7^6 is the number of cases in which the 6 tickets chosen are numbered between 1 to 7 i.e., where there is no 8, 9 or 10.

6^6 is the number of cases in which the 6 tickets are numbered between 1 to 6 i.e. where there is no 7, 8, 9 or 10.

7^6 - 6^6 is the no of cases in which 7 is there and all are numbered between 1 to 7.

So, Probability is (7^6 - 6^6)/10^6

- 5 Likes

Lava Kumar Dukanam
@MaskedMenace
4.1
k

Posted 04 Sep '11

> can u post the solution for this prob of why it is 5 ?
i am getting just two solutions
y takes value 44 and 60
Here are the solutions
(19,684)
(21,252)
(27,108 )
(45,60)
(99,44)

can u post the solution for this prob of why it is 5 ?

i am getting just two solutions

y takes value 44 and 60

Here are the solutions

(19,684)

(21,252)

(27,108 )

(45,60)

(99,44)

- 4 Likes

0Commentsnew comments