Number System - Questions & Discussions

MBA CAT 2022
81
@Ashu,
plz explain this part: "-2*-3*-4*....*-41*41! mod 83 = 1"

What happened to (-2*-3*-4*-5*.....*-1) part (here -1 is extracted from -41)?

Explanation for "-2*-3*-4*....*-41 this part ...

number of terms from 2 to 41 is 39 ,and including 2 it is 40.

40 is even number so all the minus sign canceled

so the series become 2*3*...41 = 1 * 2* ...41 = 41!

So 41! * 41! = 41! ^ 2

41! ^ 2 mod 83 = 1

hence the answer

hope this help..


82
@Ashu,
plz explain this part: "-2*-3*-4*....*-41*41! mod 83 = 1"

What happened to (-2*-3*-4*-5*.....*-1) part (here -1 is extracted from -41)?


81mod 83 could be written as = -2 or 81
80 mod 83 could be written as = -3 or 80
so on and upto -41...
now there are even number of total numbers from -2 to -41...
which makes it 41!...hence..proceed
83

In an infinite G.P every term is 7 time of the sum of all terms that follow.
t1+t2=9. Find the sum..

84
In an infinite G.P every term is 7 time of the sum of all terms that follow.
t1+t2=9. Find the sum..


here we have
t1+t2 = 9
t3 = 7(t1+t2)
t4 = 7(t1+t2+t3) = 7*8(t1+t2)
t5 = 7(t1+t2+t3+t4) = 7*8^2*(t1+t2)
t6 = 7(t1+t2+t3+t4+t5) = 7*8^3(t1+t2)...and so on

so series becomes=>
(t1+t2) + (t3+t4+t5.....infinity)
(t1+t2) + 7
proceed.......
85
Explanation for "-2*-3*-4*....*-41 this part ...

number of terms from 2 to 41 is 39 ,and including 2 it is 40.

40 is even number so all the minus sign canceled

so the series become 2*3*...41 = 1 * 2* ...41 = 41!

So 41! * 41! = 41! ^ 2

41! ^ 2 mod 83 = 1

hence the answer

hope this help..




@fuzzyl0gik
thanx buddy. u have explained it very lucidly..
86
here we have
t1+t2 = 9
t3 = 7(t1+t2)
t4 = 7(t1+t2+t3) = 7*8(t1+t2)
t5 = 7(t1+t2+t3+t4) = 7*8^2*(t1+t2)
t6 = 7(t1+t2+t3+t4+t5) = 7*8^3(t1+t2)...and so on

so series becomes=>
(t1+t2) + (t3+t4+t5.....infinity)
(t1+t2) + 7
proceed.......


then ans comes as 18 right?
87
@Ashu,
plz explain this part: "-2*-3*-4*....*-41*41! mod 83 = 1"

What happened to (-2*-3*-4*-5*.....*-1) part (here -1 is extracted from -41)?

Seems you have already got your answer. I was awake till 5am so was catching some much needed sleep.
88
series 22 , 34 , 48 , ? , 79, 96
what is d ans.
fast
.


Is it 63? let us know the answer
89
tac007 Says
then ans comes as 18 right?


The question is to find the sum till how many terms?
90

Maan it feels like eternity if u dont access this thread ....I opened the old thread there I found out that a new has startd and that too has completed a couple of pages.....maan I missed so much of action!!!!!

91
tac007 Says
then ans comes as 18 right?



Well I am confused about the question.

I am getting the following ..

(t1+t2)+7*(t1+t2)* (1+8+8*8+8*8*8+8*8*8*8+--- inf)

here the common ration is 8
since 8>1 ..the series is divergent .. then how do we get any answer for this?
We cant calculate the sum of infinite GP series when common ratio is >=1



correction invited..
92
in an infinite g.p every term is 7 time of the sum of all terms that follow.
T1+t2=9. Find the sum..



ans :64 (am i correct ???)
93
In an infinite G.P every term is 7 time of the sum of all terms that follow.
t1+t2=9. Find the sum..

its infinity....since common ratio is greater than 1
94

hey guys... got an interesting one...

There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.

Cheers!

95

hey friends i have few doubts..i will post some of them 2day. any help will be appreciated ...thanks in advance!!!
1)Find the number of negative integral solutions of
 x+1
x +|x + 1=(x+1)2
|x
a) 0 b) 1 c) 2 d) 3 e) 4

2)Amrita hosted a birthday party and invited all her friends and asked them
to invite their friends.There are n people in the party.Only Satyam is not
known to Amrita.Each pair that does not include Amrita or Satyam has exactly
2 common friends.Also,Satyam knows everyone except Amrita.If only
2 friends can dance at a time,how many dance numbers will be there at the
party?
a)3n 7 b)2n + 5 c)4n 3 d)none of these.

3)There is a circle of radius 1 cm. Each member of a sequence of regular
polygon P1(n), n = 5, 6...., Where n is the number of sides of polygon is
circumscribing the circle and each member of a sequence of regular polygon
P2(n), n = 4, 5.. where n is the number of sides of a polygon, is incribed
in the circle. Let L1(n) & L2(n) denote the perimeter of the corresponding
polygons P1(n)& P2(n). let X=L1(13)+2
L2(17) Then
a)4 1 b)1 2 c)X > 2 d)X 2

In the X Y plane, the area of the region bounded by the graph of
x + y + x = 4 is
a. 8 b. 12 c. 16 d. 20
 the 3rd one was confusing and 4th one i think should be a square so 16 but answer says otherwise please come up with your views

96
hey guys... got an interesting one...

There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.

Cheers!

hey the answer is 7here is how
lets take the numbers as c1>c2>c3>c4>c5
now there will be 10 combinations of two numbers i.e. c1+c2,c1+c3...etc add them all up you will get
4(c1+c2+c3+c4+c5)=120
so c1+c2+c3+c4+c4=30
now take out (c1+c5)+(c3+c4)
so the asnwer comes as 30-16-7=7
hope it helps
97
hey guys... got an interesting one...

There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.

Cheers!

is b = 4...?
98
hey the answer is 7here is how
lets take the numbers as c1>c2>c3>c4>c5
now there will be 10 combinations of two numbers i.e. c1+c2,c1+c3...etc add them all up you will get
4(c1+c2+c3+c4+c5)=120
so c1+c2+c3+c4+c4=30
now take out (c1+c5)+(c3+c4)
so the asnwer comes as 30-16-7=7
hope it helps



Not clear.. !!

can you explain it further.. ?
99
hey guys... got an interesting one...

There are 5 distinct integers, whose sum, taken 2 at a time, yields the results 15, 8, 7, 3, 0, 16, 18, 11, 23, 19. Find the second highest integer.

Cheers!


answer should be 4

the numbers are -4 ,4,7,11,12
100
find the remainder when (38^16!)^1777 is divided by 17

a. 1
b. 16
c. 8
d. 13



My take : remainder is 1.