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Learn to Speak Fluent English

September 24th, 2009 Posted in Uncategorized

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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?

12 Responses to “Learn to Speak Fluent English”

  1. Mearaj Says:

    Question for the day:
    According to cyclicity the answer is it is composite.

  2. Hitesh Says:

    Answer of Cube puzzle
    = 1/(27* 26C6 * 6^6 * 20C12 * 12^12 * 8^8)

  3. Shweta Subhedar Says:

    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.

  4. Hitesh Says:

    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..

  5. Hitesh Says:

    @shweta Subhedar…

    its not that much simple…

  6. thiyoshines Says:

    ya hitesh explain is correct.

  7. Rajiv Says:

    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?

  8. ashok Says:

    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………..

  9. admin Says:

    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.

  10. admin Says:

    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)

  11. SMITA Says:

    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.

  12. SMITA Says:

    I got the concept.

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