- We calculate the volume of the region which is in the first octant
and above
{(
*x*,*y*, 0) |*x*__>__*y*}; this is 1/16 of the required volume. The volume is above*R*, where*R*is the region in the*xy*-plane and bounded by*y*= 0,*y*=*x*and*y*= , and below*z*= . This volume is*dzdydx*+*dzdydx*= d *ydx*+ d*ydx*= *x*d*x*+ (1 -*x*^{2}) d*x*= [- (1 - *x*^{2})^{3/2}/3]_{0}^{1/}+ [*x*-*x*^{3}/3]_{1/}^{1}= 1/3 - 1/(6) + 2/3 + 1/(6) - 1/ = 1 - 1/.

Therefore the required volume is 16 - 8. - Let the circle with radius 1 have center
*P*, the circle with radius 2 have center*Q*, and let*R*be the center of the third circle, as shown below. Let =*a*, =*b*, and let the radius of the third circle be*r*. By Pythagoras on the triangles*PAR*and*QRB*, we obtain(1 -Therefore*r*)^{2}+*a*^{2}= (1 +*r*)^{2}, (2 -*r*)^{2}+*b*^{2}= (2 +*r*)^{2}.*a*^{2}= 4*r*and*b*^{2}= 8*r*. Also (*a*+*b*)^{2}+ 1 = 9, so 2 + 2 = and we deduce that*r*= 6 - 4.

- Let
*m*,*n*be a positive integers where*m*__<__*n*. For each*m*X*m*square in an*n*X*n*grid, replace it with the (*m*- 1) X (*m*- 1) square obtained by deleting the first row and column; this means that the 1 X 1 squares become nothing. Then these new squares (don't include the squares which are nothing) are in a one-to-one correspondence with the squares of the (*n*- 1) X (*n*- 1) grid obtained by deleting the first row and column of the*n*X*n*square. Therefore*S*_{n - 1}is the number of squares in an*n*X*n*grid which have size at least 2 X 2. Since there are*n*^{2}1 X 1 squares in an*n*X*n*grid, we deduce that*S*_{n}=*S*_{n - 1}+*n*^{2}.Thus

*S*_{8}= 1^{2}+ 2^{2}+ ... + 8^{2}= 204. - If
*p*__<__*q*< (*p*+ 1)^{2}, then*p*divides*q*if and only if*q*=*p*^{2},*p*^{2}+*p*or*p*^{2}+ 2*p*. Therefore if*a*_{k}=*p*^{2}, then*a*_{k + 3}= (*p*+ 1)^{2}. Since*a*_{1}= 1^{2}, we see that*a*_{10000}= (1 + 9999/3)^{2}= 3334^{2}= 11115556. - Let
*a*_{n}=*n*^{n}*x*^{n}/*n*!. First we use the ratio test:*a*_{n + 1}/*a*_{n}= ((*n*+ 1)^{n + 1}*x*^{n + 1}*n*!)/(*n*^{n}*x*^{n}(*n*+ 1)!) = (1 + 1/*n*)^{n}*x*._{n - > }(1 + 1/*n*)^{n}=*e*, we see that the interval of convergence is of the form { - 1/*e*, 1/*e*}, where we need to decide whether the interval is open or closed at its two endpoints. By considering*dx*/*x*, we see that1/(*n*+ 1) < ln(1 + 1/*n*) < (1/*n*+ 1/(*n*+ 10))/2 < 3/(3*n*+ 1)because 1/

*n*is a concave function, consequently (1 + 1/*n*)^{n + 1/3}<*e*< (1 + 1/*n*)^{n + 1}. Therefore when |*x*| = 1/*e*, we see that |*a*_{n}| is a decreasing sequence. Furthermore by induction on*n*and the left hand side of the last inequality, we see that |*a*_{n}| < 1/. Thus when*x*= - 1/*e*, we see that lim_{n - > }*a*_{n}= 0 and it follows that the given series is convergent, by the alternating series test. On the other hand when*x*= 1/*e*, the series is S*n*^{n}/(*e*^{n}*n*!). By induction on*n*and the above inequality, we see that*n*^{n}/(*e*^{n}/*n*!) > 1/(*en*) for all*n*> 1. Since S1/*n*is divergent, we deduce that the given series is divergent when*x*= 1/*e*, consequently the interval of convergence is [- 1/*e*, 1/*e*). - Let
*A*=( 3 1 1 3 ). If we can find a matrix

*B*=( *a**b**c**d*) such that

*B*^{2}=*A*or even 4*A*and set*f*(*x*) = (*ax*+*b*)/(*cx*+*d*), then*f*(*f*(*x*)) = (3*x*+ 1)/(*x*+ 3). So we want to find a square root of*A*. The eigenvalues of*A*are 2 and 4, and the corresponding eigenvectors are (1,-1) and (1,1) respectively. Thus if*X*=( 1 -1 1 1 ), then

*XAX*^{-1}=( 2 0 0 4 ). Set

*C*=( 0 0 2 ). Then

*C*^{2}=*XAX*^{-1}, so if we let*B*= (*X*^{-1}*CX*)^{2}, then*B*^{2}=*A*. Since2 *B*=( 2 + 2- 2- 2+ ), we may define (multiply top and bottom by 1 + /2)

*f*(*x*) = ((3 + 2)*x*+ 1)/(*x*+ 3 + 2).*f*still maps**R**^{+}to**R**^{+}. Of course there are many other solutions and answers. - Choose
*x**A*and*y**B*so that*f*(*xy*) is as large as possible. Suppose we can write*xy*in another way as*ab*with*a**A*and*b**B*(so*a*=/=*x*). Set*g*=*ax*^{-1}and note that*I*=/=*g**G*. Therefore either*f*(*gxy*) >*f*(*xy*) or*f*(*g*^{-1}*xy*) >*f*(*xy*). We deduce that either*f*(*ay*) >*f*(*xy*) or*f*(*xb*) >*f*(*xy*), a contradiction and the result follows.