thescienceofdeductionandmore

[thescienceofdeductionandmore]

Anna's Special Birthday

Anna thought to herself:

“If you add up the digits in my birthday (stated in month/date format) and then add the digits in the result until you have a one-digit number, you get 3. If you do that with John Lennon���s birthday, you get 1. If you do it with Paul McCartney’s birthday, you get 5.

"Oh, wow! My birthday is the average of John and Paul’s birthdays. That’s amazing. I have like the best birthday ever. Yay, for my special birthday.”

Anna’s birthday is in the first half of December. What is it?

[thescienceofdeductionandmore]

A Very Special Poem

This

Poem

Is special

Why is that?

What’s so special about it?

It’s not filled with rhymes. That’s not it.

Maybe it has a very special meter. But, on second thought, it doesn’t.

There’s something special about this poem, but these lines are getting way too long so I think I’ll end it here.

[thescienceofdeductionandmore]

The Ellie and Her 2% Milk Puzzle

Poor Ellie. Everyone in her house drank a different kind of milk. Her baby brother drank whole milk. Her mother and father drank skim milk. Her two sisters drank 1% milk. Ellie liked 2% milk but she was the only one who did.

At least two other people drank both skim milk and 1% milk, so those milks made the shopping list. True, only the baby drank whole milk, but babies always come first, so he got his whole milk. But poor Ellie’s beloved 2% milk did not make the shopping list.

“There’s no room in the refrigerator for another milk,” her father would say.

“Just mix whole milk with skim milk and call it 2% milk,” her mother would say.

“That’s not the same!” said Ellie. “I want my milk to be exactly 2%.

“So calculate it,” said her dad. “You’re good a math. Make your own exactly 2% milk.”

“Okay,” said Ellie.

Ellie was good at math. She knew a glass of whole milk has 8 grams of fat, a glass of skim milk has 0 grams of fat, and a glass of 2% milk has 5 grams of fat. So she knew exactly how much skim milk to mix with a glass of whole milk to make her beloved 2% milk.

How much?

[thescienceofdeductionandmore]

Anna Turns 20!

Anna woke up on her 20th birthday. “Wee,” she exclaimed. “I’m 20. Yesterday I was 19 but today I’m 20.” 

“Wait! How did I gain an entire year in one day? That doesn’t make sense.” she thought to herself. Then she drank some coffee, got the cobwebs out of her brain, and remembered that that isn’t how birthdays work.

She looked around her room, wondering what 20-year-olds are supposed to do. After all, she had never been one before. She ran over to the full-length mirror on her wall and looked at herself. “Nope. I didn’t get taller since yesterday,” she thought to herself. “Oh well.”

So she sat down on her bed and decided to come up with as many fun facts as she could about the number 20.

This is what she came up with:

20 is a pronic number. [Who knows what that means?]

20 is a tetrahedral number as 1, 4, 10, 20. [Who knows what that means?]

20 is the basis for vigesimal number system.[Who knows what that means?]

20 is the third composite number to be the product of a squared prime and a prime, and also the second member of the (22)q family in this form.

20 is the smallest primitive abundant number.

An icosahedron has 20 faces.

 A dodecahedron has 20 vertices.

20 can be written as the sum of three Fibonacci numbers uniquely, i.e. 20 = 13 + 5 + 2.

20 is the most number of moves required to optimally solve a Rubik's Cube in the worst case.

20 is the number of parallelogram polyominoes with 5 cells.

20 is the atomic number of calcium.

20 is the third magic number in physics.

20 is the number of legal opening moves in chess.

20 is the age - following the Babylonian captivity - of the Levites who were assigned "to oversee the work of the house of the Lord"

20 is the length of the period of a hockey game

20 is the telephone country code to dial Egypt

20 is the number of questions in 20 questions

When she finished coming up with her list, Anna was exhausted from all that thinking and fell back asleep. Then she had cake.

[thescienceofdeductionandmore]

Solution to the Apples for Sale puzzle

To remind you, here’s the puzzle again:

A version of this clever problem first appeared in England a century ago.

Two women were selling apples at the market. They each had an equal number of apples, but they were selling them at different prices. Mrs. Jones sold her apples at a price of 2 for a penny. Mrs. Smith sold hers at a price of 3 for a penny.

Mrs. Smith had to leave, so Mrs. Jones agreed to sell her apples for her. Mrs. Jones combined the two equal supplies of apples, and sold them at a price of 5 apples for 2 cents, which led to a slight reduction in the total proceeds.

When Mrs. Smith returned, the apples had all been sold. The proceeds were exactly 7 cents less than they would have been if Mrs. Jones had sold each supply of apples at its original price.

