ENTER TO WIN: Baby Einstein 9-Piece Discovery Kit Collection!

We all know them.  We’ve (mostly) all used them.  If you’ve used them, you love them! Baby Einstein Discovery Kits (kits that are uniquely packaged to include a DVD, a music CD and either a book or Discovery Cards in a single package) are designed to help provide parents with new ways to introduce their little ones to the world around them in playful and engaging ways. The kits not only promote discovery and inspire baby’s natural curiosities but they also encourage parents to look to their everyday experiences as a way to extend content beyond just the kit.

Discovery Kits align with a baby’s path from infant to toddler by offering three broad levels of discovery — Experience (Level 1), Explore (Level 2) and Expression (Level 3) and the suggested Retail Price: $19.99 U.S.

Take a look at this great video!!

Be sure to check out Baby Einstein online:

One (1) lucky BabyRazzi reader will receive the entire 9-Piece Discovery Kit collection!

How do I enter? You have a few options AND EACH COUNTS AS ONE ENTRY. There is NO LIMIT to the number of times a reader can enter (via any of the entry channels) this contest. Please leave a comment for each entry, and include your email in each comment as well:

  1. Please fill out the contest entry form.
  2. Become a fan of  BabyRazzi’s Facebook Page. Then leave a comment on the wall saying you love the Baby Einstein 9-Piece Discovery Kit Collection giveaway.
  3. Fan Baby Einstein on Facebook and mention that you entered the Baby Einstein 9-Piece Discovery Kit Collection giveaway on babyrazzi.com.
  4. Tweet this message: “Hope I win @BabyEinstein 9-Piece Discovery Kit Collection giveaway from babyrazzi.com(@babyrazzi). RT to enter #contest #giveaway”.
  5. Leave a comment on the Baby Einstein YouTube Channel saying you love the Baby Einstein 9-Piece Discovery Kit Collection giveaway from babyrazzi.com.
  6. Leave a comment on the Baby Einstein Blog saying you love the Baby Einstein 9-Piece Discovery Kit Collection giveaway from babyrazzi.com.
  7. Post a link to this contest on your blog (and then leave a comment on this post with your link).
  8. Leave a comment on the original contest post here.
  9. Follow BabyRazzi on NetworkedBlogs.com.
  10. Stumble this post (not the site) and leave me your StumbleUpon Name in your comment.
  11. Comment on other BabyRazzi posts and reference this contest. Be sure to add that the comment is acting as an entry to the contest. Each comment counts as one entry. Please be sure to include you email when asked otherwise I won’t be able to contact you.

Contest ends Midnight, Eastern Time, 12/27/11

Winner(s) will be selected at random.

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Disclosure: I received these products through a LBi/Baby Einstein program. However, all thoughts and opinions expressed are my own and no other monetary compensation took place.

Comments

  1. trantancuong says

    (N+).
    Two formulas
    z^3=[z(z+1)/2]^2 – [z(z-1)/2]^2
    And :
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a)^3+(x+a+1)^3+……..+(y-1)^3].

    Because:
    (x+a)^3=[(x+a)(x+a+1)/2]^2 – [(x+a)(x+a-1)/2]^2.
    So:
    x^3+y^3=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    Suppose:
    z^3=x^3+y^3.
    So:
    [z(z+1)/2]^2 – [z(z-1)/2]^2=[y(y+1)/2]^2 – [x(x-1)/2]^2 – [(x+a)(x+a+1)/2]^2 + [(x+a)(x+a-1)/2]^2 – [(x+1)^3+(x+2)^3+……..+(x+a-1)^3+(x+a+1)^3+……..+(y-1)^3]

    Attention!
    This equation have four invariant square integers . The rest will change over integers (a).
    Corresponding to a value of (a) will have a corresponding equation.
    Impossible found integers to satisfy too many equations.

