- View Sourceanswer(145)let the number is xas we know that number- 13x+2 or x=13-2=11 so that,number=divider*divident+remainder or11*13+2=145so that in case of 2nd term as same like above17x+9 or x = 17-9 =8after that, 8*17+9=145so,145 is ans.
**From:**"ascent4cat@yahoogroups.com" <ascent4cat@yahoogroups.com>**To:**ascent4cat@yahoogroups.com**Sent:**Tuesday, 25 August, 2009 3:14:39 PM**Subject:**[2IIM CAT Prep] Digest Number 752# Messages In This Digest (8 Messages)

- 1a.
- Re: Numbers From: ponmalargs_cse
- 1b.
- Re: Numbers From: prakash awasthy

- 3a.
- Re: Numbers ques From: ashok_rrsa
- 3b.
- Re: Numbers ques From: sunil singla

- 4a.
- Re: (unknown) From: Rahil Sahu

- 5a.
- Re: Quant Numbers ques From: Rahul Jain

# Messages

- 1a.
## Re: Numbers

### Posted by: "ponmalargs_cse" ponmalargs@... ponmalargs_cse

#### Mon Aug 24, 2009 3:03 am (PDT)

hi,

answer for 1 question is 8

Number of ways a number can be expressed as a product of two co-primes is 2^(n-1) (^ - power, n - different prime factors of d number).

here 24700=13*19* 5*5*2*2

n=4

so 2^3 = 8

--- In ascent4cat@yahoogro ups.com, vimlesh patelia <vimlesh_21mech@ ...> wrote:

>

> Hi Maha Lakshmi,

>

> Answer for your second question is 87(13*6+9,17* 5+2).

>

>

> Regards,

> Vimlesh

>

>

>

>

> ____________ _________ _________ __

> From: Maha lakshmi <maha.b.lakshmi@ ...>

> To: ascent4cat@yahoogro ups.com

> Sent: Tuesday, 5 May, 2009 1:45:52 PM

> Subject: [2IIM CAT Prep] Numbers

>

>

>

>

>

> Hi Friends,

>

> Can any one help in sloving the following problems

>

> 1) Find the Number of ways in which 24700 can be expressed as a product of two co-primes?

> 2) Find the smalles number which when divided by 13 leaves 9 as the remainder and when divided by 17 leaves 2 as the remainder?

>

> Thanks & Regards,

> --

>

> Mahalakshmi. B

>

>

>

>

>

> Own a website..Get an unlimited package.Pay next to nothing.*Go to http://in.business. yahoo.com/

>

- 1b.
## Re: Numbers

### Posted by: "prakash awasthy" prakash.awasthy@... prak_awasthy

#### Mon Aug 24, 2009 3:03 am (PDT)

hi maha lakshmi

as vimlesh said answer 87 is right method to get is...

Number ca be written as 13d1+9 = 17d2+2

or 13d1+7 = 17*d2......

so we have to find a smallest value of d1 such that 13d1+7 ..is copletly

divisible by 17...

we get this value for d1=6...that is 13*6+7==85

but this 85 == 13d1+7..and our number was 13d1+9....so the required smallest

no. is 87....

On Mon, Jun 15, 2009 at 12:56 PM, vimlesh patelia

<vimlesh_21mech@ yahoo.com>wrote:

>

>

> Hi Maha Lakshmi,

> Answer for your second question is 87(13*6+9,17* 5+2).

>

> *Regards,*

> *Vimlesh*

>

>

> ------------ --------- ---------

> *From:* Maha lakshmi <maha.b.lakshmi@ gmail.com>

> *To:* ascent4cat@yahoogro ups.com

> *Sent:* Tuesday, 5 May, 2009 1:45:52 PM

> *Subject:* [2IIM CAT Prep] Numbers

>

> Hi Friends,

>

> Can any one help in sloving the following problems

>

> 1) Find the Number of ways in which 24700 can be expressed as a product of

> two co-primes?

> 2) Find the smalles number which when divided by 13 leaves 9 as the

> remainder and when divided by 17 leaves 2 as the remainder?

>

> Thanks & Regards,

>

> --

>

> Mahalakshmi. B

>

>

> ------------ --------- ---------

> Cricket on your mind? Visit the ultimate cricket website. Enter now!<http://in.rd. yahoo..com/ tagline_cricket_ 1/*http:/ /beta.cricket. yahoo.com>

>

>

--

Prakash Awasthy

- 2a.
## Re: doubt

### Posted by: "Rahul Jain" championishere007@... championishere007

#### Mon Aug 24, 2009 3:03 am (PDT)

The correct answer is 741. You need to remove all the points on the

hypotenuse so formed, and the other two sides also.

