[kapildev]

Hi fellas,
I went through a fair part of the message board, and could not find a section which has topic wise database of Math questions (which is the nemesis for most of us). And the questions I find aretoo tough for me to answer, as have just started preparation.
So I suggest that CAT 2004 takers put in the questions what they do in their instittutes classes, and this builds up like the best of all institutes course material, from basics to CAT level.

I have finished most of Arithmetic in class, and would like to start this thread with chapter 1 - Number Systems. Please give your methods to answer the questions and short-cuts too.

Students of IMS, Time, Career Launcher........please postyour questions too from your classes. I am posting some interesting questionsm, and some doifficult ones.

I am sure that CAT 2003 takers will find these problems easy and can give some really cool methods.

1. Note that 1^2 + 1 + 41, 2^2 + 2 + 41, 3^2 + 3 + 41, 4^2 + 4 + 41........ and so on are all prime. Hence, x^2 + x + 41 (where x is a natural number) is always prime. True or false


2.What should you divide 11979 by, to get a perfect cube ?
1] 3 2] 9 3] 11 4] 11³


3.How many numbers are there between 100 and 200 (both inclusive), and divisible by 2 or 3?
1] 67 2]68 3] 84 4] 100


4. What is the highest power of 5 that divides 75 ! ?
1] 13 2] 15 3] 18 4] 20

5. What is the highest power of 6 that divides 152 ! ?
1] 25 2] 37 3] 50 4] 72

6.What is the last (unit’s place) digit in 7^285?
1] 9 2] 3 3] 1 4] 7

7.State whether the following statements are true or false?

(i) If remainder is 0 when x is divided by y, remainder is 0 when x5 is divided by y.

(ii) If remainder is 1 when x is divided by y, remainder is 1 when x5 is divided by y.

(iii) If remainder is 2 when x is divided by y, remainder is 2 when x5 is divided by y.


8. What is the number of prime numbers between 0 and 100 ?
1] 23 2] 24 3] 25 4] 26


9. If 7x + 3y = 240, which of these statements is true?
1] x and y are both odd. 2] Sum of x and y is even.
3] If x is odd, y is even. 4] None of these.


10.In a system of 6, what is 342 + 131 ?
1] 513 2] 504 3] 603 4] 473

11.The product of two prime numbers is
1] always even 2] always odd 3] could be odd or even

[PUNIT]

the product of 2 prime no.s or any no of prime nos. is even because it has 2 which is the only even prime no. & usually product of n prime no.s has the unit digit 0 because 5 is also prime

[vipingupta]

Hi there Kapil,

I am one of the few who just blundered CAT this year, when I was expecting all six calls throughout the year as per all the mock CATs, on the doomsday, bad luck just overran me . Anyway I have some time to kill , so................
Here are some answers:

Q1: see n^2 + n + 41 is actually n(n+1) + 41, the first term here is even as t is a product of two consecutive numbers. Now I checked with a C++ Program and this number is not a prime for n = 1681 where it becomes divisible by 41. In fact there are infinite such numbers, because if the first term n(n+1) has n or n+1 as a multiple of 41 then we can take out 41 common from the whole term and hence it would prove number is not prime as it would have 41 as a factor. But if you go on checking manually for 1, 2,..... it would be true till 1680.

Q2:This is a very simple problem. Two things, you get the hint from the options regarding the factors, secondly you need to know the divisibility rule for 11, now just check the number by 11 div. rule and you will see that the no. is divisible, go on dividing and you will find 11979 = 11X11X11X9 = 11^3 X 9, so ANswer is 9 because after dividing by 9 we will get 11^3

Q3: Again a simple question on SET theory, I hope you know that :
A Union B = A + B - A Intersection B
when you say numbers div by 2 or 3 you are actually calculating 2 Union 3, so
A = Nos. div by 2 = 51
B = Nos. div by 3 = 33
A Intersection B = Nos. div by 2 and 3 both i.e nos. div by 6 which is equal to 17

Hence Answer is 51 + 33 - 17 = 67

[vipingupta]

Continuing from previous post.....

Q4 and Q5 : In such questions just see the number whose highest power you have to calculate in the given factorial, is prime or not.
First case, lets say it is prime as in Q4 :
Highest Power of 5 in 75! Now the answer is sum of Integral Portions of 75/5,
75/(5^2) , 75/(5^3),................ till the denominator becomes larger than numerator. So in this case answer is 15 + 3 = 18

Now in Q5 the number 6 is not prime
Highest power of 6 in 152!
First find the prime factors of 6 and select the largest of them i.e. 3
Now the Answer is Sum of Integer portions of 152/3, 152/(3^2), 152/(3^3), 152/(3^4), which is 50 + 16 + 5 + 1 = 72


Q6: Again a standard problem, last digits of power of any number repeat for a period of 4, this means that 7^1 ,7^5, 7^9, 7^13................. will have the same last digit so the power of 7 here is 285, lets divide it by 4 and check the remainder it is 1, so it will have the same last digit as 7^1 i.e 7.

Q7: (i) ofcourse if x is div by y then x^5 is div by y because it is simply x multiplied 5 times . its true.
(ii) if remainder is one when x divides y then x = ny + 1 now x^5 is same as (ny+1)^5 Now if you know Binomial Expansion all terms will have y except the last term which is 1, so again this whole expansion is something like ky + 1 and therefore remainder again is 1.

(iii) as in above question we will take x = ny + 2 and calculate (ny +2)^5 and expand binomially, we will get 32 as only term without y, now check the remainder of 32 when divided by y. It may or may not be 2, but whatever it would be that would be the answer.

[sorabh]

hi
the above answer is not correct, the correct answer is it can be even or odd, it depends, now 3 * 5 = 15 which is odd as both the prime no.s are odd, but if u do 2*5=10 it comes as even so as the only prime no. which is even is 2.hence u can get even or odd depending on what u take !
another trivia about prime no.s: a no. if can be represented as 6n+1 or 6n -1 then it can be a prime no. remember this is only a sufficient condition and not a necessary one. i e there will be no.s of 6n+1 or 6n-1 form but they wont be prime.