Tuesday, October 23, 2012

Full Solution to the Three -Question Quiz Problem

First a re-statement of the brainstretcher:

Thirty students took a three-question quiz with the following results:

20 got the first question correct
16 the second question
10 the third question
11 got the first and second correct
7 the first and third
5 the second and third
4 got all three correct

How many missed all three?

For this we will eschew set theory and tempting as it is a Venn diagram (plus I don't know how to draw a Venn diagram in my blog).  We will just be very good little accountants.

Start with the original 30 students and subtract out those that got each of the problems correct.

30 - (20 + 16 + 10), of course this is negative but hold your horses.

Now add back in the ones who solved two problems because they were counted twice in the first number that was subtracted.

30 - (20 + 16 + 10) + (11 + 7 + 5)

But wait, you say!!!!  What about the ones who got all three questions correct?

AHA!!! this number was accounted for all over the place in our last expression: three times in the first parenthesis (subtracted), and three times in the second parenthesis (added).  They cancel each other out.  So we only have to subtract out the last 4 to make our expression complete.

30 - (20 + 16 + 10) + (11 + 7 + 5) - 4 = 3

Therefore only 3 students failed to solve any of the problems

Friday, October 19, 2012

Wine, Wine and More Wine solution.

Okay Puzzlers,

It seems my little experiment where I scheduled a post from my blog to Facebook last Friday worked.  So here is the solution to the wine container problem.

First a re-statement of the problem: 

3 wine containers:   Container C - 7 quarts, full
                             Container B - 13 quarts, full
                             Container A - 19 quarts, empty

I want to eventually have 10 quarts in one of the containers.


As I said before, the best I could do was 15 pourings:  I will show the pourings and what is in each of the 3 containers (A, B, C)

1.  Pour contents of 7 quart can into 19 quart can  (7, 13, 0)
2.  From the 13 quart can fill up the 19 quart can (19, 1, 0)
3.  From the 19 quart can fill the 7 quart (12, 1, 7)
4.  7-quart into 13-quart (12, 8, 0)
5.  From 19 quart fill 7 quart (5, 8, 7)
6.  From 7 quart fill 13 quart (5, 13, 2)
7.  Pour contents of 13 quart into 19 quart (18, 0, 2)
8.  Pour contents of 7 quart into 13 quart (18, 2, 0)
9.  From 19 quart fill 7 quart (11, 2, 7)
10. Pour contents of 7 quart into 13 quart (11, 9, 0)
11. Fill 7 quart from 19 quart (4, 9, 7)
12. Fill 13 quart from contents of 7 quart (4, 13, 3)
13. Pour content of 13 quart into 19 quart (17, 0, 3)
14.  Pour content of 7 quart into 13 quart (17, 3, 0)
15.  Finally from 19 quart pour into 7 quart (10, 3, 7)

You now have the 19 quart container holding 10 quarts!!!!!

Friday, October 12, 2012

Puzzler Problem for Friday, October 12, 2012

Okay Puzzlers,

I am trying something different this week in that I am away from computer (and like it that way) but want to post my brainteaser anyway.  I have scheduled this one and we'll see if it shows up on time.

Wine, Wine, and More Wine

This puzzle is a slightly harder version of the Two Container Problem.  This time we have three containers of wine: 

7 quart - full of Merlot

13 quart - full of Merlot

19 quart - empty

Your job is to end up with one container holding exactly 10 quarts.  As with all the other multiple container problems, there are no markings on the containers, so you must achieve  the desired result by clever pouring back and forth ffrom one container to another.

Hint: The best I could come up with was 15 pourings.

Monday, October 8, 2012

Solution to a Dicey Problem

Okay, let me reiterate the problem.  Three dice (red, blue, and green) are tossed and the values noted.  Then we perform the following operations on the values.  Start with red: multiply the red value by 2 then add 5 then then multiply this result by 5.  Now add in the value of the blue and multiply by 10.  Now add in the green.

Your result is 484.

What are the original values of the dice?

Solution Red = 2, Blue = 3, Green = 4

Why that's amazing you say.  How did you ever figure that out?

Funny you should ask.  

Assign variables to the values of the dice; Red = a, Blue = b, Green = c.  Now just follow the directions.

Multiply red by 2     2a

Add 5    2a + 5

Multiply this by 5    5(2a + 5) = 10a + 25

Add in Blue     10a + 25 + b

Multiply by 10     10(10a + 25 + b) = 100a + 250 + 10b

Add in green 100a + 10b + c +250

This equals 484    100a + 10b + c + 250 = 484

Subtract 250 from both sides    100a + 10b + c = 234

But the left hand side is the natural form of a three digit number (hundred, tens, ones)

Sooooooooo a = 2, b = 3, c = 4

And we could do this for any three values on the dice.  Try it for yourself.

Friday, October 5, 2012

Teddy Roosevelt

In the past few weeks, I've been using this blog and other social networking to get out some feelings I have about my mother's heath.  She is seriously ill, but has an excellent chance of recovery - the doctor say the course of medicine she is on has a 70% recovery record.  Truth is though, my mom is 82 and the medicine is supposed to make her sick along with making her better.  She is this little southern woman who lately has seen even smaller as her health has failed.

