NUMBERS

[pooja talwar]

ques: Find the sum of all the factors of 324, which are multiples of 3.


A) 846
B) 840
C) 648
D) 852

Ans is B) 840


Reply with solution please.

[Abhilash Chandra]

324= 3^4 * 2^2

so no of factors are 5*3=15

out of 15 factors  the multiple of 3 are => 3,9,27,81 -----------(powers of 3)

                                                            6,18,54,81-----------(powers of 3*2)

                                                            12,36,108,324---------(powers of 3*4)

add all n ull get 840

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[pooja talwar]

Ques: If k(N) denotes the number of ways of expressing N as a
difference of two perfect squares, then which of the follwing is the
least?

A) k(187)
B) k(120)
C) k(110)
D) k(105)


Ans is C)


Kindly reply with answer please.

[pooja talwar]

Ques: There are N houses in a colony, numbered 1 to N. All but 5 of
the houses were destroyed in an earthquake. The numbers on the houses
which were not destroyed are consecutive. If the sum of the numbers on
the destroyed houses is 1085, then the least of the numbers on the
houses which were not destroyed can be,
A) 36
B) 26
C) 25
D) EITHER A) & B)


Ans is: D)

Please give the solution.

[Abhilash Chandra]

question 2--

it will take some effort to check all solutions .

try 36 => it means houses from 1 to 35 are destroyed .. so sum till 35 =35*36/2=630, remaining is 1085-630=455

so u have to check if u can make sum from 41 to N as 455. if u take 41+42+43....50 =(40+1)+(40+2)........(40+1)=40*9+(1+2+3...10)=455

similarly,

for 26.....    1+2.......25=25*26/2=325, remaing =760,

 so from 31 to N check if u can make sum as 760.. if u take 31+32......49=(30+1)...(30+19)=30*19 +(1+2+3..19)=570+190=760

so option D.

[Abhilash Chandra]

* 49=40+9, i wrote it wrong

[Abhilash Chandra]

*50 = 40+10 ,  i wrote it wrong

On Mon, Feb 10, 2014 at 7:08 PM, Abhilash Chandra <abhi1...@gmail.com> wrote:

[smilin...@gmail.com]

Please remove me from the group.

Sent on my BlackBerry® from Vodafone

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From: Abhilash Chandra <abhi1...@gmail.com>

Sender: suii_g...@googlegroups.com

Date: Mon, 10 Feb 2014 19:09:24 +0530

To: suii_g...@googlegroups.com<suii_g...@googlegroups.com>

ReplyTo: suii_g...@googlegroups.com

Subject: Re: [suii-generis: 17486] NUMBERS

[Abhilash Chandra]

question 1..

110 cant be written as diff of any two perfect square ,

lets say a^2-b^2=110, => (a+b)(a-b)=110

here 110=2*5*11 .......so possible pairs can be {(110=2*55), (110=5*22) , (110=10*11) } 

so (a+b)(a-b)=2*55 or (a+b)(a-b)=5*22 or (a+b)(a-b)=10*11 

in any case u cant make integral val of a and b.

IF U TRY FOR OTHERS ULL GET SOME INTEGRAL VALUES .