    Example:
    62^2 -32^2=25^2-14^2 – c^2+d^2 – [e^3+f^3+g^3+h^3+t^3+r^3]
    62^2 -32^2=25^2-14^2 – p^2+q^2 – [e^3+f^3+g^3+h^3+k^3+r^3]
    62^2 -32^2=25^2-14^2 – t^2+n^2 – [e^3+f^3+g^3+h^3+t^3+s^3]
    62^2 -32^2=25^2-14^2 – h^2+k^2 – [v^3+f^3+g^3+h^3+t^3+r^3]
    Too many equations.
    How satisfied them by only integers !!!!!!!!
    So:
    z^3=/x^3+y^3
    Similar.
    z^n=/x^n+y^n.
    ADIEU.

  2. trantancuong says

    Pierre De Fermat ‘s last Theorem.
    The conditions:
    x,y,z,n are the integers and >0. n>2.
    Proof:
    z^n=/x^n+y^n.
    We have;
    z^3=[z(z+1)/2]^2-[(z-1)z/2]^2
    Example;
    5^3=[5(5+1)/2]^2-[5(5-1)/2]^2=225-100=125
    And
    z^3+(z-1)^3=[z(z+1)/2]^2-[(z-2)(z-1)/2]^2
    Example;
    5^3+4^3=[5(5+1)/2]^2-[(5-2)(5-1)/2]^2=225-36=189
    And
    z^3+(z-1)^3+(z-2)^3=[z(z+1)/2]^2-[(z-3)(z-2)/2]^2
    Example
    5^3+4^3+2^3=[5(5+1)/2]^2-[(5-3)(5-2)/2]^2=225-9=216
    And
    z^3+(z-1)^3+(z-2)^3+(z-3)^3=[z(z+1)/2]^2-[(z-4)(z-3)/2]^2
    Example
    5^3+4^3+3^3+2^3=[5(5+1)/2]^2-[(5-4)(5-3)/2]^2=225-1=224
    General:
    z^3+(z-1)^3+….+(z-m)^3=[z(z+1)/2]^2-[(z-m-1)(z-m)/2]^2
    We have;
    z^3=z^3+(z-m-1)^3 – (z-m-1)^3.
    Because:
    z^3+(z-m-1)^3=[z^3+(z-1)^3+….+(z-m-1)^3] – [(z-1)^3+….+(z-m)^3]
    So
    z^3=[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 – [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 – (z-m-1)^3.
    Similar:
    z^3=z^3+(z-m-2)^3 – (z-m-2)^3.
    So
    z^3=[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 – [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 – (z-m-2)^3.
    ….
    ….
    Suppose:
    z^n=x^n+y^n
    So
    z^(n-3)*z^3=x^(n-3)^n*x^3+y^(n-3)*y^3.
    So
    z^(n-3)*{[z(z+1)/2]^2-[(z-m-2)(z-m-1)/2]^2 – [z(z-1)/2]^3+[(z-m-1)(z-m)/2]^2 – (z-m-1)^3}=x^(n-3)*{[x(x+1)/2]^2-[(x-m-2)(x-m-1)/2]^2 – [x(x-1)/2]^3+[(x-m-1)(x-m)/2]^2 – (x-m-1)^3}+y^(n-3)*{[y(y+1)/2]^2-[(y-m-2)(y-m-1)/2]^2 – [y(y-1)/2]^3+[(y-m-1)(y-m)/2]^2 – (y-m-1)^3}
    Similar:
    z^(n-3)*{[z(z+1)/2]^2-[(z-m-3)(z-m-2)/2]^2 – [z(z-1)/2]^3+[(z-m-2)(z-m-1)/2]^2 – (z-m-2)^3=x^(n-3)*{[x(x+1)/2]^2-[(x-m-3)(x-m-2)/2]^2 – [x(x-1)/2]^3+[(x-m-2)(x-m-1)/2]^2 – (x-m-2)^3+y^(n-3)*{[y(y+1)/2]^2-[(y-m-3)(y-m-2)/2]^2 – [y(y-1)/2]^3+[(y-m-2)(y-m-1)/2]^2 – (y-m-2)^3.
    ….
    ….
    Because it is codified .
    So
    Impossible all are the integers.
    So:
    z^n=/x^n+y^n.
    ISHTAR.

