ascent4cat : Messages : 3955-3984 of 4167

1. 7^45 = 7(7^44) = 7(2401)^11 = 7(01)^11 = 7(01) = 07.
2. 3^90 + 5^90 = 9^45 + 25^45 is always divisible by (9 + 25) = 34. So remainder is zero.
3. 4 values - 1, 5, 7, 11.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Mon, November 8, 2010 10:57:25 PM
Subject: [2IIM CAT Prep] CAT Number Theory - Questions on Remainders

Hi all,

Here are a few more questions on Number Theory based on remainders. They have also been posted at ou blog - http://iimcat.blogspot.com/2010/11/questions-on-number-theory-remainder.html.

1. What are the last two digits of the number 7 ⁴⁵?

2. What is the remainder when we divide 3⁹⁰ + 5⁹⁰ by 34?

3. N²leaves a remainder of 1 when divided by 24. What are the possible remainders we can get if we divide N by 12?

Cheers all,
Rajesh
99626 48484
New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3956

From: srinivas reddy <srinivasmettureddy@...>
Date: Wed Nov 10, 2010 6:01 am
Subject: Re: [2IIM CAT Prep] mixture alligattion

srinivasmett...

the ratio in both the buckets will progressively get closer to 50-50

#3957

From: kamal lohia <kamallohia@...>
Date: Wed Nov 10, 2010 7:25 am
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM HCF

kamallohia

1. Phi(31) = 30.
2. 8 + 4 + 4 = 16.
3. 1050/2 = 525.
4. HCF(a, b) cannot be greater than a and b.

Please confirm the answers. I am sure that there can be some silly error which i a reluctant to recheck.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, November 10, 2010 9:24:37 AM
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM HCF

Hi all,

Given below are a few questions on LCM and HCF. As ever, they are also available on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf.html

1. How many pairs of positive integers x,y exist such that HCF of, x,y = 35 and sum of x and y = 1085?

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

4. Sum of two even numbers a and b = 240. If HCF (a,b) is greater than a and b, what is the minimum value of HCF (a,b)?

Happy cracking.

Cheers,

Rajesh

99626 48484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3958

From: sambit pradhan <samba_east@...>
Date: Wed Nov 10, 2010 10:31 am
Subject: number Series..

samba_east

Hii...

find the next no....

1,2,4,8,16,31,57,___

With Regards,

Sambit Pradhan

#3959

From: Uma Maheswaran <umamaheswaran.b@...>
Date: Wed Nov 10, 2010 12:16 pm
Subject: Re: [2IIM CAT Prep] number Series..

umamaheswaran_b

answer is 99!!

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

�

Hii...

�

find the next no....

�

1,2,4,8,16,31,57,___

�

�
With Regards,

Sambit Pradhan
�

�

Thank You.

Reards,

Uma Maheswaran B

#3960

From: divakar narayan <divakar_narayan@...>
Date: Wed Nov 10, 2010 2:29 pm
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM HCF

Ans1: If HCF is 35 then the numbers should be in the form 35k, where k is a prime number. Since it is given x+y=1085, we can write 35(a+b)=1085

or a+b = 31. Now only one pair of prime number satisfy this condition(2,29)..hence answere is one pair.(though my gut feeling tells me i am overlooking something so please correct me if i am wrong)

ANS2:- 16 pairs

Ans3 :- 525

Ans4:...I dont know whether it was to be solved or not but HCF cannot be greater than the numbers..please check and send us the question back.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Wed, 10/11/10, Rajesh Balasubramanian <rajesh@...> wrote:

From: Rajesh Balasubramanian <rajesh@...>
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM HCF
To: ascent4cat@yahoogroups.com
Date: Wednesday, 10 November, 2010, 9:24 AM

Hi all,

Given below are a few questions on LCM and HCF. As ever, they are also available on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf.html

1. How many pairs of positive integers x,y exist such that HCF of, x,y = 35 and sum of x and y = 1085?

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

4. Sum of two even numbers a and b = 240. If HCF (a,b) is greater than a and b, what is the minimum value of HCF (a,b)?

Happy cracking.

Cheers,

Rajesh

99626 48484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3961

From: nitin kaushik <nitinkaushik32@...>
Date: Wed Nov 10, 2010 12:02 pm
Subject: Re: [2IIM CAT Prep] number Series..

nitin_be_32

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,�

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

�

Hii...

�

find the next no....

�

1,2,4,8,16,31,57,___

�

�
With Regards,

Sambit Pradhan
�

�

#3962

From: Rajesh Balasubramanian <rajesh@...>
Date: Wed Nov 10, 2010 4:33 pm
Subject: Solutions to questions on Number Theory - LCM and HCF

ascentcatprep

Hi all,

�

Have given below solutions to questions on LCM and HCF. Apologies for the error in the 4th question that had been posted earlier (the question has now been removed from the blog). Had been thinking about this question for a while, and in my haste typed something wrong. Many thanks to the guys on yahoogroups who have pointed this out.

The solutions are as follows

1. How many pairs of positive integers x,y exist such that HCF of x,y = 35 and sum of x and y = 1085?

Let HCF of (x,y) be h. Then we can write x = h*a and y = h*b. Furthermore, note that HCF (a,b) = 1. This is a very important property. One that seems obvious when it is mentioned but a property a number of people overlook.

