# Vandal Science News Puzzlers

We love our Puzzlers.

We hope you do too.

#### Puzzler - (April 2018)

The puzzler this time is very simple to state. There are many ways to solve it, but there’s one particularly short and clever way.

*How many positive integers less than one million include at least one digit 4?*

To clarify, all 100,000 numbers between 400,000 and 499,999 are in the list. So, of course, is 539,344 and 539,349. But 539,350 is not. Can you count them all?

### Solution:

Here is a very simple way to find the answer: to find how many numbers less than one million do include a digit 4, first find the number which do not include a digit 4. There are six digits to determine, and each can be any of nine different possibilities (any digit except 4). (Note that this accounts for all numbers up to 999,999 because numbers like 3,357 which are less than six digits long would show up as 003,357.) So, there are

(9)(9)(9)(9)(9)(9) = 531,441

such numbers. There are 1,000,000 numbers total (000,000 through 999,999), so the count of those which include a digit 4 must be

1,000,000 - 531,441 = **468,559**.

#### Puzzler - (Oct. 2017)

Our puzzler for this issue uses some arithmetic, but mostly logic. Have fun!

A company runs two locations in the same town. Between the two they employ 225 workers.

- One week, exactly 1/5 of the workforce at one location is transferred across town to the other location.
- Then, the next week, exactly 1/4 of the then-present workers at one location are transferred to the other location.
- Finally, the week after that, exactly 1/3 of the workers then stationed at the Westside location are transferred to the Eastside location.
- After this last transfer, the difference between the two workforces is 21 workers.

How many workers were originally employed at the Westside location?

#### Solution

The Westside factory started with **105 workers**.

We'll get to that answer by using a little algebra and a little reasoning about divisibility. To begin with, let's call our two factories A and B and suppose that A is the one making the first transfer. Then to begin with, A has 5*x* workers and B has 225-5*x* workers. The first transfer is to move *x* workers from A to B, so the effect is:

Before: A has 5*x*, B has 225-5*x*

After: A has 4*x*, B has 225-4*x*

Now the number 225-4*x* is not divisible by 4, so factory B can't make the second transfer! So, the second transfer must consist of moving another *x* workers (one-fourth of A's current total) from A to B:

Before: A has 4*x*, B has 225-4*x*

After: A has 3*x*, B has 225-3*x* = 3(75-*x*)

Now there are two cases to consider.

**CASE 1:** If the third transfer also goes from A to B then it also amounts to moving *x* workers from A to B:

Before: A has 3*x*, B has 225-3*x*

After: A has 2*x*, B has 225-2*x*

Then, the difference between the two must be 21, so:

2*x* - (225-2*x*) = 21 or 2*x* - (225-2*x*) = -21

This means that 4*x* is either 246 or 204. The first of these is not a multiple of 4, so it must be that 4*x* = 204. This gives us *x* = 51. The only problem with this is that factory A started with 5*x* workers, and 5*x* would be greater than 225. So, Case 1 has led us to a dead end!

**CASE 2:** We are left with the conclusion that the last transfer was from B to A. (Note that this means that factory B is the Westside factory -- we know the last transfer went from west to east.) This transfer must have consisted of moving 75-*x* workers (one-third of B's current workforce) from B to A:

After: A has 3*x*, B has 3(75-*x*)

Before: A has 3*x*+(75-*x*) = 75+2*x*, B has 2(75-*x*) = 150-2*x*

The difference between the two must be 21, so:

(75+2*x*) - (150-2*x*) = 21 or (75+2*x*) - (150-2*x*) = -21

This means that 4*x* is either 96 or 54. But 54 is not a multiple of 4, so it must be that 4*x* = 96, or *x* = 24.

The Westside factory, as we mentioned, is factory B. So, it started with 225-5*x* = 225-5*(24) = 105 workers.

#### Puzzler - (Feb. 2016)

This issue’s Puzzler is an aeronautical story problem with a geometric flavor:

A jet flies at a constant speed in a circular path over a city. At exactly noon it is observed flying due north directly above a park at the eastern edge of town. One minute later it is directly above a point exactly 3 miles north and 1 mile west of the park. How long will it take the plane to complete an entire circle?