  3. trantancuong says

    Suppose
    Z^3=X^3+Y^3.
    =>
    Z=(X^3+Y^3)^1/3.
    Define F(X,Y).
    F(X,Y)= (X^3+Y^3)^1/3 – [(X-X-1)^3+(Y-X-1)^3]^1/3.
    =>
    F(X,Y)=Z- [(Y-X-1)^3-1]^1/3.
    Because.
    (Y-X-1) is an integer so
    [(Y-X-1)^3-1]^1/3. and [(Y-X-1)^3-1]^2/3. are two irrational numbers.
    =>
    { [(Y-X-1)^3-1]^1/3 – [ Z – F(X,Y)] }^3=0.
    =>
    [(Y-X-1)^3-1] +3[(Y-X-1)^3-1]^2/3*Z +V=0.
    Because
    3[(Y-X-1)^3-1]^2/3*Z is an irrational number So
    V is an irrational number.
    Because
    3Z[(Y-X-1)^3-1]^2/3 = – {[(Y-X-1)^3-1] +.V}
    So
    an integer * an irrational number=an integer+an irrational number.
    Unreasonable.
    So
    Z is an irrational number.
    So
    Z^3=/X^3+Y^3.
    Similar.
    Z^N=/X^N+Y^N.

  4. trantancuong says

    Suppose
    Z^3=X^3+Y^3.
    =>
    Z=(X^3+Y^3)^1/3.
    Define F(X,Y).
    F(X,Y)= (X^3+Y^3)^1/3 – [(X-X-1)^3+(Y-X-1)^3]^1/3.
    =>
    F(X,Y)=Z- [(Y-X-1)^3-1]^1/3.
    Because.
    (Y-X-1) is an integer so
    [(Y-X-1)^3-1]^1/3. and [(Y-X-1)^3-1]^2/3. are two irrational numbers.
    =>
    { [(Y-X-1)^3-1]^1/3 – [ Z – F(X,Y)] }^3=0.
    =>
    [(Y-X-1)^3-1] +3[(Y-X-1)^3-1]^2/3*Z +V=0.
    Because
    3[(Y-X-1)^3-1]^2/3*Z is an irrational number So
    V is an irrational number.
    Because
    3Z[(Y-X-1)^3-1]^2/3 = – {[(Y-X-1)^3-1] +.V}
    So
    an integer * an irrational number=an integer+an irrational number.
    Unreasonable.
    So
    Z is an irrational number.
    So
    Z^3=/X^3+Y^3.
    Similar.
    Z^N=/X^N+Y^N.
    ISHTAR.

  5. trantancuong says

    I have stupid string.
    !^4, 2^4, 3^4, 4^4, 5^4, 6^4….
    and1^2, 2^2, 3^2, 4^2, 5^2, 6^2…..
    Write again:
    1, 16, 81, 256, 625, 1296, 2401,…..(X)
    and
    1, 4, 9, 16,25, 36, 49……..(Y)
    X-Y:
    0, 12, 72, 240, 600, 1260, 2352……
    big number subtract small number;
    12, 60, 168, 360, 660, 1092……
    big subtract small:
    48, 108, 192, 300, 432,…….
    big subtract small;
    60, 84, 108, 132,…..
    big subtract small:
    24, 24, 24, 24,….= 4 !
    the end is zero.
    This is always right exponent m and n. Any(m,n)belong N^2.
    Happy&Peace.
    Cuong mad.

  6. trantancuong says

    The equation : X exponent n + Y exponent n + Z exponent n = T exponent n.
    Have only one solution :
    3 cube + 4 cube + 5 cube = 6 cube.
    With ( X,Y,Z,T,n ) belong set of natural numbers N exponent 5. This equation have not ever solved with n is not equal 3.
    I dream to see Sir. Pythagore send to me this equation. And Sir. Pierre De Fermat said to me that ‘“I have discovered a truly marvelous proof of this theorem, which this doodle is too small to contain’.
    Happy New Year to Everybody.
    I am a Mathematical and Musical genius. But i say with anybody who can understand the magical;math. So i never say to anybody who is stupid. Please understand me.

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