1. Matching pennies.
Well, let x be the number of tosses. Adolph won 13 of them, so Bertha won the other x-13. Then Bertha's net gain was (x-13)-13 pennies. Thus, x-26=8, and x=34.

ALTERNATE SOLUTION

Since Adolph won on 13 of the tosses, Bertha had to win on 13 others to break even, and on 8 more to be 8 ahead. Thus Bertha won on 21 of the tosses, and there were 13+21=34 tosses altogether.
 

2. Base b quadratic.
The answer is n=36b.  Since (x-9)(x-5)=x2-11bx+n, we know that 9+5=(11)b=b+1, so  b=(13)10. Then, since  (9)(5)=3b+6, we have n=36b.
 

3. Hyperbolics.
The answer is  16 (2)1/2.  From the definitions of sinh and cosh we have

Then


4. A relatively prime sequence.
Note first that all an are odd (by an easy induction). Assume that m<n. Then

and by induction, for every Thus  an=ram+2  or  ram-2 for some integer r, and therefore every common factor of am and anis a divisor of 2. Since both am and anare odd, they are relatively prime.
 

5. Sum the series.
We will show that the sum is  ln(6). Using well-known properties of the logarithm, we write the partial sum up to N (with N>4)

The terms

in (1) cancel out, leaving




6. A circle and some points.
Let the chosen points be P1, P2, . . . , P1998.  Choose any two diametrically opposite points Q and Q' on the circle, subject only to the restriction that not all Pi lie on the segment QQ', and let di and di' be the distances from Q and Q', resp. to Pi, 1i1998.  By the triangle inequality, di + di'2, with equality if and only if Pi is on the segment QQ'. Since some of the Pi are not on this segment we have

so at one of   is greater than 1998.
 

7. Equal integrals.
For simplicity we translate the origin to the midpoint of the interval (a,b). Let h=(b-a)/2,  F(x)=f(x+(a+b)/2) and G(x)= g(x+(a+b)/2). Then

F(-h)=f(a)=g(a)=G(-h),

and
F(h)=f(b)=g(b)=G(h).

Also,

and
so it suffices to show that

Let P(x)=F(x)-G(x). Then P(x) is a polynomial of degree 3 with zeros at -h, 0 and h, so for some constant c,
From (1) we see that P is an odd function, and therefore

q.e.d.

Note: This fact may also be seen as a consequence of the fact that in Simpson's Rule the error is bounded by a multiple of the fourth derivative. Simpson's Rule with n=2 approximates

where f(x) is a quadratic polynomial fitted to g(x) at a, (a+b)/2and b. The fourth derivative of g(x) is, of course, identically zero.
 
 

8. Reciprocal inequality.
We shall repeatedly use the fact that for all real numbers u and v,

which follows at once from (u-v)20. We apply (1) to obtain
Thus
Another application of (1) tells us that
and from (2) and (3) we conclude that
Applying (4) to 1/a, 1/b, 1/c,  and 1/d we have
Thus we have
so that
and therefore
Then from (5) it follows that
as desired.
 
 

9. Order of an element.
The order of y is 31. From xyx-1=y2 we have xy=y2x, and a key observation is that for every positive integer n,

as we show by induction. We already know it for n=1, and if xyk=y2kx, then

xyk+1 = y2kxy = y2ky2x = y2k+2x,

so we have (1) by induction. Since x5=e, we have

i.e., y=y32, so y31=e. Since 31 is prime, it is the order of y.
 
 

10. Not a fifth power.