Its forms a triangle with a total of [(1/2 * 41*41)] = 840.5 area or approx

841 points. Now, the hypotenuse is of length = sqrt(41^2 + 41^2) . Then

subtract the other sides and you get the final answer as 741 :)

On Thu, Jul 16, 2009 at 8:32 PM, sharada sahoo <rockingpinku@ yahoo.co. in>wrote:

>

>

> 741

>

> --- On *Mon, 15/6/09, Haneef Mohammad <haneef0786@gmail. com>* wrote:

>

>

> From: Haneef Mohammad <haneef0786@gmail. com>

> Subject: Re: [2IIM CAT Prep] doubt

> To: ascent4cat@yahoogro ups.com

> Date: Monday, 15 June, 2009, 11:45 AM

>

> Half of points in square formed by (0,0) (41,0) (0,41) (41,41)

>

> excluding the points on the boundary, 40*40 / 2

>

> Answer: 800

>

>

>

> On Mon, May 11, 2009 at 10:55 PM, satbir singh r wasu <satbirsinghr@

> yahoo.co. in<http://in.mc941.. mail.yahoo. com/mc/compose? to=satbirsinghr@ yahoo.co. in>

> > wrote:

>

>>

>>

>>

>>

>> consider a triangle drawn on the x-y plane with its three vertices of

>> (41,0),(0,41) and (0,0) each vertex being represented by its (x,y)

>> coordinates. The number of points with integer coordinates inside the

>> triangle (excluding all the points on the boundary)is

>> 1)780

>> 2)800

>> 3)820

>> 4)741

>>

>>

>> please explain with brief explanation.

>>

>> Regards,

>> Satbir singh.

>>

>>

>> ------------ --------- ---------

>> Cricket on your mind? Visit the ultimate cricket website. Enter now!<http://in.rd. yahoo..com/ tagline_cricket_ 1/*http:/ /beta.cricket. yahoo.com>

>>

>

>

> ------------ --------- ---------

> Looking for local information? Find it on Yahoo! Local<http://in.rd. yahoo.com/ tagline_local_ 1/*http:/ /in.local. yahoo.com/>

>

>

--

------------ --------- ---------

RAHUL JAIN

rahul_jain@daiict. ac.in

Ph.no.: 9328067371- 2b.
## Re: doubt

### Posted by: "ManMohanPuri" mmpuri@... purimmpuri

#### Mon Aug 24, 2009 3:04 am (PDT)

The integer point plotted in a square on this x and y axis are ;

x=( 1 to 40)

y =(1 to 40)

total point plotted in a square on this x and y axis are =40 X 40

=1600

integer point plotted o the diagnol of this square :

x= (1 ---40)

y= (40----1)

integer point plotted on the diagnol of this square :(x,y) are (1,41),(2,39) -----

40 integer point plotted on the diagnol of this square

The integer point plotted in a square on this x and y axis less 40 integer point plotted on the diagnol of this square are the integral points in the square to be divided by 2 will give integer point plotted on the triangle in the question

1600-40=1560

1560/2=780=Answer

THE INTEGER POINTS PLOTTED IN H

On Fri, 17 Jul 2009 18:11:28 +0530 wrote

>The integer points which can be plotted are (1,1),(1,2), (1,3).... ......... ....(1,39) ,(2,1),(2,2) ,(2,3)... ......... ......(2, 38)...... ........last would be (39,1)

>The total sum is equal to 39+38+...... .1

>39*40/2=780

>

>

>--- On Mon, 15/6/09, Haneef Mohammad wrote:

>

>From: Haneef Mohammad

>Subject: Re: [2IIM CAT Prep] doubt

>To: ascent4cat@yahoogro ups.com

>Date: Monday, 15 June, 2009, 11:45 AM

>

>

>

>

>

>

>

>

>

>

>

>

> Half of points in square formed by (0,0) (41,0) (0,41) (41,41)

>

>excluding the points on the boundary, 40*40 / 2

>

>Answer: 800

>

>

>

>On Mon, May 11, 2009 at 10:55 PM, satbir singh r wasu wrote:

>

>

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>consider a triangle drawn on the x-y

>plane with its three vertices ),(0,41of (41,0) and (0,0) each vertex being

>represented by its (x,y) coordinates. The number of points with integer

>coordinates inside the triangle (excluding all the points on the boundary)is

>

>

>1)780

>

>2)800

>

>3)820

>

>4)741

>

>

>

>

>

>

>please explain with brief explanation.

>

>

>

>Regards,

>

>Satbir singh.

>

>

>

>

>

>

>

> Cricket on your mind? Visit the ultimate cricket website. Enter now!

>

>

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> Love Cricket? Check out live scores, photos, video highlights and more. Click here http://cricket. yahoo.com

- 3a..
## Re: Numbers ques

### Posted by: "ashok_rrsa" ashok_rrsa@... ashok_rrsa

#### Mon Aug 24, 2009 3:04 am (PDT)

--- In ascent4cat@yahoogro ups.com, gagan kumar <gagan_ks@.. .> wrote:

>

> How to get sum of all multiple of 7 between 1 to 1000. Any formula??

