Solutions to Mensa Math Brain Busters

1) We note here that each specific divisor x (3, 4, 5, 6 or 7) of the desired integer N yields a remainder of (x - 2). Therefore, each specified divisor x of N + 2 yields a remainder of 0.  In other words, N +2 is a multiple of 3, 4, 5, 6 and 7. The smallest such number is:

420 (2 · 2 · 3 · 5 · 7)

If N + 2 = 420 then the desired number N is: N = 420 - 2 =  418

Following on from this:

418/ 3 =   139 r1

418/4 = 104 r2

418/5 = 83 r3

418/6 = 69 r4

418/7 = 59 r5

2) Write the formula for the volume of a cone:

V = 1/3 (pr² h)  

Also, for this situation:  r²  +  h ²  =   1: V= 1/3 pr² Ö(1 - r²) 

We see the volume V is zero when r = 0 or 1.  Hence, the maximum ratio must be found for some r between 0 and 1.  How to obtain it? We first introduce the formula for the surface area of a cone:

A = pr h _s    where  h _s  denotes the slant height of the cone for which:

h _s =  Ö (r²  +  h ²)

Now, because in our case:  h _s =  1  then A = pr.

Then the ratio V/ A  = 1/3 pr² Ö(1 - r²)/ pr

Or, more simply:

V/ A  =  1/3 r Ö(1 - r²)

Differentiation of a function must be used at this point. The maximum or minimum of a differentiable function occurs at critical points where the derivative is zero.  Then let (V/A) be a differentiable function, e.g.

(V/ A)'  = 1/3 r Ö(1 - r²)

 By the product rule:

1/3 r Ö(1 - r²)  =  1/3 Ö(1 - r²) +  1/3 r Ö(1 - r²)

Using the chain rule for derivatives:

(V/ A)'  = 1/3  Ö(1 - r²) + 1/3 r (½) Ö(1 - r²) (2r) 

=  1/3  Ö(1 - r²) - r²/3 Ö(1 - r²)

We then set the derivative to 0 and solve for r:

1/3  Ö(1 - r²) - r²/3 Ö(1 - r²)  =  0

Multiply both sides by: 3 Ö(1 - r²):

0= (1 - r²)  - r²   =   1 -    2r²

Or:  2r² =  1

è

r  =  Ö1/2  =   Ö2/2

 Then   Max (V/A)  = (1/3) (Ö2/2) (Ö2/2) = 2/ 12  =  1/6

 Thus the maximum ratio (V/A) occurs when:

r =  Ö2/2   and h  = Ö2/2

This can be achieved when one full quadrant of a circle (sector angle 90 deg) is removed from a paper circle and the straight edges as re-connected.