#### Solution

Consider the diagram at right. Because the plane was flying due north at noon, we know that the center of its circular path must be directly west of the park. We know two points along that circular path, and the arrangement helps us form a right triangle with sides R, R-1, and 3 (where R is the radius of the circle).

Using the Pythagorean Theorem:

(R-1)^{2} + 3^{2} = R^{2}

R^{2} - 2R + 10 = R^{2}

10 = 2R

so R=5. But this means that the right triangle here is a 3-4-5 right triangle, so the angle inscribing the plane’s arc over that one minute time period is Arctan(3/4), or about 36.87 degrees. Now 360 divided by 36.87 is about 9.76, so it takes roughly **9.76 minutes** for the plane to complete the entire circle.

#### Puzzler - (June 2016)

Our puzzler for this issue deals with probability. Suppose that a bin contains 12 balls – 5 red, 4 blue, and 3 yellow. We draw 3 balls at random from the bin, and we “win” this game if the three balls we choose represent exactly two colors. That is, we win if we choose two balls of the same color with the third ball being a different color. What is the probability of winning this game?

#### Solution

We first need to know how many possible outcomes there are to this game. But that’s just a matter of some basic mathematics. At one time or another you probably met C(n,k), the number of ways to choose k objects out of a set of n objects. You might even remember the formula C(n,k) = n!/k!(n-k)!. Our game consists of choosing three balls from a set of 12, so the number of possible outcomes is C(12,3) = 12!/3!9! = (12*11*10)/(3*2) = 220.

Now all we need to do is figure out how many of those outcomes are “winning” outcomes. Well, it turns out that in this case it’s actually easier to count the number of outcomes that don’t win. To not win we would need to choose either all red, all blue, all yellow, or one ball of each color. Let’s take those one at a time:

- There are 5 red balls in the bin, so there are C(5,3) = 5!/3!2! = 10 ways to choose three of them.
- There are 4 blue balls in the bin, so there are C(4,3) = 4!/3!1! = 4 ways to choose three of them.
- There are only 3 yellow balls in the first place, so there’s only one way to choose all of them!
- There are 5*4*3 = 60 ways to choose one ball of each color.

This gives us a total of 75 ways to choose three balls that don’t win. That means the other 220-75=145 outcomes are all winning draws. So, the probability of winning is 145/220 = 29/44 ≈ 0.65909.

#### Puzzler - (Feb. 2015)

Anything can inspire a good puzzle – even something as simple as a tile floor. You've probably seen a pattern of floor tiles like the one pictured here. It consists of squares and regular octagons (all eight sides are the same length and all eight angles measure 135 degrees). Suppose the area of each green octagon is 6 square inches. What then is the area of a pink square? Express your answer in the form a + squareroot(b) where a and b are integers.

#### Solution

The area of the pink square is sqrt(18) – 3, or 3(sqrt(2) – 1).

The first thing to do is calculate the area of the octagon in terms of its side length x. We can cut the octagon into a square, four triangles, and four rectangles as shown. The triangles are right triangles with hypotenuse of length x, so their legs must be length [sqrt(2)/2]x. So each of the triangles has area x2/4. Adding everything together, we see that the octagon has area

x2 + 4[x2/4] + 4x[sqrt(2)/2]x = 2x2[1 + sqrt(2)]

This gives us the equation

2x2[1 + sqrt(2)] = 6

(since we know the octagon has area 6). So

x2 = 3/[1 + sqrt(2)]

But x2 is exactly the area of the square tile, so all we need to do is simplify this number. A little work at that yields our final answer.

#### Puzzler - (June 2015)

The puzzler for this issue of the Vandal Science News is a really fun logic puzzle. At first reading, it doesn't seem like there's enough information given – but there is! Can you put it together?

Jeff takes Abby and Brian to a lot where a dozen autos are parked, one of them being his. The autos in the lot are as follows:

- 2 Chevrolets (red and tan)
- 1 white Ford
- 3 Jeeps (black, white, and tan)
- 1 red Toyota
- 2 Nissans (red and tan)
- 1 white Hyundai
- 2 Buicks (blue and white)

He tells Abby the color of his automobile and tells Brian its make, and challenges the two of them to figure out which one belongs to him without revealing the make or color directly. The following conversation then occurs:

Abby: “I don’t know which one belongs to Jeff.”

