CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

implex

Quote:

Originally Posted by srikar2097

Thanks Impex. But yaar, still not clear. I had seen the equality sign being added, but how and why exactly does addition of 1 leads to this? Can you elaborate on this point?

let me explain once more
suppose I say there are 10 soldiers in an army camp! Now there is en enemy army camp, which has more soldiers than the first army camp.

What can we say now! that there must be at least (10+1) soldiers in the enemy army!!

hope that helps

srikar2097

Quote:

Originally Posted by implex

let me explain once more
suppose I say there are 10 soldiers in an army camp! Now there is en enemy army camp, which has more soldiers than the first army camp.

What can we say now! that there must be at least (10+1) soldiers in the enemy army!!

hope that helps

Oh man. How could I have not understood this! Thanks anyway.

gir

@ Aarav,

Can you help with the gcd concept I am not clear with it.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)
Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, b) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1]. [if x divides q then x divides 2q => it can not divide 2q+1]

Aarav

Quote:

Originally Posted by gir

@ Aarav,

Can you help with the gcd concept I am not clear with it.

Now ps = q^2 => gcd(p, s) = 1 (the explanation is below)
Because if x divides gcd (p, s) and x is prime (or it will be product of two or more primes, but we assume the base case which covers other case as well), then x would divide q [because p = ax, q = bx => ps = abx^2 = q^2 and gcd(a, b) = 1) and thus x dividing 2q+1 [= p+s = x(a+b)] is a contradiction => each of p and s is a perfect square [gcd(p, s) = 1]. [if x divides q then x divides 2q => it can not divide 2q+1]

I suggest you to take some examples and understand this. If still unclear, please write in steps what is written above and ask me which step i not clear.

ps = q^2 has been proved by us; we are now showing that both p and s don't have any common factor i.e. e.g. if ps = 36 then p is 9 and q = 4 only and not something else.

rafatsxc2

is de correct answer to ds question mark3

rafatsxc2

x)square ends in 1,4,9,7.hence here s=81 i.e. p=64
y)50<p the only 2 values above 50 which are square 64 and 81 hence p=64

siddooba

Kudos for a great solution to yesterday's problem !!

Could someone post today's question?

Aarav

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Quantitative Question # 020
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For all integers x, y, f(x, y) is defined as f(x+2, y+1) = f(f(x+1, y), f(x, y)) and f(x+1, 0) = f(x, 1), then f(f(2, 3), f(2, 2)) =

(1) f(4, 5) (2) f(3, 3) (3) f(3, 4) (4) f(4, 3) (5) none of these

srikar2097

Has anyone got today's question? Can you please post it...

naga25french

aarav , i did not get quant question today.. can u help me out? pls send it as soon as possible..

thanx in advance..