They divided the money equally. How much did Mrs. Jones lose compared to what she would have made if she had sold her own apples at their original price?

Solution:

The problem gives you 3 prices. The original prices charged by Mrs. Smith and Mrs. Jones, and the price all the apples were actually sold at. Start by converting all the prices into “cents per apple” format, so you can compare the prices on an “apples to apples” basis (so to speak).

Mrs. Jones sold them at 2 for a penny, so her price was 1/2 cent/apple.

Mrs. Smith sold them at 3 for a penny, so her price was 1/3 cent/apple.

The price they were actually sold at was 2 for 5 cents, so that price was 2/5 cent/apple.

   Jones – 1/2 cent/apple.

   Smith – 1/3 cent/apple.

   Actual price – 2/5 cent/apple.

To compare fractions with different denominators we need to determine the least common denominator and convert them. In this case, the least common denominator is 30. Once we convert them, the prices are:

   Jones – 15/30 cent/apple.

   Smith – 10/30 cent/apple.

   Actual price – 12/30 cent/apple.

Now that we made the 3 prices easy to compare, we immediately see what’s going on. The actual price they were sold at is higher than the Smith price and lower than the Jones price. So on every apple sold, Smith made extra money and Jones lost some. The next step is to calculate how much.

We can do this by calculating the price all the apples should have been sold at to collect the exact amount of money that would have earned at the original prices. Since there are an equal number of “Smith apples” and “Jones apples” the correct price to have sold them at would have been the average of the two prices (10/30 and 15/30). To make it easy to average the two numbers, let’s convert the denominator from 30 to 60. Now we take the average of 20/60 and 30/60 and quickly determine that the apples should have sold at a price of 25/60 cents/apple.

One final step and then we can solve it. Let’s convert the price the apples were actually sold at – 12/30 cent – to the same denominator. The price the apples were sold at was 24/60 cent per apple.

Once we take that step, the key to the solution now jumps off the page and whacks us on the head. The apples should have sold at 25/60 cent/apple. They were sold at 24/60 cent/apple. So for every apple sold, 1/60th of a penny was lost.

The rest is easy. We’re told in the problem that they lost a total of 7 cents. Since they lost 1/60th of a penny per apple, we divide 1/60th into 7 and determine that they started with a total of 420 apples.

Since they had an equal number of apples, we know Mrs. Jones had 210 apples. Now all we need to do is calculate how much money she lost. Since her original price was 30/60th of a cent, and they were sold at a price of 24/60th of a cent, we know she lost 6/60th (or 1/10th) of a cent per apple. Multiply 1/10th of a cent by her 210 apples, and we see that Mrs. Jones lost 21 cents! (Since overall they lost 7 cents, then Mrs. Smith made 21-7 or 14 cents.)

 Answer: Mrs. Jones lost 21 cents.

[thescienceofdeductionandmore]

Take another look at what the loss was on each apple sold -- by calculating the AVERAGE price an apple should have been sold at at the original prices, and subtracting from it the price each apple was actually sold at.

The Apples for Sale puzzle

A version of this clever problem first appeared in England a century ago.

Two women were selling apples at the market. They each had an equal number of apples, but they were selling them at different prices. Mrs. Jones sold her apples at a price of 2 for a penny. Mrs. Smith sold hers at a price of 3 for a penny.

Mrs. Smith had to leave, so Mrs. Jones agreed to sell her apples for her. Mrs. Jones combined the two equal supplies of apples, and sold them at a price of 5 apples for 2 cents, which led to a slight reduction in the total proceeds.

When Mrs. Smith returned, the apples had all been sold. The proceeds were exactly 7 cents less than they would have been if Mrs. Jones had sold each supply of apples at its original price.

They divided the money equally. How much did Mrs. Jones lose compared to what she would have made if she had sold her own apples at their original price? 

[thescienceofdeductionandmore]

The Apples for Sale puzzle

A version of this clever problem first appeared in England a century ago.

Two women were selling apples at the market. They each had an equal number of apples, but they were selling them at different prices. Mrs. Jones sold her apples at a price of 2 for a penny. Mrs. Smith sold hers at a price of 3 for a penny.

Mrs. Smith had to leave, so Mrs. Jones agreed to sell her apples for her. Mrs. Jones combined the two equal supplies of apples, and sold them at a price of 5 apples for 2 cents, which led to a slight reduction in the total proceeds.

When Mrs. Smith returned, the apples had all been sold. The proceeds were exactly 7 cents less than they would have been if Mrs. Jones had sold each supply of apples at its original price.

They divided the money equally. How much did Mrs. Jones lose compared to what she would have made if she had sold her own apples at their original price? 

Click here for the solution.

Kudos to @jfyfractal2 for solving it.