So, we can write x = 35a; y = 35b

x + y = 1085 => 35( a + b) = 1085. => (a+b) = 31. We need to find pairs of coprime integers that add up to 31. (Another way of looking at it is to find out integers less than 31 those are coprime with it or phi(31) as one of the replies had mentioned. More on this wonderful function in another post).

Since 31 is prime. All pairs of integers that add up to 31 will be coprime to each other. Or, there are totally 15 pairs that satisfy this condition.

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

This has become one of my favourite questions. Had not thought much about it when I had written down the question - but came to realise that there are perhaps many more interesting questions that can be created in this genre.

Again let us write x = h * a; y = h * b,

a and b are coprime. So, LCM of (x,y) = h*a*b

So, in essence h + h*a*b = 91. Or h(ab + 1) = 91

Now, 91 can be written as 1*91 or 7*13

Or, we can have HCF as 1, LCM as 90 - There are 4 pairs of numbers like this (2,45), (9,10), (1,90) and (5,18)

We can have HCF as 7, ab +1 as 13 => ab =12 => 1*12 or 4*3

Or, the pairs of numbers are (7,84) or (21,28)

The third option is when HCF = 13, ab+1 = 7 => ab=6

Or (a,b) can be either (1,6) or (2,3)

The pairs possible are (13,78) and (26,39)

There are totally 8 options possible - (2,45), (9,10), (1,90), (5,18), (7,84), (21,28), (13,78) and (26,39)

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

x=525 y =525 works best.

If the question states x,y have to be distinct, then the best solution would be x=350 y =700, HCF = 350

�

Will try to send the correct version for Qn 4 as soon as possible. Have been thinking about it, but am somehow not able to place it.

�

Cheers all,

Rajesh

Reply | Messages in this Topic (3)

#3963

From: divakar narayan <divakar_narayan@...>
Date: Wed Nov 10, 2010 6:04 pm
Subject: Re: [2IIM CAT Prep] Solutions to questions on Number Theory - LCM and HCF

Hello Rajesh Balasubramanian

I wanted to know about this phi function and more about how and when to use it nuber system questions...If you can tell me a little it would be of great help.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Wed, 10/11/10, Rajesh Balasubramanian <rajesh@...> wrote:

From: Rajesh Balasubramanian <rajesh@...>
Subject: [2IIM CAT Prep] Solutions to questions on Number Theory - LCM and HCF
To: ascent4cat@yahoogroups.com
Date: Wednesday, 10 November, 2010, 10:03 PM

Hi all,

Have given below solutions to questions on LCM and HCF. Apologies for the error in the 4th question that had been posted earlier (the question has now been removed from the blog). Had been thinking about this question for a while, and in my haste typed something wrong. Many thanks to the guys on yahoogroups who have pointed this out.

The solutions are as follows

1. How many pairs of positive integers x,y exist such that HCF of x,y = 35 and sum of x and y = 1085?

Let HCF of (x,y) be h. Then we can write x = h*a and y = h*b. Furthermore, note that HCF (a,b) = 1. This is a very important property. One that seems obvious when it is mentioned but a property a number of people overlook.

So, we can write x = 35a; y = 35b

x + y = 1085 => 35( a + b) = 1085. => (a+b) = 31. We need to find pairs of coprime integers that add up to 31. (Another way of looking at it is to find out integers less than 31 those are coprime with it or phi(31) as one of the replies had mentioned. More on this wonderful function in another post).

Since 31 is prime. All pairs of integers that add up to 31 will be coprime to each other. Or, there are totally 15 pairs that satisfy this condition.

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

This has become one of my favourite questions. Had not thought much about it when I had written down the question - but came to realise that there are perhaps many more interesting questions that can be created in this genre.

Again let us write x = h * a; y = h * b,

a and b are coprime. So, LCM of (x,y) = h*a*b

So, in essence h + h*a*b = 91. Or h(ab + 1) = 91

Now, 91 can be written as 1*91 or 7*13

Or, we can have HCF as 1, LCM as 90 - There are 4 pairs of numbers like this (2,45), (9,10), (1,90) and (5,18)

We can have HCF as 7, ab +1 as 13 => ab =12 => 1*12 or 4*3

Or, the pairs of numbers are (7,84) or (21,28)

The third option is when HCF = 13, ab+1 = 7 => ab=6

Or (a,b) can be either (1,6) or (2,3)

The pairs possible are (13,78) and (26,39)

There are totally 8 options possible - (2,45), (9,10), (1,90), (5,18), (7,84), (21,28), (13,78) and (26,39)

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

x=525 y =525 works best.

If the question states x,y have to be distinct, then the best solution would be x=350 y =700, HCF = 350

Will try to send the correct version for Qn 4 as soon as possible. Have been thinking about it, but am somehow not able to place it.

Cheers all,

Rajesh

Reply | Messages in this Topic (3)

#3964

From: Rajesh Balasubramanian <rajesh@...>
Date: Thu Nov 11, 2010 1:35 am
Subject: Questions on Number Theory - LCM and HCF

ascentcatprep

�

Hi all,

�

Have given below three more questions on HCF and LCM on Number Theory. The questions can also be found on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf_10.html�.

�

1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) = 1080?

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) = n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This cannot be determined").

�

Happy cracking.

