Oct
31
Number Series #1
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The Problem
What are the next two numbers in this series? This is a purely arithmetic series.
30, 33, 42, 50, 55, 80, 88, 152, 162, 174, ?, ?
The Solution
202, 206. To figure out the next number in the series multiply each of the non-zero digits of the previous number and add them to that previous number.
Oct
31
Bridge Crossing
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The Problem
There are 4 women who want to cross a bridge. They all begin on the same side. You have 17 minutes to get all of them across to the other side. It is night. There is one flashlight. A maximum of two people can cross at one time. Any party who crosses, either 1 or 2 people, must have the flashlight with them. The flashlight must be walked back and forth, it cannot be thrown, etc. Each woman walks at a different speed. A pair must walk together at the rate of the slower woman’s pace.
- Woman 1: 1 minute to cross
- Woman 2: 2 minutes to cross
- Woman 3: 5 minutes to cross
- Woman 4: 10 minutes to cross
For example if Woman 1 and Woman 4 walk across first, 10 minutes have elapsed when they get to the other side of the bridge. If Woman 4 then returns with the flashlight, a total of 20 minutes have passed and you have failed the mission. What is the order required to get all women across in 17 minutes? Now, what’s the other way?
The Solution
Let’s think about this. How many different ways can we add the numbers 1, 2, 5 and 10 together to get 17? There’s only one combination that works: 10, 2, 2, 2, 1. This tells us a few things.First, ten and five cross exactly once and they cross together (ten “masks” the five because ten is larger/slower). We also know that the exercise will take place in five steps. We know that two crosses three times and one crosses at least once (because it could cross with two and therefore be “masked”). I use the term masking, but it’s kinda like big-O notation in a way.
That all said, here are the two solutions.
Send over women one and two. Send back woman two. Woman two sends women three and four across the bridge together. Woman one then returns for woman two. Women one and two then cross back over together. Total time == 10 + 2 + 2 + 2 + 1 == 17 minutes.
Send over women one and two. Send back woman one. Woman one sends women three and four across the bridge together. Woman two then returns for woman one. Women one and two then cross back over together. Total time == 10 + 2 + 2 + 2 + 1 == 17 minutes.
Oct
31
Four Jars of Pills
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The Problem
You have 4 jars of pills. Each pill is a certain weight, except for contaminated pills contained in one jar, where each pill is weight + 1. How could you tell which jar had the contaminated pills in just one measurement?
The Solution
We have to assume a unit weight of one for uncontaminated pills to solve this. Take one pill from the first jar, two pills from the second, three pills from the third and four pills from the fourth. You now have ten pills. Put all ten pills on a scale. Subtract ten from the weight and you have the number of the jar with poison pills.
Oct
31
The Problem
You have two jars, 50 red marbles and 50 blue marbles. A jar will be picked at random, and then a marble will be picked from the jar. Placing all of the marbles in the jars, how can you maximize the chances of a red marble being picked? What are the exact odds of getting a red marble using your scheme?
The Solution
Fill the bottom half of each jar with blue marbles. Then fill the top half of each jar with red marbles. You then have a 100% chance of picking a red marble.
Oct
30
San Francisco Scramble
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The Problem
The San Francisco Chronicle has a word game where all the letters are scrambled up and you have to figure out what the word is. Imagine that a scrambled word is 5 characters long:
- How many possible solutions are there?
- What if we know which 5 letters are being used?
- Develop an algorithm to solve the word.
The Solution
This is a fantastic way to phrase the question. There are 265 solutions (each letter can be one of 26 possible values). If we know which five letters are being used there are 55 possible solutions.
Oct
30
Rope Burns
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The Problem
A rope burns non-uniformly for exactly one hour. How do you measure 45 minutes, given two such ropes?
The Solution
Let’s call the ropes A and B. Light rope A at one end and rope B at both ends. B will burn in exactly 1/2 hour. When B burns out light the unlit end of rope A. Rope A will burn out 15 minutes later. Total elapsed time, 45 minutes.
Oct
30
Two Trains
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The Problem
One train leaves Los Angeles at 15mph heading for New York. Another train leaves from New York at 20mph heading for Los Angeles on the same track. If a bird, flying at 25mph, leaves from Los Angeles at the same time as the train and flies back and forth between the two trains until they collide, how far will the bird have traveled?
The Solution
At first blush this seemed like a limit problem. But it’s not. It’s a vector problem. Sum the train vectors. The trains will collide in the amount of time it takes to travel the distance between them (d) at 35mph. So, we end up with a term t = d/35. The only important part about the bird is that it travels at a constant speed, 25mph. So the amount of distance covered by the bird will be the how far it can go in time t at 25mph. So, the solution to this, d’, the distance traveled by the bird is 25 * t. To put all that together, d’ = 25d/35.
Oct
30
Six Shoes and Twenty-Four Socks
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The Problem
You are in the dark, and on the floor there are six shoes of three colors, and a heap of twenty-four socks, black and brown. How many socks and shoes must you take into the light to be certain that you have a matching pair of socks and a matching pair of shoes?
The Solution
Let c := the number of colors of socks and h := the number of colors of shoes. The number of shoes you have to take into the light is h + 1 (which is 4). The number of socks you have to take into the light is c + 1 (which is 3).
Oct
30
Ten Gnomes
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The Problem
Ten gnomes are about to be executed. Although they don’t like the idea, they are each selfless and want to do anything (even by sacrificing themselves) in order to help their fellow gnomes. They are told what will happen to them: They will each be lined up single file, so that they are each facing the gnome in front of them. Each of them will be given a red or a white hat on their head, and from the back of the line (the gnome who can see everyone else) they will ask him to state his hat color, ‘red’ or ‘white.’ If he can state it correctly (he cannot see his own hat, only those in front of him), he is allowed to live. Knowing what is going to happen to them, they are allowed to devise a strategy beforehand. How many people can they guarantee to save, and what strategy will ensure this? (There are NOT five reds and five whites necessarily!)
The Solution
The fact that they’re willing to sacrifice themselves is a hint. There is a 50/50 chance of saving 10. 9 can be saved for sure. Each gnome will shout the color of the hat on the head of the gnome in front of them. If the color of the hat on the gnome at the back of the line is the same as the color of the hat on the head of the gnome in front of him all 10 gnomes will be saved. Otherwise only the front-most 9 will live.
Oct
30
One From Five and Three
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The Problem
A milkman has two empty jugs: a three gallon jug and a five gallon jug. How can he measure exactly one gallon without wasting any milk?
The Solution
Fill the three gallon jug. Empty it into the five gallon jug. Fill the three gallon jug again. Fill the five gallon jug from the three. You’re left with one gallon of milk in the three gallon jug and five gallons in the five gallon jug.