Brian: “Even knowing that you don’t know, I still don’t know.”

Abby: “Oh! Now I know.”

Which auto is Jeff’s?

#### Solution

Jeff owns the white Jeep.

The table here helps to organize the logic.

- Abby knows the color, so when she announces that she doesn’t know which car is Jeff’s, everyone can eliminate the blue and black cars – that is, the pink-shaded columns are thrown out.
- Now Brian only has to consider the three remaining columns, so when he says that knowing the model still isn’t enough, that means we can throw out the rows that have only a single car still in consideration, shaded here in blue. (Note that we can throw out the Buicks because even though there are two of them, one of them was already eliminated by Abby’s statement – if Jeff’s car were the white Buick, Brian would know it.)
- So now Abby only needs to consider the six remaining cars, highlighted here in yellow. She says that she
*does*now know which car is Jeff’s, so it must be the only remaining car of its color. But there are still two red cars and three tan cars under consideration. The only choice with a unique color is the white Jeep!

#### Puzzler - (Nov. 2015)

We’ll play a simple game for this Puzzler. Suppose we start with a pile of 13 pebbles. Two players take turns making moves, where each move consists of removing 1, 3, or 4 pebbles from the pile. (You must take away at least one pebble in your turn, and you cannot take away any number other than 1, 3, or 4.) The player to remove the last pebble wins. There is an easy-to-describe winning strategy for this game that can be discovered by some analysis. Figure out the strategy and then answer this question: if you are the first player, what should be your initial move?

#### Solution

Your move should be to **remove four pebbles, leaving 9** – here's why.

The status of the game at any time is indicated by the number of pebbles remaining, which can be anything from 0 to 13. Each of our moves will land on one of those numbers, and we'll keep track of the ones we want to land on with a pink circle.

- Now clearly we want to land on zero – that's how we win. We can do that if our opponent lands on either 1, 3, or 4 – so we'll put green squares around those – that's the first row in the diagram at right.
- Since position 2 isn't spoken for yet (that is, it isn't one we want to make our opponent land on), it must be one that we want to land on. (And notice: if we land on 2, then our opponent's only choice for his move is to remove one pebble, which means we would win in our next move!) But we can land on 2 for our move if we force our opponent to land on either 3, 5, or 6. Now 3 already has a green square, but we'll add squares to 5 and 6, which gives us the 2nd row.
- This means we want position 7 (the lowest available “unclaimed” spot) as one of our pink circles. This means we put green squares on 8, 10, and 11 as positions we want to force for our opponent. This gives the 3rd row.
- Position 9 is then left as one of our desired landing spots, and this then adds green squares to 12 and 13. But more importantly, since we know that landing on 9 is a winning move, we can see that our first move should be to go to that position by removing four pebbles!

So the game strategy becomes this: we always move to land on one of the pink circle positions. Our opponent will then have no choice but to land on a green square position, which in turn lets us get back to a pink circle. Since we're the only player that can get to a pink circle position, we're the only one who has a chance to win!

#### Puzzler - Folding Paper (Feb. 2014)

Take a square piece of paper measuring 12 inches on each side. Call the corners (in cyclic order) A, B, C, and D. Make four folds in the paper as follows:

- Fold from the midpoint of AB to the midpoint of CD.
- Fold from the midpoint of BC to the midpoint of DA.
- Fold diagonally from A to C.
- Fold from D to the midpoint of AB.

These folds should divide the square into nine regions. Find the area of the smallest of these regions.

#### Solution

The four folds and nine resulting regions are shown in the diagram. The area of the (pink shaded) smallest region is 3.

You can find that area lots of different ways – probably the most common approach would be to put the square on the coordinate plane with the origin at point A, find the equations of the various lines, and work it out in coordinate geometry. Here’s a different way that’s fairly simple and uses only a little geometric reasoning and elementary algebra.