�

Cheers,

Rajesh

9962648484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3965

From: divakar narayan <divakar_narayan@...>
Date: Thu Nov 11, 2010 4:36 am
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF

Ans 1: If we split 1080 it comes out to be 2^3X3^3X5. Since the product of a,b and HCF(a,b) has to have a cube(since it can be written as k*xk*yk=k^3*x*y. The value of K can be 2^3 or 3^3 or (2*3)^3....and after making pairs with 5, we can make total 9 pairs. I hope I am correct.

Ans2: I m not getting hold of this one..sorry no answer.but don't worry I have a number...2519..i am almost certain its not the least number...but yeah it fulfills the criteria!!

Ans3: I think a =l*n, b=l*m, c=n*m where l,m and n are prime numbers(or co-prime)...so the LCM will be l^2*m^2*n^2.

Please correct me if I am wrong, I am a little shaky in number system.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Thu, 11/11/10, Rajesh Balasubramanian <rajesh@...> wrote:

From: Rajesh Balasubramanian <rajesh@...>
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF
To: ascent4cat@yahoogroups.com
Date: Thursday, 11 November, 2010, 7:05 AM

Hi all,

Have given below three more questions on HCF and LCM on Number Theory. The questions can also be found on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf_10.html .

1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) = 1080?

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) = n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This cannot be determined").

Happy cracking.

Cheers,

Rajesh

9962648484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3966

From: divakar narayan <divakar_narayan@...>
Date: Thu Nov 11, 2010 4:51 am
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF

Sorry the LCM should be l*m*n instead of what i wrote in my previous mail. Very sorry for the blunder.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Thu, 11/11/10, Rajesh Balasubramanian <rajesh@...> wrote:

From: Rajesh Balasubramanian <rajesh@...>
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF
To: ascent4cat@yahoogroups.com
Date: Thursday, 11 November, 2010, 7:05 AM

Hi all,

Have given below three more questions on HCF and LCM on Number Theory. The questions can also be found on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf_10.html .

1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) = 1080?

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) = n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This cannot be determined").

Happy cracking.

Cheers,

Rajesh

9962648484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3967

From: kamal lohia <kamallohia@...>
Date: Thu Nov 11, 2010 6:11 am
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF

kamallohia

1) We need to find ordered pairs (x, y) such that xy*HCF(x, y) = 1080.
Let x = ha and y = hb where h = HCF(x, y) and HCF(a, b) = 1.
So h^3(ab) = 1080 = (2^3)(3^3)(5).
Three cases: I - h = 1, number of ordered pairs = 8
II - h = 2, number of ordered pairs = 4
III - h = 3, number of ordered pairs = 4
IV - h = 6, number of ordered pairs = 2
Hence total ordered pairs of (x, y) are = 18.

2) LCM(5, 6, 7, 8, 9) - 1 = 2519.

3) Let a = lnx, b = lmy, c = mnz.
Now given that HCF(a, b) = l, that means HCF(nx, my) = 1. Also given that HCF(n, m) = 1, that means HCF(x, y) = 1 and HCF(m, x) = HCF(n, y) = 1.
Similarly it can also be shown that HCF(y, z) = HCF(z, x) = 1 and others also.
So in general it can be written any two of the set {l, m, n, x, y, z} are co-prime.
Now LCM(a, b, c) = LCM(lnx, lmy, mnz) = lmnxyz = abc/lmn.

Please verify the answers/solutions as usual.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, November 11, 2010 7:05:37 AM
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM and HCF

Hi all,

Have given below three more questions on HCF and LCM on Number Theory. The questions can also be found on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf_10.html .

1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) = 1080?

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) = n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This cannot be determined").

Happy cracking.

Cheers,

Rajesh

9962648484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3968

From: Hemal Patel <hemalpatel55@...>
Date: Thu Nov 11, 2010 12:11 pm
Subject: Re: [2IIM CAT Prep] number Series..

hemalpatel55

Hello ..

i want some of the question bank which is latest with anwer please kindly send me

HEMAL M PATEL
Email : info@...
Contact : (1) 97141 65362
(2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

Hii...

find the next no....

1,2,4,8,16,31,57,___

With Regards,

Sambit Pradhan

#3969

From: kamal lohia <kamallohia@...>
Date: Thu Nov 11, 2010 5:52 am
Subject: Re: [2IIM CAT Prep] Solutions to questions on Number Theory - LCM and HCF

kamallohia

Bhaisaab

1) When you say how many pairs of positive integers x, y exist such that......, then there you need to find number of ordered pairs (x, y) satisfying the given conditions. But if you say how many pairs of positive integers exist such that their HCF is .... and sum is ....., then you will find unordered pairs as you have found out in your solution.
So for given question, answer will be number of ordered pairs of (x, y) which is = phi(31) = 30.

2) Again with same reasoning as above, number of ordered pairs = 8 + 4 + 4 = 16.

