CAT4MBA

Q1. If n is a positive integer then
n(n+1)(n+2)(n+3) + 1 is
i. Always an even number
ii. Always a perfect square
iii. Both (i) and (ii)
iv. Cant be determined.

Ans) put n=1, u'll get 25= 5^2

         put n=2, u'll get 121= 11^2

         put n=3, u'll get 361 = 19^2

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n so on....u'll always get a perfect sq, so the ans is option (ii)

Q2. How many times a fair die must be thrown to get all the possible values at least once?

Ans) I think it depends on is it min or max??

if min no of times, then ans = 6

if max no of times, then ans = infinite

so according to me it cannot be determined !!

Q1 ans is perfect

but i think there must be some better way to prove it.

wat abt Q2??? is dat correct??? m trying other qns as well..vil tell u if i reach to a solutn

let see Q1:

 the product of  4 consecutive +ve integers is divisible by 4!

thus the given number  n(n+1)(n+2)(n+3)+1  is odd. eliminate 1 and 3.

n(n+1)(n+2)(n+3) = n(n+3)(n+1)(n+2)   i just rearranged the terms and it works like magic!

let n²+3n = x  , should be obvious why i did this.

x(x+2) + 1 = x² + 2x + 1 which is definitely a perfrct sqaure.

so Option 2 is definitely correct.

Q2. the problem looks funny but it is not , believe me!

The problem is to find the EXPECTED NUMBER OF TRIALS for the given condition to happen. I am not very good at these questions , but i hope someone is and understands what i mean.

QUESTION 4 is textbook : The big pain being to fing the prime factors of 1200193 which i wont take, who knows the number could be prime and iwould have wasted half  a day.

let me give the formula generally :

if p = a^m  * bⁿ  * c^o  * ..... where a,b,c are prime.

thrn the number of divisors = (m+1)(n+1)(o+1).....

the sum of divisors = [(a^m  - 1)(bⁿ  - 1)(c^o  - 1)......]/[(a-1)(b-1)(c-1).....]

That is it.
I don’t think such higher-level questions are asked in CAT but you never know. I am trying my best to explain it in simple terms.

What is Expected Value

What is the expected value when a dice is thrown ?
we can get 1, 2 , 3, 4, 5 or 6 and the probability of getting each value is 1/6.
So the expected value when the dice is thrown only once
 = 1 x (1/6) + 2 x (1/6) + 3 x (1/6) + 4 x (1/6) + 5 x (1/6) + 6 x (1/6)
= 3.5
Which means every a time a dice is thrown we expect to get a value of 3.5.

In mathematical terms
The expected value of a random variable indicates its average or central value.

Now coming back to the question

How many times a fair die must be thrown to get all the possible values at least once ?

A fair dice has 6 faces and when we throw it for the first time the probability that we ‘ll get a face (any face and lets consider face with value 6) = 1
=> we have to throw the dice only once to get any face

When the dice is thrown for the second time, the probability of a getting a different face (not 6 , any other value say 5) = 5/6
=> we have to throw the dice (1 + 6/5 ) times to get any 2 different faces

. . . . .

Similarly to get 6 different faces we have to throw the dice 1 + 6/5 + 6/4 + 6/3 + 6/2 + 6 = 14.7 times.

For an n-sided die, the number of rolls needed, on average, is = 1 + n/(n-1) + n/(n-2)  =  S(N/I)  for I = 1 to N.

For large n, this is approximately n log n.

i totally agree with nishit. i dunno why everybody is trying so hard at the quant thing. that question would give a tough time to guys studyying for iit jee.

hey buddy the formula to calculate the sum of allt he divisors that u gave is a bit wrong

it should be  a^m+1 rather than a^m in the formula.

like try it for 8, it can be written as 2^3

and there sum is =[ {2^(3+1)} - 1]/ {(2-1)}

regards

dinesh

my GP isn't so strong then :). anyways good you pointed that out .