## 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:

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

3. Shweta Subhedar Says:

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:

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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………..

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.

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)

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.