Please verify.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, November 10, 2010 10:03:03 PM
Subject: [2IIM CAT Prep] Solutions to questions on Number Theory - LCM and HCF

Hi all,

The solutions are as follows

1. How many pairs of positive integers x,y exist such that HCF of x,y = 35 and sum of x and y = 1085?

So, we can write x = 35a; y = 35b

Again let us write x = h * a; y = h * b,

a and b are coprime. So, LCM of (x,y) = h*a*b

So, in essence h + h*a*b = 91. Or h(ab + 1) = 91

Now, 91 can be written as 1*91 or 7*13

Or, we can have HCF as 1, LCM as 90 - There are 4 pairs of numbers like this (2,45), (9,10), (1,90) and (5,18)

We can have HCF as 7, ab +1 as 13 => ab =12 => 1*12 or 4*3

Or, the pairs of numbers are (7,84) or (21,28)

The third option is when HCF = 13, ab+1 = 7 => ab=6

Or (a,b) can be either (1,6) or (2,3)

The pairs possible are (13,78) and (26,39)

There are totally 8 options possible - (2,45), (9,10), (1,90), (5,18), (7,84), (21,28), (13,78) and (26,39)

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

x=525 y =525 works best.

If the question states x,y have to be distinct, then the best solution would be x=350 y =700, HCF = 350

Will try to send the correct version for Qn 4 as soon as possible. Have been thinking about it, but am somehow not able to place it.

Cheers all,

Rajesh

Reply | Messages in this Topic (3)

#3970

From: kamal lohia <kamallohia@...>
Date: Thu Nov 11, 2010 6:28 am
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM HCF

kamallohia

(http://westofsunset.blogspot.com/)

Hi Divakar
a + b = 31 such that HCF(a, b) = 1. Now it is not compulsory that a and b are prime. eg. (a, b) can be (1, 30), (2, 29), (3, 28), .....(30, 1) i.e. total 30 pairs.
Now observe closely that HCF of all these pairs is 1 and their sum is 31.
Hope it is clear.

From: divakar narayan <divakar_narayan@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, November 10, 2010 7:59:52 PM
Subject: Re: [2IIM CAT Prep] Questions on Number Theory - LCM HCF

Ans1: If HCF is 35 then the numbers should be in the form 35k, where k is a prime number. Since it is given x+y=1085, we can write 35(a+b)=1085

or a+b = 31. Now only one pair of prime number satisfy this condition(2,29)..hence answere is one pair.(though my gut feeling tells me i am overlooking something so please correct me if i am wrong)

ANS2:- 16 pairs

Ans3 :- 525

Ans4:...I dont know whether it was to be solved or not but HCF cannot be greater than the numbers..please check and send us the question back.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Wed, 10/11/10, Rajesh Balasubramanian <rajesh@...> wrote:

From: Rajesh Balasubramanian <rajesh@...>
Subject: [2IIM CAT Prep] Questions on Number Theory - LCM HCF
To: ascent4cat@yahoogroups.com
Date: Wednesday, 10 November, 2010, 9:24 AM

Hi all,

Given below are a few questions on LCM and HCF. As ever, they are also available on our blog - http://iimcat.blogspot.com/2010/11/number-theory-questions-lcm-hcf.html

1. How many pairs of positive integers x,y exist such that HCF of, x,y = 35 and sum of x and y = 1085?

2. How many pairs of positive integers x,y exist such that HCF(x,y) + LCM (x,y) = 91?

3. Sum of two numbers x,y = 1050. What is the maximum value of the HCF between x and y?

4. Sum of two even numbers a and b = 240. If HCF (a,b) is greater than a and b, what is the minimum value of HCF (a,b)?

Happy cracking.

Cheers,

Rajesh

99626 48484

New CAT2011 batches starting @ Chennai
Nov 13th @ Velachery, Nov 20th @ Mylapore

#3971

From: Ravjeet Singh <rav_sr@...>
Date: Thu Nov 11, 2010 3:46 pm
Subject: Re: [2IIM CAT Prep] number Series..

rav_sr

can u calculate one ques. 6^83+8^83/49 gives remainder...??? even trying with euler number. bt logic is not buliding up

From: Hemal Patel <hemalpatel55@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 5:41:47 PM
Subject: Re: [2IIM CAT Prep] number Series..

Hello ..

i want some of the question bank which is latest with anwer please kindly send me

HEMAL M PATEL
Email : info@...
Contact : (1) 97141 65362
(2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

Hii...

find the next no....

1,2,4,8,16,31,57,___

With Regards,

Sambit Pradhan

#3972

From: Ravjeet Singh <rav_sr@...>
Date: Thu Nov 11, 2010 5:04 pm
Subject: Re: [2IIM CAT Prep] Number Theory - Factorial

rav_sr

its 3!/4 u r right..its an exception

From: Naveenan Ramachandran <6NaveenanR@...>
To: ascent4cat@yahoogroups.com
Sent: Sat, 6 November, 2010 12:43:53 PM
Subject: Re: [2IIM CAT Prep] Number Theory - Factorial

Hi Hari,

I guess you have counted those cases where n+1 is NOT a prime number.
However, there is ONE exception. There is one case where (n+1) is not
a factor of n! even though (n+1) is NOT a prime number ... It is a
fairly small number ...

