Subcontests
(21)Rounding Necessarily Increases the Answer
Amelia needs to estimate the quantity ba−c, where a, b, and c are large positive integers. She rounds each of the integers so that the calculation will be easier to do mentally. In which of these situations will her answer necessarily be greater than the exact value of ba−c?<spanclass=′latex−bold′>(A)</span>She rounds all three numbers up.<spanclass=′latex−bold′>(B)</span>She rounds a and b up, and she rounds c down.<spanclass=′latex−bold′>(C)</span>She rounds a and c up, and she rounds b down.<spanclass=′latex−bold′>(D)</span>She rounds a up, and she rounds b and c down.<spanclass=′latex−bold′>(E)</span>She rounds c up, and she rounds a and b down. Modulo Ackermann
For every positive integer n, let mod5(n) be the remainder obtained when n is divided by 5. Define a function f:{0,1,2,3,…}×{0,1,2,3,4}→{0,1,2,3,4} recursively as follows:
f(i,j)=⎩⎨⎧mod5(j+1)f(i−1,1)f(i−1,f(i,j−1))if i=0 and 0≤j≤4if i≥1 and j=0, andif i≥1 and 1≤j≤4
What is f(2015,2)?<spanclass=′latex−bold′>(A)</span>0<spanclass=′latex−bold′>(B)</span>1<spanclass=′latex−bold′>(C)</span>2<spanclass=′latex−bold′>(D)</span>3<spanclass=′latex−bold′>(E)</span>4 Recursion of Circles
A collection of circles in the upper half-plane, all tangent to the x-axis, is constructed in layers as follows. Layer L0 consists of two circles of radii 702 and 732 that are externally tangent. For k≥1, the circles in ⋃j=0k−1Lj are ordered according to their points of tangency with the x-axis. For every pair of consecutive circles in this order, a new circle is constructed externally tangent to each of the two circles in the pair. Layer Lk consists of the 2k−1 circles constructed in this way. Let S=⋃j=06Lj, and for every circle C denote by r(C) its radius. What is C∈S∑r(C)1?[asy]
import olympiad;
size(350);
defaultpen(linewidth(0.7));
// define a bunch of arrays and starting points
pair[] coord = new pair[65];
int[] trav = {32,16,8,4,2,1};
coord[0] = (0,73^2); coord[64] = (2*73*70,70^2);
// draw the big circles and the bottom line
path arc1 = arc(coord[0],coord[0].y,260,360);
path arc2 = arc(coord[64],coord[64].y,175,280);
fill((coord[0].x-910,coord[0].y)--arc1--cycle,gray(0.78));
fill((coord[64].x+870,coord[64].y+425)--arc2--cycle,gray(0.78));
draw(arc1^^arc2);
draw((-930,0)--(70^2+73^2+850,0));
// We now apply the findCenter function 63 times to get
// the location of the centers of all 63 constructed circles.
// The complicated array setup ensures that all the circles
// will be taken in the right order
for(int i = 0;i<=5;i=i+1)
{
int skip = trav;
for(int k=skip;k<=64 - skip; k = k + 2*skip)
{
pair cent1 = coord[k-skip], cent2 = coord[k+skip];
real r1 = cent1.y, r2 = cent2.y, rn=r1*r2/((sqrt(r1)+sqrt(r2))^2);
real shiftx = cent1.x + sqrt(4*r1*rn);
coord[k] = (shiftx,rn);
}
// Draw the remaining 63 circles
}
for(int i=1;i<=63;i=i+1)
{
filldraw(circle(coord,coord.y),gray(0.78));
}[/asy]<spanclass=′latex−bold′>(A)</span>35286<spanclass=′latex−bold′>(B)</span>70583<spanclass=′latex−bold′>(C)</span>73715<spanclass=′latex−bold′>(D)</span>14143<spanclass=′latex−bold′>(E)</span>1461573