The good news is that women in our family live to a ripe old age.  My grandmother lived almost to 100 and mom has 4 sisters all of whom are either in their eighties or close to it. With any luck and some providence, I should get to enjoy my mother's company, if not for many, at least for a few years to come.

Which strangely enough, brings up the subject of Teddy Roosevelt.

Recently I had lunch with a friend, and a fan of Bonnie Pinkwater.  We'd been corresponding and trading lunches since last March when we'd run into each other at Left Coast Crime in Sacramento.  While enjoying a meal at the Garden of the Gods Trading Post, he shared the titles of a few good books he'd enjoyed in the past year.  One of them was 'River of Doubt'by Candice Millard.  The book is non-fiction and relates the 1000 mile journey the past president took on an unexplored tributary of the Amazon.  I am currently reading - actually listening to - the book in my car.

Now, I am a student of history, minored in the subject in college.  I know the usual stuff about Teddy and maybe some not so usual stuff.

Face on Mount Rushmore

"Talk softly, but carry a big stick"  His response to the Barbary pirates who were plaguing American ships.  Think 'From the Halls of Montezuma, to the shores of Tripoli.'

Twenty-Sixth president of the United States of America.  Youngest in history.

Leader of the 'Rough Riders' in the Spanish American War.  Led the famous charge up San Juan Hill.

Came from a prominent New York family, was Chief of Police of New York City.  Was instrumental in cleaning up the corruption that had become rampant in the police ranks.

Was asthmatic and did rigorous exercise to get over his infirmities.  A practice he continued all the days of his life.

Boxed in college and had a boxing ring set up in the White House.

Came out west to clear his head when tragedy struck him or the people he loved.  We'll visit this theme again in a minute.

Was one of the premier naturalists of his time and donated to the collections of the famous museums of his day.

Tried and failed to run for a third term of presidency (losing to Woodrow Wilson) in the Progressive or "Bull Moose Party'.

Was instrumental in setting up the National Park system we enjoy to this day.

All of which is fascinating, but it isn't the reason for this post.  I began it speaking of my mother and her recent struggles with her health.  While I am confident she will beat this round of cancer, She is eighty-two and my father died at eighty-one.  Someday, I will have to live in a world where this sweet women doesn't call me weekly on the phone.

So here's the reason for this post.  In 1884 Teddy Roosevelt lost not only his wife but also his beloved mother.  And not just in the same year but in the same day.  I can only imagine the grief he felt at this monumental loss.

Here was a man who played out life on a national stage, was the hero of millions, was one of the most popular presidents ever to hold the office and he was powerless in face of these twin tragedies.

So here's the thing.  Life throws hard things at you sometimes.  Sometimes those hard things threaten to overwhelm.  Sometimes, you just feel like going off to a dark place to weep.  And I suppose that's okay.

For a while.

But in the end, Roosevelt remarried, raised four children, and got on with his life.  I don't know about you but this makes me hopeful that all of us can do the same.

Life is good.

Solution to Liar, Liar

Dear Puzzlers,

If you haven't been here before, welcome to my blog.  Please, if you are so inclined, feel free to to follow me.  I promise to entertain and hopefully edify you.

And now onto the solution.  Please remember, this particular puzzle had nothing to do with solving for a a number.  Your task was to identify the liar.  Lets' restate the problem:

Four friends make statements about a number

Andrew: It has two digits

Barbara: It goes evenly into 150

Cindy: It is not 150

Daniel: It is divisible by 25.

One more thing there is exactly one liar, no more, no less.

Our Solution:
 By cases - Andrew: Suppose Andrew is the liar and the number has either one digit or more than two.  If it has one digit, then Daniel is a liar also since it cannot be divisible by 25.  If it is three or more digits then Barbara is a liar since it can't go evenly into 150

Therefore Andrew is not a liar!!!!

Barbara: Suppose Barbara is the liar, then our number does not go evenly into 150.  Then either Andrew or Daniel is a liar.  The two digit multiples of 25 are 25, 50, and 75 all of which  go evenly into 150.  

Therefore Barbara is not a liar!!!!

Cindy:  Suppose Cindy is the liar.  Then the number is 150.  Then Andrew is a liar since the number has more than two digits.

Therefore Cindy is not a liar!!!!

Daniel:  Suppose Daniel is the liar.  Then our number is not divisible by 25, say 15.  It is easily shown that each of the other three are telling the truth.

Daniel is a lying Punkbaby!!!

Monday, October 1, 2012

Liar, Liar, Pants on Fire

Four friends: Andrew, Barbara, Cindy, and Daniel were shown a number.  They then made the following startements:

Andrew:  It has two digits

Barbara:  It goes evenly into 150

Cindy:  It is not 150

Daniel:  It is divisible by 25

Exactly one of the four friends is lying.  Which one is it?  You are not responsible for figuring out the actual number, but if you want to come up with a number that works, go for it!