On 11/4/10, hari laxman <harilaxmanm@...> wrote:
> 3.n can take 19 values..
> Are they right?
> On Thu, 04 Nov 2010 17:19 EST rajesh@... wrote:
>>3. Given N <31, how many values can n take if (n+1) is a factor of n!?
>>Rajesh B (B.Tech IITM, PGDM IIMB)

--
Regards,
Naveenan
2IIM - Mumbai
Batch of 20 Starts Nov 20th

Office: +91-80970-48484

Reply | Messages in this Topic (7)

#3973

From: "sanjay_852000" <sanjay4c@...>
Date: Thu Nov 11, 2010 3:26 pm
Subject: Re: number Series..

sanjay_852000

answer is 99
--- In ascent4cat@yahoogroups.com, sambit pradhan <samba_east@...> wrote:
>
> Hii...
>
> find the next no....
>
> 1,2,4,8,16,31,57,___
>
>
> �With Regards,
>
>
> Sambit Pradhan
>

#3974

From: Rajesh Balasubramanian <rajesh@...>
Date: Fri Nov 12, 2010 4:03 am
Subject: Re: [2IIM CAT Prep] number Series..

ascentcatprep

I think we can think of this as (7-1)^43 + (7+1) ^43/ 7^2

�

Binomial expansion could give the solution.

�

(a-1)^43 + (a+1)^43 divided by a^2

�

-1 and +1 will get cancelled. Any term higher than a^2 will be a multiple of a^2. We will be left with 43a + 43a or 86a.

�

We need to think of remainder left by 86 * 7 �when divided by 49. = -12 *7 = -84 = +14

�

I think the remainder will be 14.

�

I need to send replies to 2 more mails as well. And also need to post solutions (though I think K Kohia's solutions are more or less spot on). Will do so by end of day today. This qn was interesting so got tempted to have a crack. Will reply to the rest of the messages by end of day.

�

Cheers,

Rajesh

On Thu, Nov 11, 2010 at 9:16 PM, Ravjeet Singh <rav_sr@...> wrote:

�

can u calculate one ques.�� 6^83+8^83/49 gives remainder...???�� even trying with euler number. bt logic is not buliding up

From: Hemal Patel <hemalpatel55@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 5:41:47 PM

Subject: Re: [2IIM CAT Prep] number Series..

�

�
Hello ..

i want some of the question bank which is latest with anwer please kindly send me��

HEMAL� M� PATEL
Email : info@...
Contact : (1) 97141 65362
�� (2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,�

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

�

Hii...

�

find the next no....

�

1,2,4,8,16,31,57,___

�

�
With Regards,

Sambit Pradhan
�

�

#3975

From: kamal lohia <kamallohia@...>
Date: Fri Nov 12, 2010 5:05 am
Subject: Re: [2IIM CAT Prep] number Series..

kamallohia

Sir
It is Kamal Lohia not Kohia. Please do the correction for future reference.

Ravjeet
Besides binomial expansion you can find the remainders of individual numbers and then add them to get your answer.
As 6 and 49 are co-prime we can apply Euler's theorem which says that 6^42 = 1 mod49
Also 6^84 = 1 mod49.
or 6*6^83 = 1 mod49.
or 6x = -48 mod49. [Let 6^83 = x mod49]
or x = -8 mod49.

Similarly 8 and 49 are co-prime. Using Euler's theorem in the same manner, we get
8^84 = 1 mod49.
8*8^83 = 1 mod49.
8y = -48 mod49. [Let 8^83 = y mod49]
or y = -6 mod9.

So the required answer is: 6^83 + 8^83 = (x + y) mod49 = (-8 - 6) mod49 = -14 mod49 = 35 mod49.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Fri, November 12, 2010 9:33:51 AM
Subject: Re: [2IIM CAT Prep] number Series..

I think we can think of this as (7-1)^43 + (7+1) ^43/ 7^2

Binomial expansion could give the solution.

(a-1)^43 + (a+1)^43 divided by a^2

-1 and +1 will get cancelled. Any term higher than a^2 will be a multiple of a^2. We will be left with 43a + 43a or 86a.

We need to think of remainder left by 86 * 7 when divided by 49. = -12 *7 = -84 = +14

I think the remainder will be 14.

Cheers,

Rajesh

On Thu, Nov 11, 2010 at 9:16 PM, Ravjeet Singh <rav_sr@...> wrote:

can u calculate one ques.       6^83+8^83/49 gives remainder...???    even trying with euler number. bt logic is not buliding up

From: Hemal Patel <hemalpatel55@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 5:41:47 PM

Subject: Re: [2IIM CAT Prep] number Series..

Hello ..

i want some of the question bank which is latest with anwer please kindly send me

HEMAL M PATEL
Email : info@...
Contact : (1) 97141 65362
               (2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

Hii...

find the next no....

1,2,4,8,16,31,57,___

With Regards,

Sambit Pradhan

#3976

From: sambit pradhan <samba_east@...>
Date: Fri Nov 12, 2010 7:13 am
Subject: Re: [2IIM CAT Prep] Re: number Series..

samba_east

hii....

how u got it..???

With Regards,

Sambit Pradhan

From: sanjay_852000 <sanjay4c@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 8:56:51 PM
Subject: [2IIM CAT Prep] Re: number Series..

answer is 99
--- In ascent4cat@yahoogroups.com, sambit pradhan <samba_east@...> wrote:
>
> Hii...
>
> find the next no....
>
> 1,2,4,8,16,31,57,___
>
>
> ï¿½With Regards,
>
>
> Sambit Pradhan
>

how

#3977

From: Uma Maheswaran <umamaheswaran.b@...>
Date: Fri Nov 12, 2010 12:17 pm
Subject: Re: [2IIM CAT Prep] number Series..

umamaheswaran_b

Hi�Kamal Lohia

amazing way to arrive at the answer (35 is correct).. never thought abt Euler's thm though...

here is one more way...

two basic rules...