- Label all of the points as in the figure, and let x represent the area of the small triangle PQR.
- Since the square is 12-by-12, the area of the whole is 144. The first two folds (between midpoints of sides) divide the square into four squares, each with area 36.
- It’s clear that Q must be the midpoint of HP, so HQD is a right triangle with legs of length 3 and 6. That makes its area 9.
- But triangle PQE is congruent to HQD, so its area must also be 9.
- Now notice that the triangle AER is similar to PQR, but double the scale. That means its area will be four times as large as the area of PQR, so area(AER) = 4x.
- The areas of AER and PRE must add to be 18 – half of the square AEPH. So:

area(AER) + area(PRE) =18 4x + (9 – x) =18 3x =9 x =3

#### Puzzler - Physics, a Bowling Ball Sized Rock and a Pool (June 2014)

The puzzler for this issue comes from Physics, and is appropriate to summer recreational pursuits.

Suppose you are floating on a rubber raft in a swimming pool. For some reason, you have a bowling ball sized rock with you on the raft. You decide you've had enough of holding the rock, so you put it overboard and watch it sink to the bottom of the pool. What happens to the water level of the pool?

A correct answer must include a short explanation. Have fun thinking this one through while you're poolside this summer!

#### Solution

As Archimedes would have known, the solution is all about displacement. The water level of the pool will drop a very small amount when the rock is thrown overboard.

The reason is this:

- While the rock is floating with the raft, its weight is displacing an amount of water with weight equal to that of the rock.
- Once it sinks, however, it displaces an amount of water with volume equal to that of the rock.

Now the rock is more dense that water (after all, we're told that it sinks), so the amount of water weighing the same as the rock will have a volume greater than the volume of the rock. This means more water was being displaced when the rock was in the raft, so the water level will be higher at that time. Eureka!

#### Puzzler - (Oct. 2014)

This issue's puzzler is a fun counting problem. An athletic league has ten teams. Each week these ten teams pair off into five matches. The league's scheduler wants to avoid having two weeks with the exact same slate of games. How long can he go without repeating? That is, how many different weekly schedules are possible?

#### Solution

There are 945 unique weekly schedules possible.

- Choose a team – there are 9 possible opponents for them to play.
- This accounts for two teams, so choose one of the remaining teams and there will be 7 possible opponents to which they can be matched.
- Similarly, choosing one of the unmatched teams, there will be 5 possible opponents for them.
- There are only 3 possible opponents for the next unmatched team we select.
- Finally, there are only two remaining unmatched teams, so they must be assigned to play each other.

The total number of choices we could make is then 9*7*5*3 = 945.

#### Puzzler - The Chemistry Class (Feb. 2013)

The students in a chemistry class compare notes and find that all of them have a math, biology, or geography class in addition to chemistry.

- Everyone with a geography class also has a math class, but nobody has both a geography class and a biology class.
- One out of every three students with a math class also has a geography class, and half of those with a biology class also have a math class.
- There are 14 students with a geography class, and 26 students with a biology class.

How many students are in the chemistry class?

#### Solution

There are 55 students in the Chemistry class. Getting that answer is a matter of organizing the information, and probably the easiest and best way to do that is with a Venn diagram such as the one shown here. From the clues tell us:

- The set of Geography students lies entirely inside the set of Math students.
- The Biology and Geography sets are disjoint.
- The 26 Biology students are split with 13 inside the Math set and 13 outside.
- The Geography set contains 14 students.
- The portion of the Math set outside of Geography has to contain 28 (two times as many as Geography), and 13 of them are already known to be inside the Biology set – so the remaining region must contain 15 students.

Adding it all up gives us our total of 55.

#### Puzzler - Paths on a Grid (June 2013)

Consider the five-by-six grid of squares shown here. We'll say that the length of each square side in this grid is one. There are lots of length 11 paths in the grid that start at the green dot in the lower left corner and end at the red dot in the upper right corner – each of these paths will consist of a sequence of eleven "up" or "right" moves. What is the probability that such a path, generated at random, will pass through the purple dot?

