[jaidev_iyer]

All the natural numbers which are multiples of 7 are written one beside the other. What will be the 5000 th digit?

[bishoo123]

Originally Posted by protoplasm

Find n^th term of the series.
4 7 11 17 27 45...

Did anyone give a try on this ?

[jaidev_iyer]

There are 777 houses in a society. The houses were numbered in the ascending order of various number systems stating from binary and exhausting each system and moving into next. The number plate has place for only 3 digits. So what will be the number of number system between 33 rd and 777 th houses?

[shanks4mba]

Originally Posted by Lakshya13@@@

Heyyy... anybody help me wid disss....

If x and y are positive real numbers such that

21xy + 15x + 28y = 16

Then find the minimum value of 21(x + y)?
[A]7
[B]17
[C]21
[D]27
[E]11

21xy + 15x + 28y = 16
21x(y + 5/7) + 28(y + 5/7) - 20 = 16
(21x + 2(y + 5/7) = 36 = 2^2*3^2
(21x + 2 = 36 , (y + 5/7) = 1
x = 8/21 y = 2/7
x+y = (8+ 6)/21 = 14/21
21(x+y) = 14 not in options...

[shanks4mba]

Originally Posted by jaidev_iyer

There are 777 houses in a society. The houses were numbered in the ascending order of various number systems stating from binary and exhausting each system and moving into next. The number plate has place for only 3 digits. So what will be the number of number system between 33 rd and 777 th houses?

total = 777 houses..
now...binary system - > total possible 111 = 7
n= 3-->222 = 26
total = 33
n=4-->333 = 63
n=5--> 444 = 124
n =6-->555 = 215
n= 7 -->666 = 342
total = 777
hence there are 4,5,6,7 a total of 4 no. systems

[jaidev_iyer]

Originally Posted by shanks4mba

21xy + 15x + 28y = 16
21x(y + 5/7) + 28(y + 5/7) - 20 = 16
(21x + 2(y + 5/7) = 36 = 2^2*3^2
(21x + 2 = 36 , (y + 5/7) = 1
x = 8/21 y = 2/7
x+y = (8+ 6)/21 = 14/21
21(x+y) = 14 not in options...

(x+y) will be minimum when the two factors are equal.
Refer to previous posts buddy, has been solved.

[jaidev_iyer]

Originally Posted by shanks4mba

total = 777 houses..
now...binary system - > total possible 111 = 8
n= 3-->222 = 26
total = 34
n=4-->333 = 63
n=5--> 444 = 124
n =6-->555 = 215
n= 7 -->666 = 342
total = 778
hence there are 3,4,5,6,7 a total of 5 no. systems

But the OA says 4.
Dont think they are counting n=3.

[jaidev_iyer]

If 3 AMs, 3 GMs and 3 HMs are inserted between 1 and 5, then find the cubic equation whose roots are 1st AM, 2nd GM and 3rd HM between 1 and 5?

[bishoo123]

Originally Posted by Vineetverma

11/3 frm the top ???

ps see my approach just put it in my previous post

thnxxx

Yes, 11/3 is the correct one....me too getting same answer.

[divishth]

Originally Posted by bishoo123

Did anyone give a try on this ?

Originally Posted by protoplasm

Find n^th term of the series.

4 7 11 17 27 45...

f(1) = 4
f(2) = 2*f(1) - 1 = 7
f(3) = 2*f(2) - 3 = 11
f(4) = 2*f(3) - 5 = 17
f(5) = 2*f(4) - 7 = 27
f(6) = 2*f(5) - 9 = 45
f(7) = 2*f(6) - 11 = 79

So, f(n) = 2*f(n-1) - (2n-1)