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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?

MearajSays:September 24th, 2009 at 10:40 pm

Question for the day:

According to cyclicity the answer is it is composite.

HiteshSays:September 24th, 2009 at 10:58 pm

Answer of Cube puzzle

= 1/(27* 26C6 * 6^6 * 20C12 * 12^12 * 8^8)

Shweta SubhedarSays: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.

HiteshSays: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..

HiteshSays:September 24th, 2009 at 11:12 pm

@shweta Subhedar…

its not that much simple…

thiyoshinesSays:September 24th, 2009 at 11:17 pm

ya hitesh explain is correct.

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ashokSays: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………..

adminSays: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.

adminSays: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 Â· 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 Â· 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 Â· 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 Â· 8! Â· 2^12 Â· 12! Â· 4^6 Â· 6! Â· 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 Â· 27! possibilities in all.

Therefore, the probability that the outside of the reassembled cube is completely black is

a/b = 1/(2^56 Â· 3^22 Â· 5^2 Â· 7 Â· 11 Â· 13^2 Â· 17 Â· 19 Â· 23)

SMITASays: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.

SMITASays:September 28th, 2009 at 10:42 pm

I got the concept.