Official Quant thread for CAT 2013

MBA CAT 2024
24128
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
this occurs only when x=y for all positive integers.
24129
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
at x=y...............
infinite solution ho jayega.........
24130
@iLoveTorres said:
@scrabbler bhai mujhe bhi thode graph wale classes de do so that i am able to save time while solving such questions

See attached

@Logrhythm @ChirpiBird

regards
scrabbler
24131
@saurabhlumarrai said:
this occurs only when x=y for all positive integers.
That's a trivial solution. I want unique.. :)
24132
@saurabhlumarrai said:
this occurs only when x=y for all positive integers.
And at 2^4 = 4^2

regards
scrabbler
24133
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
x=y
and
x=4 and y=2

??
24134
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
2^4 = 4^2 and x = y ?
24135
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
Distinct +ve integers you mean? (2,4) or (4,2)

regards
scrabbler
24136
@scrabbler said:
See attached@Logrhythm@ChirpiBirdregardsscrabbler
24137
@scrabbler said:
Distinct +ve integers you mean? (2,4) or (4,2)regardsscrabbler
Nai...only 1 solution ..we have to consider only non-trivial unordered solution
24138
Given that a, b, and c are positive integers, solve the following equation
a!*b! = a! + b! + c^2
24139
@ScareCrow28 said:
Given that x and y are positive integers, solve x^y = y^x
x = y is an obvious answer

x^y = y^x
x^(1/x) = y^(1/y)

We know that x^(1/x) is an increasing function from x = 0 to e and a decreasing function from x = e to inf

and limit x tends to inf x^(1/x) = 1

So, for all 1 ‰¤ x

Only integer between 1 and e is 2

and 2^(1/2) = 4^(1/4)

So, apart from x = y, (2, 4) and (4, 2) are the only integral solutions
24140
@ScareCrow28 said:
Given that a, b, and c are positive integers, solve the following equation a!*b! = a! + b! + c^2
2!*3! = 2!+3!+2^2
24141
@ScareCrow28 said:
Given that a, b, and c are positive integers, solve the following equation a!*b! = a! + b! + c^2
2 x 6 = 2 + 6 + 4
So a = 2, b = 3, c = 2?

regards
scrabbler
24142
@chillfactor said:
x = y is an obvious answerx^y = y^xx^(1/x) = y^(1/y)We know that x^(1/x) is an increasing function from x = 0 to e and a decreasing function from x = e to inf and limit x tends to inf x^(1/x) = 1So, for all 1 ‰¤ x Only integer between 1 and e is 2and 2^(1/2) = 4^(1/4)So, apart from x = y, (2, 4) and (4, 2) are the only integral solutions
The GOD has arrived! Sir, next wala b dekho please!
24143
@ScareCrow28 said:
Nai...only 1 solution ..we have to consider only non-trivial unordered solution
Why? x and y poocha hai to automatically ordered pair...

regards
scrabbler
24144
@scrabbler said:
Why? x and y poocha hai to automatically ordered pair...regardsscrabbler
You are right.

What if the numbers were rational??
24145
@ScareCrow28 said:
You are right. What if the numbers were rational??
I would have given up 😃 But Chillfactor's solution covers it all...

regards
scrabbler
24146
Given that n is a natural number, when is n^4 + 4 prime?
24147
@ScareCrow28 said:
Given that n is a natural number, when is n^4 + 4 prime?
1^4 = 1
2^4 = x6
3^4 = x1
4^4 = x6
5^4 = x5
6^4 = x6
7^4 = x1
8^4 = x6
9^4 = x1
0^4 = x0

only 1 solution at n=1

i tried n=5, but did not get a prime...so aage try nahi kara for 5k form of numbers...

so my money is on n=1 :p