#### Solution

The probability of a path going through the purple dot will be the fraction

[# of paths passing through the purple dot] / [total number of paths from green dot to red dot]

The blue path shown above is a typical corner-to-corner path. Note that we can associate it with a sequence of “R” (right) or “U” (up) moves – 11 moves in all, of which six must be “R” and five must be “U”. So the total number of such paths is the number of ways of arranging six R’s and five U’s into a sequence. You might remember from algebra class that the answer to this is the “binomial coefficient” C(11,6). (It’s called a binomial coefficient because it would be the number in front of the x^{6}y^{5} term in (x+y)^{11} – the 11^{th} power of the binomial x+y.) You can find this number from the famous “Pascal’s Triangle”, or you can simply use the formula C(n,r) = n!/[r!(n-r)!]. In our case, C(11,6) = 11!/[6!5!] which comes out to be 462.

A typical path through the purple dot is shown in orange. Note that it really consists of two paths – one from green dot to purple dot and the other from purple dot to red dot. Using the same logic as we did above, the number of paths from green to purple is C(6,4) (since it corresponds to a sequence of six moves, four of which must be “R”) and the number of paths from purple to red is C(5,2). Computing these, we have C(6,4) = 6!/[4!2!] = 15 and C(5,2) = 5!/[2!3!] = 10. Each of the 15 paths from green to purple could be continued in any of the 10 ways from purple to red – so the total number of paths through the purple dot will be 15*10 = 150.

So now we can conclude that the probability of a path going through the purple dot is **150/462 = 25/77, or just slightly less than one-third**.

#### Puzzler - "Landlocked" Country (Oct. 2013)

This puzzler is an intriguing geography challenge. As you know, a country that has no ocean shoreline is said to be "landlocked". You need to cross at least one international border to get from a landlocked country to the Ocean. We'll say that a country is "doubly landlocked" if you have to cross at least two international borders to get from that country to an ocean. Another way to say this would be to say that a doubly landlocked country is landlocked and all of its neighbors are landlocked as well.

There are two doubly landlocked countries in the world. Name at least one of them.

#### Solution

The two doubly-landlocked countries are **Uzbekistan** and **Liechtenstein**. The map here shows Uzbekistan in red, and all of its neighbors in other colors. A quick scan verifies that no neighbor touches an ocean. Liechtenstein, on the other hand, is a tiny country (just 62 square miles) having only two neighbors, Austria and Switzerland, both of which are landlocked

#### Puzzler - A Cannon and Cart (March 2012)

Suppose a cannon is mounted on a cart, pointed perpendicular to the cart’s bed. Now suppose the cannon shoots a ball upward while the cart is traveling at a constant velocity in a straight line on a horizontal surface, as in the left half of the figure. If we disregard friction, most people would guess (correctly) that the ball will eventually land back in the cart.

But now suppose the cart is actually rolling freely down a hill subject to the influence of gravity. If the cannon is again fired, will be ball land *ahead *of, *behind*, or *exactly on* the cart? This isn’t multiple choice: credit will be given only if the answer is accompanied by an explanation!

#### Solution

The ball will land *in* the cart, the same as when the cart was moving on a level surface. Brian Hill (Chemistry, 1965) knew that the key is to look at the problem a bit askew – adopt a new axis system in which the “vertical” axis is perpendicular to the slope the cart is on. In this axis system, gravity now provides a force somewhat off of vertical, but it’s clear that this force will be acting on both the cart *and* the ball.

#### Puzzler - A Clock Face (July 2012)

Consider an ordinary circular clock face with the numbers 1 through 12 marking the hour positions. It is possible to draw two straight lines through the face, neither touching one of the twelve numbers, so that each region created has the same sum of hour numbers. Describe how to do this.

#### Solution

The answer is to draw one line from between the 10 and 11 to between the 2 and 3, and the other from between the 8 and 9 to between the 4 and 5, as shown here. You could try lots of “guess and check” work to find this, but the elegant method uses some mathematical detective skills:

- First, do we want two lines that cross (determining four regions) or non-crossing lines (determining three regions)? Well, the sum of the clock numbers is 1+2+3+…+11+12=78, which is not a multiple of four, but is a multiple of three – so the non-crossing lines is what we need.
- Then, since 78 divided by three is 26, we know we’re looking for sums of 26, and two of them will need to be with consecutive clock numbers. It isn’t hard to find that 11+12+1+2 and 5+6+7+8 work.

