Logic Q

[Amina-DZ]

Originally posted by phylay
n = 52921
thanks to Gauss

OK correct but if you were to find the smallest number that satisfies the conditions, 52921 wouldn't be the one!

[phylay]

Originally posted by Amina-DZ

Originally posted by phylay
n = 52921
thanks to Gauss

OK correct but if you were to find the smallest number that satisfies the conditions, 52921 wouldn't be the one!

Lol, 25201

[Amina-DZ]

Originally posted by phylay
How can the group cross the bridge in 17 minutes?

Come on

Annie + Bob (2 mins)
Annie returns (1 min)
Volodia Mitlin + Dorothy (10 mins)
Bob returns (2 mins)
Annie + Bob (2 mins)

2 + 1 + 10 + 2 + 2 = 17 mins

[phylay]

Originally posted by Amina-DZ

Originally posted by phylay
How can the group cross the bridge in 17 minutes?

Come on

Annie + Bob (2 mins)
Annie returns (1 min)
Volodia Mitlin + Dorothy (10 mins)
Bob returns (2 mins)
Annie + Bob (2 mins)

2 + 1 + 10 + 2 + 2 = 17 mins

hahaha, I must admit I am not that good at finding hard ones

[phylay]

Here are two so called “thought problems” for you. It is said that Albert Einstein was fond of “thought experiments” and used it as a powerful tool (or rather tried to use, I should say) against protagonists of the then new “quantum theory”. We are nowhere near that league but, who knows, which Einstein is hiding among our viewers?

One can use, of course, pen and paper, to remove any cobwebs in the mind.

A man, say, Tom, climbs a mountain through a fixed route on a weekend- Saturday. He starts at 06.00 hrs at the bottom of the mountain and climbs with totally haphazard speed (totally unpredictable varying speeds) and reaches the top of the mountain by noon-12.00hrs. Having done enough mountaineering for a day, he decides to enjoy the scenery and take the remaining day off on the peak.

The next day, exactly at 06.00hrs sharp, he starts to climb down again using the same path, which he had taken the previous day; again the speed of climbing down is also totally haphazard and totally unpredictable. He manages to reach the bottom from where he started a day earlier exactly at 12.00 hrs noon.

Now, given that he climbed both up and down the mountain at totally erratic speeds but with the time of commencement and completion being the same, will there be any point of elevation (or to say in other words, a location in the path) he was present at the same time on both the days?

Obviously there are possible answers:

a) No, can never be; erratic speeds preclude that.

b) May be, one can never be sure.

c) Yes, definitely.

What is the answer you will vote for?

[Amina-DZ]

Differentiation Method

Let a^b mean a "raised to" b

1^2 = 1
2^2 = 2 + 2
3^2 = 3 + 3 + 3

so for any X > 0

X^2 = X + X + X + ..... (X times)

Differentiate both sides with resepct to X

2X = 1 + 1 + 1 + ... (X times)

2X = X

2 = 1 .... Cancelling X from both sides as X <> 0

[Amina-DZ]

Originally posted by phylay
a) No, can never be; erratic speeds preclude that.

b) May be, one can never be sure.

c) Yes, definitely.

What is the answer you will vote for?

c) Yes, definitely.