official quant thread for cat08

[mohitjha]

have anyone used the QA book by Nishit sinha. pearson publishers. i have got Arun Sharma's book with me. just want to add to my prep mat.

[selebratinglife]

Originally Posted by implex

problem 147)
Determine the number of three digits number N divisibly by 11, and N/11 is equal to the sum of squares of the digits of N

Seems like a long one..Anyway, my shot at it..we get two cases

If the number is of the form abc and divisible by 11..b=a+c and abc/11 is of the form ac..So a^2+b^2+c^2 = 10a+c which further reduces to..
10a+c=2(a^2+c^2+ac).So 'c' is even..If we substitute c= 2k..we can further prove that k is a multiple of 2 so 'c' is a multiple of 4...so c can be 0,4,8 only..Checked the values..No solution for the equations...Didn't re-check it though..Got bored..

There can be one more case where a+c=11+b..then abc/11 is of the form
10(a-1)+c....then forming the equation, we can easily prove that c is odd..
so c can take the values 3,5,7,9..Easy to check for 5 and 9..But the quadratic equations seem to have no solution..Whoa...After so much of solving, no solution eh?..Too tedious to go for a re-check..Hope the official solution is much simpler and less time consuming!!

[OperationBLACKI]

Originally Posted by implex

problem 147)
Determine the number of three digits number N divisibly by 11, and N/11 is equal to the sum of squares of the digits of N

Here's how I went about it . Before starting things I knew that the sum of squares should always be less than 90 because the highest 3 digit multiple of 11 is 990 .

Consider the following cases .

902
803 / 814
704 / 715 / 726
605 / 616 / 627 / 660 / 671
506 / 517 / 550 / 561 / 572
407 / 418 / 440 / 451/ 462 / 473
308 / 330 / 341 / 352 / 363 / 374
209 / 220 / 231 /242 / 253 / 264 / 275
110 / 121 / 132 / 143 / 154 / 165 / 176

In all these cases Sum of Squares is less than 90 and only in the bolded cases we have N/11 = Sum of squares . Please let em know if I have missed something or overlooked something.

In all these cases we have

[selebratinglife]

Originally Posted by selebratinglife

Seems like a long one..Anyway, my shot at it..we get two cases

If the number is of the form abc and divisible by 11..b=a+c and abc/11 is of the form ac..So a^2+b^2+c^2 = 10a+c which further reduces to..
10a+c=2(a^2+c^2+ac).So 'c' is even..If we substitute c= 2k..we can further prove that k is a multiple of 2 so 'c' is a multiple of 4...so c can be 0,4,8 only..Checked the values..No solution for the equations...Didn't re-check it though..Got bored..

There can be one more case where a+c=11+b..then abc/11 is of the form
10(a-1)+c....then forming the equation, we can easily prove that c is odd..
so c can take the values 3,5,7,9..Easy to check for 5 and 9..But the quadratic equations seem to have no solution..Whoa...After so much of solving, no solution eh?..Too tedious to go for a re-check..Hope the official solution is much simpler and less time consuming!!

Grave mistakes..c=0 and c=3 does give the solution...courtesy : Above post!!Patience indeed pays off!

[implex]

Originally Posted by OperationBLACKI

Here's how I went about it . Before starting things I knew that the sum of squares should always be less than 90 because the highest 3 digit multiple of 11 is 990 .

Consider the following cases .

902
803 / 814
704 / 715 / 726
605 / 616 / 627 / 660 / 671
506 / 517 / 550 / 561 / 572
407 / 418 / 440 / 451/ 462 / 473
308 / 330 / 341 / 352 / 363 / 374
209 / 220 / 231 /242 / 253 / 264 / 275
110 / 121 / 132 / 143 / 154 / 165 / 176

In all these cases Sum of Squares is less than 90 and only in the bolded cases we have N/11 = Sum of squares . Please let em know if I have missed something or overlooked something.

In all these cases we have

well done but it can be reduced by using the concept of last digits we won't need to find much ! still the method is quite crude, and answer is indeed 550 and 803
so 2

[implex]

Here comes Quant problem set II - Problems 101-150!!

This time I have attached the answer key with it . Hope people enjoyed the first set, as I received too many queries for solutions, I have attached the answer key with this set and will soon put up the solution for the other set.

Have fun!!

Rahul

[catmydream]

hi puys.. i have some material that I can share with you and some (actually a lot) that I need to sell off.. Anyone from Pune , PM me.

[selebratinglife]

Originally Posted by implex

Here comes Quant problem set II - Problems 101-150!!

This time I have attached the answer key with it . Hope people enjoyed the first set, as I received too many queries for solutions, I have attached the answer key with this set and will soon put up the solution for the other set.

Have fun!!

Rahul

Nice job..Loved doing almost all the problems(as some were really tedious), though most of them are way above the CAT standard....Solved the remaining ones..Answer for 150 is not provided..But easy to prove it's (e) none of these..But i feel after the 150Q's were posted individually, you should've posted the answers..

[selebratinglife]

@Implex:I'm assuming that in the 150th questions the choices are given for m+n..But the question says Find(m,n).Please clarify!

[implex]

Originally Posted by selebratinglife

@Implex:I'm assuming that in the 150th questions the choices are given for m+n..But the question says Find(m,n).Please clarify!

we have to find m+n in 150

I have not provided the solution for the obvious reasons, people just read teh solution. IF they have the answer key they will go back and look twice or thrice. As I have used the same question number here and there, they can come and find the answers easily here.