Problems of the Week 2 Answers

1. The answer to this is really obvious. The sum of n (1/n)'s is 1, so you need k=1000 ones to get 1000.

2. By definition, all multiples of x can be represented as Nx mod x, where N is an integer. Working backward, we find out that the code must be a multiple of 7, 8, and 9. The only three digit number with this property is 7*8*9= 504.

3. Let's replace the cost of a cone with x. Al must have (x-24) cents and Bob must have (x-2) cents. If Bob is only two cents short and the sum of their money is less than x, then Al must have 0 cents or 1 cent. Because Al took at least some money from his bank, he took out one cent; solving for x, we find that the cost is 25 cents.

4. This problem is simple replacement. We can take out the one to get 22 as a solution; and, if you take out 2, 3, and 4 and replace them with 24, you get 20. You can't add 25 because only one five follows, so those are your only solutions.

5. Two digit numbers with digits (xy) are written as 10x+y. Therefore the equation for their ages are:
10x+y+(41-14)=10y+x. One of the solutions is already (1, 4), and when we simplify we get
y=x+3. The next positive integer solution would be (2, 5),and therefore the answer is 11 years later.

6. When you graph 0<y<2, 0<x<2, and abs(y-x)<1, you get:

The green area takes up 3/4 of the 2x2 square, so the probability is 3 in 4.

7. After the first repetition, there are three digits. So after two repetitions, there are exactly four possibilities for the number: 033, 123, 213, and 303. However, 033 and 213 are null because no number simplifies to either of the two. 123 and 303 both go to 123 again.

8. For the sake of understanding what I'm saying, let's use the @ symbol in place of 'plus or minus'. abs(x) in this case simplifies to @x, abs(1@x) simplifies to @1@x, and abs(2@1@x) simplifies to @2@1@x=1. The solutions to this are @4, @2, and @0.

9. Drawing out the hexagon, it looks like this:

Notice that the corners are equilateral triangles of side lengths 9, 14, and 10. If you think about it, the other equilateral triangle will have corner triangles of 7, 11, and 6. The side length of the large triangle will consist of the side lengths of two smaller triangles and the side inbetween them. That would make the triangle's side length equal 7+14+6=6+10+11=11+9+7= 27 units.

10. Draw this out too.

Because FA is tangent to circle C at point F, radius FC is perpendicular to FA. This forms right triangle CAF, with hypotenuse CA and legs FA and FC. We know that FC is 3 units, and AC is 5 units, so therefore FA is 4 units long.

Sorry for the kind-of bad explanations I made. I tried, at least!

Stay coolio,
John

PS: Credit to SketchBookExpress and GeoGebra for the illustrations.