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The NSA has a Kangaroo Problem

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eric76
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18 Nov 2017, 2:37 pm

The October 2017 Puzzle Periodical from the NSA contains the following problem:

Quote:
Joey, the baby kangaroo, has been kidnapped and placed at 2^100 on a number line.

His mother, Kandice the Kangaroo, is at 0 on the number line, and will try to save him. Kandice normally jumps forward 6 units at a time. Guards have been placed at n^3 on the number line, for every integer n≥1. If Kandice lands on a number with a guard on it, she will be caught and her mission will fail. Otherwise, she will safely sneak past the guard. Whenever she successfully sneaks past a guard, she gets an adrenaline rush that causes her next jump (the first jump after passing the guard) to take her 1 unit farther than it normally would (7 units instead of 6). (After a single 7-unit jump, she resumes jumping 6 units at a time, until the next time she sneaks past a guard.)

Will Kandice the Kangaroo reach (or pass) her son Joey safely?


I think that some people here may enjoy it. It's not terribly hard if you think about it correctly.

Hint: don't try to test each jump individually -- you may die of old age before you finish.



naturalplastic
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22 Nov 2017, 12:57 pm

Even with out the guards it would take Kandace the lifetime of the entire Solar System (or something of that magnitude) to reach Joey anyway!

Two to the 64th is 18 and half million trillion (I know this because when I was a kid I read the story about the inventor of chess supposedly asking for payment for his invention in "one grain for the first square of the chessboard, and two for the next square, and so on...).

Two raised to the 100th power is vastly bigger than 2^64.



kokopelli
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28 Nov 2017, 4:30 am

You could consider any natural number as long it is not a cube. The point for using 2^100 was to make it very difficult to solve by just generating the sequence.



kokopelli
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17 Jan 2025, 6:49 pm

Nobody solved this yet?



Mona Pereth
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27 Jan 2025, 12:18 pm

My solution:

Show Spoiler
To try to solve this, consider the relationship between the sequence of cubes (of natural numbers) and the modulo 6 numbers (remainders of whole number division by 6).

Code:
 1^3 ==   1 == 1 (mod 6)
 2^3 =   8 == 2 (mod 6)
 3^3 =  27 == 3 (mod 6)
 4^3 =  64 == 4 (mod 6)
 5^3 = 125 == 5 (mod 6)
 6^3 = 216 == 0 (mod 6)
 7^3 = 343 == 1 (mod 6)

The pattern that seems to be emerging above is:

If

Code:
n^3 == a (mod 6)

then

Code:
(n+1)^3 == (a+1) (mod 6)

For now I will assume the above pattern holds fr all subsequent cubes. (I might try to do an algebraic proof later.)

Kandice the Kangaroo is initially at n = 0, one less than the first cube. When she jumps forward 6 units, her position mod 6 remains the same. When she jumps forward 7 units, her position mod 6 increments by 1 -- as ALSO does the position of the next cube.

Hence she never lands on cube, hence is never caught.


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