Solutions to (Mensa) Brain Buster Challenge (#3)

1) Arrange the 6 L-shaped pieces above into a 4 x 6 grid such that:

- Each row has the same number of stars

-  Each column has the same number of stars.

- No diagonal has 4 stars.

The pieces may be rotated but not reflected.

Solution:   The best approach is to cut the pieces out  and assemble them  - according to the rules - like  a jigsaw puzzle, to get;

2) Prove all square numbers (1,4,9,16, 25...) are either divisible by three OR have a remainder of one when divided by three.

Solution:  All numbers, including square numbers, are a multiple of 3 (or 3n). Or a one more than a multiple of 3, i.e. (3n + 1), or one less than a multiple of three, i.e. (3n - 1).   Then:

(3 n) ²   =   9 n ²   =   3 (3 n) ²   is divisible by 3

While:

 (3 n + 1  ) ²    =   9n ² +   6n +   1 =   3 (3 n ²  +   2n) + 1

has a remainder of 1 when divided by 3.

Lastly:

 (3 n -  1  ) ²   =   9n ² -   6n +   1   =   3 (3 n ² -   2n) + 1

has a remainder of 1 when divided by 3.

3) Find all solutions for positive integers a and b (a > 1):

(a + 2)! a! (a - 2)! = b!

Then prove there are no other solutions.

Solution:

There are two solutions:  a = 3, b = 6

I.e.  (a + 2)! a! (a - 2)! = b!

=  (3 + 2)! 3! (3 - 2)! = 6!  =   5! 3! (1!) = (5 x 4 x 3 x 2 x1) (1) (3 x 2 x 1)

=  120 (6) = 720

Also:   a = 5,  b = 10

=  (5 + 2)! 10! (5 - 2)! = 7!  5! 3!  =  362, 880

We prove there are no solutions for a > 6:

4) Given a rectangle of length  (Ö2) x  and height x, we have one circle and two semi-circles inscribed within it,  and all three intersect at two points, as shown below:

Find the area of either green crescent shape between the two arcs in terms of x. 

Solution:

We render the sketch so it conforms to:

Where u and v mark the end points where the circle and both semi-circles intersect. Then line segment uv is a diameter of the inscribed circle.   We then let w (extreme mid point of bottom line of rectangle)  be the point where the inscribed circle is tangent for the bottom side of the rectangle, which is also the diameter of the lower semi-circle.  

Note than an inscribed angle subtended by a diameter is a right angle.  Therefore, angle uvw  is a right angle.  Further,  line segment wv is a radius of the lower semi-circle, i.e.

wv =  ½(Ö2p) x

And line segment uv is the hypotenuse of right triangle uvw, so:

uv = 1/2 (Ö2) (Ö2) x  = x 

The area of the inscribed circle is then:

A(c)= p (x/2) ²   =  (p /4) x ²

The area of the upper half of the inscribed circle is:

A(c1) = (p /8) x ²

The area of sector uwv of the lower semicircle is:

A(uwv) =  (p /4) [½ (Ö2)x]²  =  (p /8)x ²

The area of the chord uv of the lower semicircle is equal to the area of the sector uwv minus the area of the right triangle uwv:

A(uv^) = A(uwv)  -  A(<uwv) =

 (p /8) x ²   -  ½ [½ (Ö2)x] [½(Ö2)x] = (p /8)x ²  - (¼) x ²

The area of the lune then is equal to the area of the upper half of the inscribed circle minus the area of the chord uv,

A(L) = A(c1)  -  A(uv^)  = (p /8)x ²  - [(p /8)x ²  - (¼) x ²]

A(L) = (p /8)x ²  - (p /8)x ² + (¼) x ²  =   x ²/4

5) We are given the equation:

A^B + C^D = E ^F + G^H + I^J + K^L

 

We obtain on appropriate replacement of letters A-L with integers 0-11:

11⁰ + 5⁶ = 0 ⁷ + 2⁸ + 9³ + 11⁴

And:

1⁷ + 5⁶ = 0 ¹⁰ + 2⁸ + 9³ + 11⁴

 

Of which there are 96 total permutations.

There are in addition 96 total permutations for:

1⁹ + 4⁷ = 0¹⁰ + 2¹¹ + 3⁸ + 6⁵

And:

1¹⁰ + 4⁷ = 0⁹ + 2¹¹ + 3⁸ + 6⁵

 

There are in addition 96 further total permutations for:

3⁹ + 7² = 0¹⁰ + 5⁶  + 11¹ + 8⁴

And:

3⁹ + 7²  =  11⁰ + 5⁶ + 10¹ + 8⁴

 

Yielding a total of: 96 + 96 + 96 = 288 answers/solutions