[thescienceofdeductionandmore]

Solution to the Two Dice Wager puzzle

To remind you, here’s the puzzle again:

I offer you a simple bet. You roll two dice. If the highest number showing is 1, 2, 3, or 4, you win. If the highest number on either die is 5 or 6, I win. Should you take the bet? Why or why not?

Solution:

If there were only one die, the odds would be in your favor. Your chances of  rolling a 1, 2, 3, or 4 would have been 4 out of 6 or 2/3.

The odds of two independent events occurring is the product of the odds of each event occurring.

So if the odds of you winning on one roll are 2/3, the odds of you winning on both rolls is 2/3 x 2/3 or 4/9.

4/9 is less than half. The odds of you losing are 5/9. There is a better chance you will lose than win.

Answer: Don’t take the bet.

[thescienceofdeductionandmore]

Noooooo!!!! Don’t take a losing bet. That’s how you go broke!

The Two Dice Wager problem

I offer you a simple bet. You roll two dice. If the highest number showing is 1, 2, 3, or 4, you win. If the highest number on either die is 5 or 6, I win. Should you take the bet? Why or why not?

[thescienceofdeductionandmore]

The Two Dice Wager problem

I offer you a simple bet. You roll two dice. If the highest number showing is 1, 2, 3, or 4, you win. If the highest number on either die is 5 or 6, I win. Should you take the bet? Why or why not?

Click here for the solution.

Kudos to @bluekino @jfyfractal2 and @itsernestok​ who solved it.

[thescienceofdeductionandmore]

Solution to the What’s My Birthday? puzzle

To remind you, here’s the puzzle again:

The day before yesterday I was 21.

Next year I will be 24.

What’s my birthday?

Solution:

This is a neat, and a bit tricky, puzzle. We need to fit four ages, 21, 22, 23, and 24, into what seems like a pretty short period of time.

One’s first thought might be that it involves February 29th and a leap year, but if you try that, it turns out it doesn’t work.

The obvious place to look is right around New Years, when the year changes.

If you try a couple of possibilities, you can quickly determine that today must be January 1st, and my birthday must be December 31st.

To check the answer against the problem:

1. The day before yesterday - December 30th - I was 21.

2. Yesterday - December 31st - I turned 22.

3. Today - January 1st - I am still 22 and will remain 22 until December 31st of this year - when I turn 23.

4. Next year - on December 31st - I will turn 24. 

 Answer: My birthday is December 31st.

[thescienceofdeductionandmore]

The What’s My Birthday? puzzle

The day before yesterday I was 21.

Next year I will be 24.

What’s my birthday?

[thescienceofdeductionandmore]

Solution to the Blue and Red Marble puzzle

To remind you, here’s the puzzle again:

You are sentenced to death. But the queen offers you a chance to live. (Yay.)

1.    She gives you 50 red marbles, 50 blue marbles, and 2 empty bowls.

2.    You must divide the 100 marbles into the 2 bowls any way you want so long as you use all the marbles.

3.    One bowl will be selected at random, and one marble will be selected at random from that bowl.

4. If the marble is blue you live.

How should you divide the marbles so that you have the greatest probability of choosing a BLUE marble?

Solution:

It’s obvious that if you divide the blue and red marbles evenly, 25 of each in each bowl, you’ll have a 50% chance of selecting a blue.

Similarly, if you put all the reds in one bowl and all the blues in the other, it’s obvious you’ll still have a 50% chance of selecting a blue.

Can you do better than 50%?  A good way to approach a problem like this is to look at the extremes and see how they affect the result. Instead of doing this in a half and half sort of way, let’s try an extreme imbalance. 

Let’s say you put a single red marble in one bowl and the rest of the marbles in the second bowl. The bowl with the single red marble will be selected 50% of the time, so a red marble will definitely be selected (and you’ll be executed) at least 50% of the time. But the bowl with 50 blue marbles and 49 red marbles will also be selected 50% of the time. And a red marble will be selected almost half of those times. (49 out of 99 times to be exact). 49/99 is roughly 50% and you’ll select that bowl 50% of the time. So a red marble from the second bowl will be selected around 50% x 50% or 25% of the time. Add the odds of each of the two events (a red marble being selected from the first bowl - 50% - and a red marble being selected from the second bowl - almost 25%)  occurring, and you realize that your probability of selecting a red marble (and unfortunately being executed) just skyrocketed from 50% to almost 75%.

That was a bad idea. But if going to that extreme made the odds skyrocket against you, going to the opposite extreme ought to do the exact opposite, right? 