>

> ~Gagan

>

>

> Yahoo! recommends that you upgrade to the new and safer Internet Explorer 8. http://downloads. yahoo.com/ in/internetexplo rer/

>

SOLUTION-

as u know minimum no. of multiple is 1 and maximum multiple is 1000/7=142 then 7(1+2+3+4+.. .....142) = 142/2[2*1+(142- 1)1]=10153, answer

- 3b.
## Re: Numbers ques

### Posted by: "sunil singla" sunil_singla2004@... sunil_singla2004

#### Mon Aug 24, 2009 3:04 am (PDT)

multiples of seven

are 7,14,21,.... ......... 994

sum is 7*(1+2+3+4+5+ ......... ..+142)

= 7*142*143/2

= 71071

--- On Mon, 7/20/09, gagan kumar <gagan_ks@yahoo. co.in> wrote:

From: gagan kumar <gagan_ks@yahoo. co.in>

Subject: [2IIM CAT Prep] Numbers ques

To: ascent4cat@yahoogro ups.com

Date: Monday, July 20, 2009, 9:46 AM

How to get sum of all multiple of 7 between 1 to 1000. Any formula??

~Gagan

See the Web's breaking stories, chosen by people like you. Check out Yahoo! Buzz.

- 4a.
## Re: (unknown)

### Posted by: "Rahil Sahu" rahilsahu@... rahilsahu

#### Mon Aug 24, 2009 3:04 am (PDT)

Hi kapil,

a+b+c+d do 1/20th of the work in a day.

1/4th work takes 5 days.

After A left, additional 6 days required. Hence b+c+d take (15+6) = 21 days for 3/4th work

b+c+d do 3/4*1/21 of the work in a day ==> b+c+d will complete the whole of work in 28 days

1/2 work now takes 5 + 7 = 12 days

B left,additional 4 days required. Hence c+d take (14 + 4) = 18 days for 1/2 work

c+d do 1/2*1/18 of the work in a day ==> c+d will complete the whole work in 36 days.

3/4 work now takes 5+7+9 = 21days

C left. addtional 2 days required. Hence d takes 9+2 = 11days to do 1/4th work

d will take 1 / (1/4 *1/11) = 44 days to complete whole work.

1/c + 1/d = 1/36

1/c + 1/44 = 1/36

c= 198 days

Rahil Sahu

Naval Architect

VEDAM Design & Techinal Consultancy Pvt. Ltd.

Ph - 91-22-6670 9192

www.vedamindia. com

____________ _________ _________ __

From: Kapil Bhatia <kapi_007@ymail. com>

Sent: Sunday, 19 July, 2009 1:31:50 PM

Subject: [2IIM CAT Prep] (unknown)

hi all.....

A+B+C+D cn together do a peice of work in 20days.

after 1/4th of work is complete A left & 6 more days r needed to complete job.

after 1/2th ,, ,, ,,, ,, B,, ,, 4 ,, ,, ,, ,, ,,, ,, ,,

after 3/4th ,, ,,, ,, C ,,,, 2 ,, ,, ,, ,,

In how many days'll C finish the job alone??????

Good Day

Kapil Bhatia

____________ _________ _________ __

See the Web's breaking stories, chosen by people like you. Check out Yahoo! Buzz.

Love Cricket? Check out live scores, photos, video highlights and more. Click here http://cricket. yahoo.com

- 5a.
## Re: Quant Numbers ques

### Posted by: "Rahul Jain" championishere007@... championishere007

#### Mon Aug 24, 2009 3:06 am (PDT)

If the sum of the digits is equal to 9, the number is a multiple of 9. Now

find the first multiple of 9 that comes after 1000. It would be 1008. Now

count the last one ending before 10000 and that has to be 9999.

The respective quotients are 112 and 1111. Hence, the total numbers having

sum of digits as 9 is (1111-112+1) = 1000. The method by Narayana Reddy can

be flawed at times wen the number just starts with a multiple of 9 and ends

with one, but the above method is full proof. :)

Cheers:)

Rahul

On Thu, Jul 16, 2009 at 7:57 PM, Narayana Reddy <email2reddy@ gmail.com>wrote:

>

>

> Ans = 1000

> Solution:

> (10000 - 1000) / 9

>

> On Sun, Jun 14, 2009 at 11:58 AM, vaibhav

> varshney<vaibhav_3758@ yahoo.co. in <vaibhav_3758% 40yahoo.co. in>> wrote:

> >

> >

> > How many integers between 1000 and 10000 have the sum of their digits

> equal

> > to 9?

> > a) 112

> > b) 132

> > c) 126

> > d) 165

> > Tell me the shortest method to solve it...

> >

> > vaibhav

> > ____________ _________ _________ __

> > Explore and discover exciting holidays and getaways with Yahoo! India

> Travel

> > Click here!

> >

>

>

--

------------ --------- ---------

RAHUL JAIN

rahul_jain@daiict. ac.in

Ph.no.: 9328067371

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