(1)---->(a+b)^2=a^2+b^2+2ab...n (a+b)^3=a^3+3(a^2)b+3a(b^2)+b^3...the power raised will be the co-eff of the minimum power of either a or b...u can verify this in pascal triangle...

(2)---->(a+b)^3-(a-b)^3=2(a^3+3 a b^2)...terms getting cancelled are (+b^3,-b^3) and (+3 (a^2), -3(a^2) b)

now to the soln....

6^83=(7-1)^83

8^83=(7+1)^83

now,wen we add we get...

2 X sum of all the odd power of 7....as the even powers of 7 cancel out each other...just like in (2)...

now take out the lowest power of 7..which is 7^1...the co-eff of this would be 83 (the power that we have raised to)...just like in (1)...

2 X 83 X 7 + all other odd powers of 7..

now 7^3/49--- can be written as 7^2 X 7/49---remainder is 0...7^5/49 --- 7^2 X 7^3/49 �---is also 0...

hence...2 X 83 X7/49 is left...

rem of 83 /49 is 34

2 X 34 X 7/49

rem of 68 /49 is 19..

19 X 7/49...

rem of 133/49 is 35..the answer..

pls comment if i am wrong in some place or my logic is not reliable or sth...

On Fri, Nov 12, 2010 at 10:35 AM, kamal lohia <kamallohia@...> wrote:

�

Sir
It is Kamal Lohia not Kohia. Please do the correction for future reference.

Ravjeet
Besides binomial expansion you can find the remainders of individual numbers and then add them to get your answer.
As 6 and 49 are co-prime we can apply Euler's theorem which says that 6^42 = 1 mod49
Also 6^84 = 1 mod49.
or 6*6^83 = 1 mod49.
or 6x = -48 mod49.�� [Let 6^83 = x mod49]
or x = -8 mod49.

Similarly 8 and 49 are co-prime. Using Euler's theorem in the same manner, we get
8^84 = 1 mod49.
8*8^83 = 1 mod49.
8y = -48 mod49.�� [Let 8^83 = y mod49]
or y = -6 mod9.

So the required answer is: 6^83 + 8^83 = (x + y) mod49 = (-8 - 6) mod49 = -14 mod49 = 35 mod49.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Fri, November 12, 2010 9:33:51 AM

Subject: Re: [2IIM CAT Prep] number Series..

�

I think we can think of this as (7-1)^43 + (7+1) ^43/ 7^2

�

Binomial expansion could give the solution.

�

(a-1)^43 + (a+1)^43 divided by a^2

�

-1 and +1 will get cancelled. Any term higher than a^2 will be a multiple of a^2. We will be left with 43a + 43a or 86a.

�

We need to think of remainder left by 86 * 7 �when divided by 49. = -12 *7 = -84 = +14

�

I think the remainder will be 14.

�

I need to send replies to 2 more mails as well. And also need to post solutions (though I think K Kohia's solutions are more or less spot on). Will do so by end of day today. This qn was interesting so got tempted to have a crack. Will reply to the rest of the messages by end of day.

�

Cheers,

Rajesh

On Thu, Nov 11, 2010 at 9:16 PM, Ravjeet Singh <rav_sr@...> wrote:

�

can u calculate one ques.�� 6^83+8^83/49 gives remainder...???�� even trying with euler number. bt logic is not buliding up

From: Hemal Patel <hemalpatel55@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 5:41:47 PM

Subject: Re: [2IIM CAT Prep] number Series..

�

�
Hello ..

i want some of the question bank which is latest with anwer please kindly send me��

HEMAL� M� PATEL
Email : info@...
Contact : (1) 97141 65362
�� (2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,�

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

�

Hii...

�

find the next no....

�

1,2,4,8,16,31,57,___

�

�
With Regards,

Sambit Pradhan
�

�

Thank You.

Reards,

Uma Maheswaran B

#3978

From: Rajesh Balasubramanian <rajesh@...>
Date: Sat Nov 13, 2010 3:23 am
Subject: Re: [2IIM CAT Prep] number Series..

ascentcatprep

Hi Kamal Lohia (correction noted. Sorry about the error)�,

�

Your solution is spot on. Sorry, I had misread the qn as 6^43 + 8^43.

�

Neat solution. Good use of Euler's phi function as well. Will try to send in a detailed post on Euler's phi function by end of this week. (But to be honest, I do not think CAT will give a question that will require students to know the phi function)

Cheers,

Rajesh

On Fri, Nov 12, 2010 at 10:35 AM, kamal lohia <kamallohia@...> wrote:

Sir
It is Kamal Lohia not Kohia. Please do the correction for future reference.

Ravjeet
Besides binomial expansion you can find the remainders of individual numbers and then add them to get your answer.
As 6 and 49 are co-prime we can apply Euler's theorem which says that 6^42 = 1 mod49
Also 6^84 = 1 mod49.
or 6*6^83 = 1 mod49.
or 6x = -48 mod49.�� [Let 6^83 = x mod49]
or x = -8 mod49.

Similarly 8 and 49 are co-prime. Using Euler's theorem in the same manner, we get
8^84 = 1 mod49.
8*8^83 = 1 mod49.
8y = -48 mod49.�� [Let 8^83 = y mod49]
or y = -6 mod9.

So the required answer is: 6^83 + 8^83 = (x + y) mod49 = (-8 - 6) mod49 = -14 mod49 = 35 mod49.

