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Today’s Puzzle
Painted Cube PUZZLE
Twenty-seven identical white cubes are assembled into a single cube, the outside of which is painted black. The cube is then disassembled and the smaller cubes thoroughly shuffled in a bag. A blindfolded man (who cannot feel the paint) reassembles the pieces into a cube. What is the probability that the outside of this cube is completely black?
Today’s Question
Is the number (2^58+1)/5 prime or composite?
Related posts:
September 24th, 2009 at 10:40 pm
Question for the day:
According to cyclicity the answer is it is composite.
September 24th, 2009 at 10:58 pm
Answer of Cube puzzle
= 1/(27* 26C6 * 6^6 * 20C12 * 12^12 * 8^8)
September 24th, 2009 at 11:05 pm
Answer to today’s puzzle-
The (2^58 + 1)/5 will always be a composite number.
It’s simple. 2^58 will always be an even number. Now adding 1 to this number makes it an odd number which, when divided by 5 results in a composite number.
September 24th, 2009 at 11:10 pm
Answer ifor second question is composite
because only in case of 2 ^ (2+4K) this expression is divisible by 5.. and if k= odd.. then result is prime. and if
k= even number then result is composite
In case of 2^58 ; 58= 2 + 4*14; k= 14
And hence result sud be composite..
September 24th, 2009 at 11:12 pm
@shweta Subhedar…
its not that much simple…
September 24th, 2009 at 11:17 pm
ya hitesh explain is correct.
September 25th, 2009 at 12:13 am
Anyone from delhi want to purchase the CD-Rom suggested for english….my suggestion is we can share..btw it seems cheap only…what say?
September 25th, 2009 at 2:04 pm
today’s ques:-
2^58+1/5 is composite
2^58 ends up with 4 after adding 1 it becomes 5 …………
and any having its end digit 5 will surely have 5 as its factor definetly……………
so no. is definetly comppsite………..
September 25th, 2009 at 11:15 pm
Answer of Today’s Question:
Hitesh’s answer is correct
here is the explanation
we factorize 2^58 + 1.
Note that (2^29 + 1)2 = (2^58 + 1) + 2^30, and so 2^58 + 1 can be written as a difference of two squares:
2^58 + 1 = (2^29 + 1)^2 – (2^15)^2 = (2^29 + 2^15 + 1)(2^29 – 2^15 + 1).
Clearly, both factors are greater than 5, and so 2^58 + 1 = 5ab, where a and b are integers greater than 1.
Therefore, the number is composite.
September 25th, 2009 at 11:21 pm
Consider the four types of cubes upon disassembly:
a. 8 cubes with three faces painted black;
b. 12 cubes with two black faces;
c. 6 cubes with one black face;
d. 1 completely white cube.
Each cube of type (a) must be oriented in one of three ways, giving 3^8 possible orientations. Next, each corner cube must be placed in one of eight corners,giving 8! possible arrangements. Thus we have 3^8 · 8! possibilities in all.
Similarly, each cube of type (b) must be oriented in one of two ways, giving 2^12 possible orientations. Then, each edge cube must go to one of 12 edges,giving 12! possible arrangements. Thus we have 2^12 · 12! possibilities in all.
Also, each cube of type (c) has four possible orientations, and may be placed in one of six positions, yielding 4^6 · 6! possibilities.
Finally, the one white cube of type (d) may be oriented in 2^4 ways. (Four ways for each face.)
Thus the total number of correct reassemblings is a = 3^8 · 8! · 2^12 · 12! · 4^6 · 6! · 2^4
To find the total number of possible reassemblings, consider that each cube may be oriented in 24 ways, and there are 27! possible arrangements of the cubes, giving b = 24^27 · 27! possibilities in all.
Therefore, the probability that the outside of the reassembled cube is completely black is
a/b = 1/(2^56 · 3^22 · 5^2 · 7 · 11 · 13^2 · 17 · 19 · 23)
September 28th, 2009 at 10:02 pm
I am replying late but would appreciate the comment on my answer
The first question could be solved this way too
2^58 will have last two digits = 44
hence when 1 will be added , last two digits will be 45
which’ll be divisible by 5 in everycase ..
this number will be prime not composite then.
September 28th, 2009 at 10:42 pm
I got the concept.