Easy | General |

Without explicitly calculating, find out which is bigger: e^Pi or Pi^e?

Answer

e^Pi

Solution

This is a common placement test question. Since e^x = 1 + x…., we have e^x>1+x. Now, conveneintly choosing x = (Pi / e -1) and solving, we get e^Pi > Pi^e.

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Another way is to observe for f(x) = x^(1/x), differentiating: f'(x) = f(x)*(1/x^2)*(1- ln x); f is decreasing for x > e so pi^(1/pi) < e^(1/e); hence e^(pi)> pi^(e)

Another approach is to assume e^Pi > Pi^e <=> Pi*lne > e*ln Pi <=> lne/e > lnPi/Pi. Now, notice that f(x) = lnx/x is decreasing function since f'(x) is negative in range (e<x<pi), hence above assumption is true!

Easy | General |

100 pirates stand in a circle. They start shooting alternately in a cycle such that 1st pirate shoots 2nd, 3rd shoots 4th and so on. They continue in circle till only one pirate is left. Which position should Captain Jack Sparrow stand to survive?

Solution

This is not a puzzle at all, this is apti/maths question, but has become quite common. You can calculate for a N=100, 73rd position is safe. For a general N, find largest m such that 2^m < N. Take t = N-2^m. Then Safe position is 2*t + 1. Please learn this formula. No need to prove it using strong induction. Proof: http://en.wikipedia.org/wiki/Josephus_problem#k.3D2

Source: Written Tests

Enable Like and Comment Easy | General |

In a dark room, there is a deck of 52 cards, with exactly 10 cards facing up, rest facing down. You need to split this into two decks with equal number of cards facing up!

Hint

The question can be solved even when up facing cards are 11. The split decks need not have equal number of up-facing cards.

Solution

Create a deck of k cards randomly from the original 52-card deck, (k=10 here) and then turn over the k-card deck. Goal is achieved!

Source: Quant Interview

Enable Like and Comment Easy | General |

You are in a game against devil, on a perfectly round table and with an infinite pile of pennies. He says, "OK, we'll take turns putting one penny down, no overlapping allowed, and the pennies must rest flat on the table surface. The first guy who can't put a penny down loses." You can go first. How will you guarantee victory?

Hint

Try the problem if coins have unit radii, and table has radius 1 & 3.

Solution

You place a penny right in the center of the table. After that, whenever the devil places a penny on the table, mimic his placement by placing a penny diametrically opposite and at same distance from center. If he has a place to place a penny, so will you. The devil will run out of places to put a quarter before you do.

Medium | General |

I guessed 3 natural numbers - x,y,z. You can ask me 2 sums of these numbers with any integer coefficients - (a,b,c). That is, you give me a, b and c and I tell you the result of the expression a*x+b*y+c*z. Seeing the answer, you then give me the 2nd triplet of (a,b,c) & I will tell a*x+b*y+c*z. Give me the algorithm to find x,y and z.

Hint

If digits are small, we can solve any number of variables by asking a=1, b=10^100, c=10^200 etc just by reading these numbers between the zeros of result.

Solution

Since they are natural numbers, if you knew the maximum number of digits any of them can have, say d, you could set a=1, b=10^d, c=10^2d, and you would be able to read the d-digit numbers directly. So, you use the first calculation to find the maximum number of digits, (a,b,c)=(1,1,1). let d = digits of this result (x+y+z)

Then, set (a,b,c) = (1, 10^d, 10^2d) Let the sum be S.

Then x = (first d digits of S), y = [d+1] to 2d-digits of S, z = [2d+1 to 3d] digits of S

Thus, we note that its posssible to solve for n natural numbers x_1,x_2,...x_n with just 2 questions.

Then, set (a,b,c) = (1, 10^d, 10^2d) Let the sum be S.

Then x = (first d digits of S), y = [d+1] to 2d-digits of S, z = [2d+1 to 3d] digits of S

Thus, we note that its posssible to solve for n natural numbers x_1,x_2,...x_n with just 2 questions.

Source: Quantnet Forums

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