Spaceship Enterprise
Spaceship Enterprise starts at 3 am from planet Offensive to target planet Defensive.
It travels for one hour and then reduces its speed to 2/3rd for the next hour. The total distance at the end of the first hour is a four-digit square and at the end of the second hour a four-digit cube having no three digits identical. Defensive is informed by Informative about all this and it fires a missile Destroyer at 4 am itself. If the separation between the two planets is twice the initial speed of Speedster, at what speed must Destroyer travel to destroy Enterprise midway at 5 am?
* Speedster is the Enterprise
It travels for one hour and then reduces its speed to 2/3rd for the next hour. The total distance at the end of the first hour is a four-digit square and at the end of the second hour a four-digit cube having no three digits identical. Defensive is informed by Informative about all this and it fires a missile Destroyer at 4 am itself. If the separation between the two planets is twice the initial speed of Speedster, at what speed must Destroyer travel to destroy Enterprise midway at 5 am?
* Speedster is the Enterprise
Labels: mathemagic, mathschallenge
posted by Rajesh Lal at
11:58 AM
10 Comments:
mmmmmmm.....was following okay until I got to the Speedster comment ...
I don't get the Speedster bit either, but maybe it doesn't matter . . .
I can work out from the squares and cubes that Enterprise has covered 2025 after 1 hour and 3375 after 2 hours, i.e. at 5 am. If this is then midway between the two planets and Destroyer, leaving at 4 am, strikes at this time, then it must have travelled at 3375 per hour.
So, what does Speedster have to do with it? And how can a separation (distance) equal a speed?
Now that it's been clarified that Enterprise and Speedster are one and the same we seem to have a contradiction.
We are told that Destroyer whacks Enterprise "midway". I assume that this is midway between the two planets (can't see what else it could mean) and that this is at 5 am, i.e. 2 hours after departure. We are also told that the "separation" between the two planets is twice the initial speed of the Enterprise. I take this to be 3375. So Enterprise reaches "midway" after one hour, not two, i.e. at 4 am yet is destroyed at 5 am.
Does relativity come into this? I'm confused.
Confused? ... oh yeah!
The space vehicle has detected an error in her logic circuitry ... causing an unfortunate identity crisis problem.
Spock unsuccessfully attempted a mind-meld with the AI logic module, hoping to locate and re-write the erroneous code ... he accidentally merged with the recreational game file. Now he wanders the ship muttering "trick of mind ... trick of mind ... trick of m ...
Being unable to resolve the identity crisis, whether she is "Enterprize" or "Speedster", resulted in a program override, with a single option ... return to port.
Since she has reversed direction, as is therfore no longer at risk from Destroyer, the delimma of possible destruction has been avoided.
However, another problem of seemingly lesser importance has surfaced. When Spock performed the mind-meld, he accidentally moved the recreational game file into a path in the computer's logic which cannot be accessed. Now ... all computer displays are showing the Trick of Mind home screen. Hopefully, the program override, which caused the vessel to head home has not been compromised.
Beam me up, Scotty - I'm outta here!
Spaceship Enterprise
Let x be the intital speed of Enterprise
Distance travelled by 4 AM is d(4AM)=x and by 5 AM is d(5AM)=x+2/3*x=5/3*x
x = 4 digit square
let y^2=x
Find the range of y
sqrt(10,000)=100
sqrt(1000)~=31.6.
y is greater than 31 and less than 100
Let z^3=d(5AM)=5/3*x=5/3*y^2
Find range of z
10,000^1/3~=21.5
1000^1/3=10
z is greater than 10 and less than 21
since z^3=5/3*y^2 OR 3/5*z^3=y^2 and both are 4 digit numbers, z^3 must be divisible by 5 (which means z must be divisible by 5) and y^2 must be divisible by 3 (which means y is divisble by 3).
this leaves z=15, or z=20. Since 20^3=8000 has 3 identical digits it can not be used, thus z must be 15
15^3= 3375
3375*3/5=2025=y^2=x
y=45
Distance between planets = 2x
Assumption: There is a solution ! Term midway used incorrectly, and we need to intepret the meaning as part way. Can't see how we can use Lorentz contraction (special relativity) to resolve i.e. the distance could become shorter but the fraction travelled by Enterprise would still be greater than half.
d(5AM)=5/3*x, so seperation between Defensive and Enterprise is 2x-5/3x=1/3*x
1/3*x=1/3*2025= 675
ANSWER:
675
Cam
nice work Anonymous .....
Beam me down again, Scotty. There's intelligent life here after all.
Cam, I think that your assumption (i.e. midway not = halfway) leads to the only way of arriving at a sensible answer.
I want to quote your post in my blog. It can?
And you et an account on Twitter?
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