If you put a single blue marble in one bowl and the rest of the marbles in the second bowl, the bowl with the single blue will be selected 50% of the time. So you’ll definitely select a blue marble 50% of the time. The second bowl with the 50 red and 49 blue marbles will also be selected 50% of the time. And you’ll select a blue marble almost half of those times. (49 out of 99 to be exact).  The math is the same; just the color is different. So your probability of selecting a blue marble (and living, yay!) just skyrocketed from 50% to almost 75%.

Since we solved the problem for the extremes, it’s obvious we have the correct answer. But just to be sure and make sure we didn’t miss anything, let’s take a tiny step back from the extreme and see what happens. Let’s say you put two blue marbles in the bowl instead of one. You’ll still have a 50% chance of selecting that bowl, so you’ll definitely select a blue marble at least 50% of the time. But what about the second bowl? Instead of the second bowl having 49/99 blue marbles, it now has 48/99. That bowl will still be selected 50% of the time, but of the times it is selected the probability a blue marble will be selected just went down a drop (from 49/99 to 48/99).  So we can see for sure that moving away from the extreme hurts us.

Answer: Put a single blue marble in one bowl and the other 99 marbles in the second bowl.     

[thescienceofdeductionandmore]

The Blue and Red Marble puzzle

You are sentenced to death. But the queen offers you a chance to live. (Yay.)

1.    She gives you 50 red marbles, 50 blue marbles, and 2 empty bowls.

2.    You must divide the 100 marbles into the 2 bowls any way you want so long as you use all the marbles.

3.    One bowl will be selected at random, and one marble will be selected at random from that bowl.

4. If the marble is blue you live.

How should you divide the marbles so that you have the greatest probability of choosing a BLUE marble?

Click here for the solution.

Kudos to @shrumpydumps​ and @this-is-trivial​ for solving it.

[thescienceofdeductionandmore]

Solution to the Let Them Eat Cake puzzle

To remind you, here’s the puzzle again:

Becky and Anelise are on a long walk and sit down to rest and eat cake. Becky had brought along in her backpack 3 cakes to eat and Anelise had brought 5.

As they are just about to eat, Ellie arrives without any cake and asks to share. Becky and Anelise agree. They cut each cake into three equal pieces, and each girl eats one piece of each cake.

After the cakefest, Ellie pays 8 coins for the cake she ate. Becky hands Anelise 5 of the coins and keeps 3 for herself. But Anelise complains, and says she should be given 7 of the coins with only 1 remaining for Becky.

Who is right and why?

Solution:

A simple way to solve this is to start by converting cakes to pieces, since every cake will be cut into 3 equal pieces. Then compare the starting point and the ending point.

At the beginning Becky has 9 pieces (3 x 3) and Anelise has 15 pieces (5 x 3).

They each eat 8 pieces of cake. Let’s look at the net movement of cake pieces.

Becky ate 1 fewer than she brought.

Anelise ate 7 fewer than she brought.

Ellie ate 8 more than she brought.

Since Ellie paid 8 coins for the 8 pieces, the pieces were valued at one coin per piece.

Becky had given up one of her pieces, so she was entitled to one coin. Anelise had given up 7 of her pieces, so she was entitled to 7 coins.

Answer: Anelise is right. (Becky had not considered the fact that she herself had eaten some of Anelise’s cake.)

[thescienceofdeductionandmore]

The Let Them Eat Cake puzzle

Becky and Anelise are on a long walk and sit down to rest and eat cake. Becky had brought along in her backpack 3 cakes to eat and Anelise had brought 5.

As they are just about to eat, Ellie arrives without any cake and asks to share. Becky and Anelise agree. They cut each cake into three equal pieces, and each girl eats one piece of each cake.

After the cakefest, Ellie pays 8 coins for the cake she ate. Becky hands Anelise 5 of the coins and keeps 3 for herself. But Anelise complains, and says she should be given 7 of the coins with only 1 remaining for Becky.

Who is right and why?

Click here for the solution.

Kudos to​ @thareqkun for solving it. ​

[thescienceofdeductionandmore]

Solution to the $97 Shirt puzzle

To remind you, here’s the puzzle again:

I borrow $50 from my mom and $50 from my dad, and buy a shirt for $97.

With the $3 change, I pay my mom back $1 and my dad back $1 and keep $1 for myself.

I owe my mom $49 and my dad $49 - that’s $98. And I have $1. That’s $99.

What happened to the extra dollar?

Solution:

Obviously, there’s no extra dollar. The problem just compared apples and oranges.

The apples were the money. The oranges were the money owed. If you don’t mix them up, they each add up perfectly.

The money: At the end, the store has $97. My mom, dad, and I each have $1. That adds up to $100.

The money owed: At the end, I owe my parents $98.

In the problem, I added the $1 I have to the $98 I owe. There was no basis for adding those two things together. It was mixing apples and oranges.