#### Puzzler - Add Another Square (Oct. 2012)

Suppose we start with a square of side length one. We then create a new larger figure by attaching four squares, each of side length 1/3, to the middle thirds of each of the four sides. This new figure now has 20 segments making up its boundary. We’ll create our third iteration by attaching small squares (side length 1/9) to the middle thirds of each of those 20 segments. The new figure is rather complicated and has 100 segments in its boundary. We could create an even more complicated figure by attaching tiny (side length 1/27) squares to each of those segments. If we continue this forever, what is the limiting area of the resulting figure? [Warning: there’s both a hard way *and* an easy way to do this!]

#### Solution

As we said, there is both an easy way and a hard way to do this one – we’ll give both here.

**Easy way** :Notice that as we add smaller and smaller squares in the various steps to create this shape, it seems to be filling in a new square, rotated 45 degrees from the original square. Also, the new square has a side length equal to the diameter of the original square. Since the original square’s diagonal is the square root of 2, **the area of the new square will be 2**.

**Hard way** :The mathematical purists among us may prefer to show analytically that the area is two. We can do this by writing the total area as an infinite series:

- The original square has area 1, so at stage 1 the area is 1.
- We create the stage two figure by adding four small squares, each with area 1/9 – thus: 1 + 4/9.
- At stage 3 we add 20 small squares, each with area (1/9)
^{2}– thus: 1 + 4/9 + 20/9^{2}. - At stage 4 we add 100 smaller squares, each with area (1/27)
^{2}= (1/9)^{3}. Thus: 1 + 4/9 + 20/9^{2}+ 100/9^{3}. - There’s an obvious pattern developing that becomes even more apparent with a little rewriting:

1 + 4/9 + 20/9^{2}+ 100/9^{3 }+ . . . = 1 + (4/9)[1 + 5/9 + (5/9)^{2}+ (5/9)^{3}+ . . . ] - If you remember infinite series from calculus, you might recognize the sum inside those brackets as a
*geometric series*. Then remembering that the sum of 1 + r + r^{2}+ r^{3}+ . . . is 1/(1-r), we can see that the sum in the brackets is 1/(1-(5/9)) = 1/(4/9) = 9/4. - So the area of the limiting figure is 1 + (4/9)[9/4] = 2.

#### Puzzler - Travel and Time (June 2011)

There is a place in the world where traveling nine miles in any direction will require you to set your watch ahead one half hour, then back one half hour, then finally ahead again one half hour (in order to keep the watch set correctly at all times). In what country is this place located?

#### Solution

The June Vandal Science News featured the following tricky Geography puzzler:

There is a place in the world where traveling nine miles in any direction will require you to set your watch ahead one half hour, then back one half hour, then finally ahead again one half hour (in order to keep the watch set correctly at all times). In what country is this place located?

The simple answer is that the place is located in India. More specifically, it is a very small enclave of India called Dahala Khagrabari that lies within an enclave of Bangladesh called Upanchowki that itself lies within an enclave of India called Balapara Khagrabari (that lies within Bangladesh). Add to that confusion the fact that India and Bangladesh maintain time zones separated by one half-hour, and you get the situation described by the puzzler.

See Cooch Behar district (Wikipedia) for more information, or an annotated map of Cooch Behar of this messy piece of the India / Bangladesh border.

#### Puzzler - Two Fuses and a Pack of Matches (Oct. 2011)

Suppose you have two fuses that each burn for exactly one hour. However, they don't burn at uniform rates, nor do they necessarily burn at the same rate. (That is, if you light the two fuses simultaneously they would both burn out exactly an hour later, but they may not always appear to be the same length at intermediate times, and would not necessarily be half their original length after a half-hour.) How can you use these two fuses (and a pack of matches) to time exactly 45 minutes?

#### Solution

You first light *both* ends of one fuse, and one end of the other. When the first fuse burns completely out you know that 30 minutes have passed and that the second fuse has 30 minutes of burn time left. At that moment you light the *second end* of that second fuse. With both ends burning, its remaining 30 minutes of burn time is shortened to 15 minutes, and when it burns completely out, a total of 45 minutes will have passed.