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Fri, November 12, 2010 9:33:51 AM

Subject: Re: [2IIM CAT Prep] number Series..

�

I think we can think of this as (7-1)^43 + (7+1) ^43/ 7^2

�

Binomial expansion could give the solution.

�

(a-1)^43 + (a+1)^43 divided by a^2

�

-1 and +1 will get cancelled. Any term higher than a^2 will be a multiple of a^2. We will be left with 43a + 43a or 86a.

�

We need to think of remainder left by 86 * 7 �when divided by 49. = -12 *7 = -84 = +14

�

I think the remainder will be 14.

�

I need to send replies to 2 more mails as well. And also need to post solutions (though I think K Kohia's solutions are more or less spot on). Will do so by end of day today. This qn was interesting so got tempted to have a crack. Will reply to the rest of the messages by end of day.

�

Cheers,

Rajesh

On Thu, Nov 11, 2010 at 9:16 PM, Ravjeet Singh <rav_sr@...> wrote:

�

can u calculate one ques.�� 6^83+8^83/49 gives remainder...???�� even trying with euler number. bt logic is not buliding up

From: Hemal Patel <hemalpatel55@...>
To: ascent4cat@yahoogroups.com
Sent: Thu, 11 November, 2010 5:41:47 PM

Subject: Re: [2IIM CAT Prep] number Series..

�

�
Hello ..

i want some of the question bank which is latest with anwer please kindly send me��

HEMAL� M� PATEL
Email : info@...
Contact : (1) 97141 65362
�� (2) 95106 89503

From: nitin kaushik <nitinkaushik32@...>
To: ascent4cat@yahoogroups.com
Sent: Wed, 10 November, 2010 4:02:41 AM
Subject: Re: [2IIM CAT Prep] number Series..

Dear Sambit,

The next digit is 99.

You just start finding the difference between digits. Finally u will find a series 1,2,3,4,5. Then do reverse. u well get to the answer:

2-1=1,4-2=2,8-4=4,16-8=8,31-16=15,57-31=26

Next Series is 1,2,4,8,15,26

2-1=1,4-2=2,8-4=4,15-8=7,26-15=11.

Next Series is 1,2,4,7,11

2-1=1,4-2=2,7-4=3,11-7=4

Final lowest Series is 1,2,3,4. Now calculate reverse.

Next series no of lowest =5,�

Add 5+11=16, Add 16+26=42, Add 42+57=99.

Hence next digit is 99.

I hope this resolve your doubt.

Regards

Nitin

On Wed, Nov 10, 2010 at 4:01 PM, sambit pradhan <samba_east@...> wrote:

�

Hii...

�

find the next no....

�

1,2,4,8,16,31,57,___

�

�
With Regards,

Sambit Pradhan
�

�

#3979

From: Rajesh Balasubramanian <rajesh@...>
Date: Sat Nov 13, 2010 3:20 am
Subject: Solutions to Number Theory Questions - LCM HCF

ascentcatprep

Hi all,

�

Have given below the solutions to the LCM HCF questions discussed in the previous post. The solutions can also be found at our blog http://iimcat.blogspot.com/2010/11/solutions-to-number-theory-questions_12.html�. As I had mentioned in the previous post, most of the solution had already been sent in by some of the contributors. From now on, I will also be more careful to make the distinction between ordered pairs and unordered pairs.

�

1. How many pairs of integers (x,y) exist such that the product of x, y and HCF (x,y) = 1080?

We need to find ordered pairs (x, y) such that xy*HCF(x, y) = 1080.

Let x = ha and y = hb where h = HCF(x, y) => HCF(a, b) = 1.

So h^3(ab) = 1080 = (2^3)(3^3)(5).

We need to write 1080 as a product of a perfect cube and another number.

Four cases:

I h = 1, ab = 1080 and b are co-prime. We gave 4 pairs of 8 ordered pairs (1,1080), (8, 135), (27,40) and (5,216) (Essentially we are finding co-prime a,b such that a*b = 1080)

II h = 2, We need to find number of ways of writing (3^3) * (5) as a product of two coprime numbers. This can be done in two ways - 1 and (3^3) * (5) , (3^3) and (5)

number of pairs = 2, number of ordered pairs = 4

III - h = 3, number of pairs = 2, number of ordered pairs = 4

IV - h = 6, number of pairs = 1, number of ordered pairs = 2

Hence total pairs of (x, y) = 9, total number of ordered pairs = 18.

The pairs are (1,1080), (8, 135), (27,40), (5,216), (2,270), (10, 54), (3,120), (24,15) and (6,30),

2. Find the smallest number that leaves a remainder of 4 on division by5, 5 on division by 6, 6 on division by 7, 7 on division by 8 and 8 on division by 9?

LCM(5, 6, 7, 8, 9) - 1 = 2519.

3. There are three numbers a,b, c such that HCF (a,b) = l, HCF(b,c) =m and HCF (c,a) = n. HCF(l,m) = HCF(l,n) = HCF(n,m) =1. Find LCM of a,b,c. (The answer can be "This cannot be determined").

a is a multiple of l and n. Also HCF (l,n) =1; => a has to be a multiple of ln, similarly b has to be a multiple of lm and c has to be a multiple of mn.

We can assume, a = lnx, b = lmy, c = mnz.

Now given that HCF(a, b) = l, that means HCF(nx, my) = 1. This implies HCF(x, y) = 1 and HCF(m, x) = HCF(n, y) = 1.

Similarly it can also be shown that HCF(y, z) = HCF(z, x) = 1 and others also.

So in general it can be written any two of the set {l, m, n, x, y, z} are co-prime.

Now LCM(a, b, c) = LCM(lnx, lmy, mnz) = lmnxyz = abc/lmn.

Quiet obviously, it is a reasonable assumption that a question in CAT will not be as tough as the last one here. However, it is a good question to get an idea of the properties of LCM and HCF.

�

Cheers all,

Rajesh

CAT batches starting @ Chennai

Nov 20th @ Mylapore

Reply | Messages in this Topic (1)

#3980

From: Rajesh Balasubramanian <rajesh@...>
Date: Sat Nov 13, 2010 4:29 am
Subject: Final few questions on Number Theory

ascentcatprep

Hi all,

�

Have given below three more questions on Number Theory. They are also there at our blog. (These are simpler than the ones that we have seen previously. Have toned down the difficulty a notch based on feedback).http://iimcat.blogspot.com/2010/11/cat-number-theory-final-few-questions.html

�

A 4-digit number of the form aabb is a perfect square. What is the value of a-b?
LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?
N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

Happy cracking

�

Rajesh

CAT batches starting @ Chennai, Nov 20th @ Mylapore

#3981

From: kamal lohia <kamallohia@...>
Date: Sat Nov 13, 2010 12:08 pm
Subject: Re: [2IIM CAT Prep] Final few questions on Number Theory

kamallohia

1. The number aabb is 7744 and a - b = 3.
2. 27 ordered triplets. 5 unordered triplets.
3. 4, 15, 26, 37, 48

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Sat, November 13, 2010 9:59:49 AM
Subject: [2IIM CAT Prep] Final few questions on Number Theory

Hi all,

A 4-digit number of the form aabb is a perfect square. What is the value of a-b?
LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?
N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

Happy cracking

Rajesh

CAT batches starting @ Chennai, Nov 20th @ Mylapore

#3982

From: Ravjeet Singh <rav_sr@...>
Date: Sat Nov 13, 2010 7:26 pm
Subject: Re: [2IIM CAT Prep] Final few questions on Number Theory

rav_sr

hi kamal g,

can u plz tell me how the unordered pair u had found out

From: kamal lohia <kamallohia@...>
To: ascent4cat@yahoogroups.com
Sent: Sat, 13 November, 2010 5:38:01 PM
Subject: Re: [2IIM CAT Prep] Final few questions on Number Theory

1. The number aabb is 7744 and a - b = 3.
2. 27 ordered triplets. 5 unordered triplets.
3. 4, 15, 26, 37, 48

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Sat, November 13, 2010 9:59:49 AM
Subject: [2IIM CAT Prep] Final few questions on Number Theory

Hi all,

A 4-digit number of the form aabb is a perfect square. What is the value of a-b?
LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?
N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

Happy cracking

Rajesh

CAT batches starting @ Chennai, Nov 20th @ Mylapore

#3983

From: divakar narayan <divakar_narayan@...>
Date: Sat Nov 13, 2010 4:18 pm
Subject: Re: [2IIM CAT Prep] Final few questions on Number Theory

Lohia Sir, can you give the detailed answers? It would be a great help.

And I wanted to ask one thing regarding the mod question...

You wrote : 6^84 = 1 mod49.
or 6*6^83 = 1 mod49.
or 6x = -48 mod49. [Let 6^83 = x mod49]

I didn't get this last line, from did you write -48mod49...can you explain. I am just asking out of curiosity. I solved the question by expansion method and answer cam to be 35 but your method is really good..I want to learn it.

Opinion is like an AssHole, everybody has one!

Divakar Narayan Singh

--- On Sat, 13/11/10, kamal lohia <kamallohia@...> wrote:

From: kamal lohia <kamallohia@...>
Subject: Re: [2IIM CAT Prep] Final few questions on Number Theory
To: ascent4cat@yahoogroups.com
Date: Saturday, 13 November, 2010, 5:38 PM

1. The number aabb is 7744 and a - b = 3.
2. 27 ordered triplets. 5 unordered triplets.
3. 4, 15, 26, 37, 48

From: Rajesh Balasubramanian <rajesh@...>
To: ascent4cat@yahoogroups.com
Sent: Sat, November 13, 2010 9:59:49 AM
Subject: [2IIM CAT Prep] Final few questions on Number Theory

Hi all,

Have given below three more questions on Number Theory. They are also there at our blog. (These are simpler than the ones that we have seen previously. Have toned down the difficulty a notch based on feedback).http://iimcat.blogspot.com/2010/11/cat-number-theory-final-few-questions.html

A 4-digit number of the form aabb is a perfect square. What is the value of a-b?

LCM of three numbers is equal to 1080; HCF of the three numbers is equal to 2. How many such triplets are possible?

N leaves a remainder of 4 when divided by 33, what are the possible remainders when N is divided by 55?

Happy cracking

Rajesh

CAT batches starting @ Chennai, Nov 20th @ Mylapore

#3984

From: Rajesh Balasubramanian <rajesh@...>
Date: Mon Nov 15, 2010 2:29 am
Subject: Euler's phi function

ascentcatprep