All-time high in Math Puzzler winners: 9

Students at St. Gabriel School were really on their game in the October Mr. Math Puzzler competition. There were nine kids who answered all six questions correctly. All the winners were from the same school in the City of St. Louis.

This is the first time we've had more winners than the allotted number of Borders' gift certificates. Under Puzzler rules, if there are more than three winners, we put all winning entries in a hat and draw out three who get the $10 gift certificates.

First, here are the names of those who got all six answers correct:

Dominic DaVasta, Leslie Ganer, Tim Hunt, Lauren Kloeppinger, Amy Lange, Tim McCann, Zane Reifsteck and Jeffrey Vaninger,

The three that survived the random drawing were Dominic DaVista, Tim McCann and Zane Reifsteck. They will receive their Border's certificates in the mail.

Congratulations to all of the St. Gabriel students who got all the answers right.

Teacher Amy Ruzicka has told her students they will get extra credit if they want to enter the Math Puzzler competition. Each month, she runs off copies of the entry blank and then mails all completed entries to YSL.com.

Why don't you ask your math teacher if he or she would do the same in your math class.

(To enter November's Math Puzzler contest, click here for questions and entry form.)

Here are the answers and explanations for the October Puzzlers:

The Math Puzzler Answers
(October, 2002)

1. If one-half of 12 were 8, what would one-third of 36 be?

Answer: 16

The explanation: This is a question that looks for the proportion between two numbers. This can be answered by setting up a simple formula. The problem says one half of 12 is 8 but in "real life" it's actually 6. One third of 36 in "real life" is 12. We're looking for a same proportion that an unknown number is to 12 that 6 is to 8. The formula:

     6   8
     — x —   becomes 6x = 96 then
     8   x

     6x   96
     —— = ——   or   x = 16
      6    6

 

2. What size square has a perimeter that is equal (in number only) to its area?

Answer: 4

The explanation: This is a problem in geometry. Remember, in a square, Area (A) equals one side (s) multiplied by itself or (A=sxs). In a square, the Perimeter (P) is obtained by adding the four sides (s) or (P = 4s). Therefore, you can use trial and error to see which side length will give you the same number if you multiple by 4 or multiple the number by itself. The number that works is 4.

 

3. a.) If the number of quarters I have is a multiple of 5, it is a number between 1 and 19.
b.) If the number of quarters I have isn't a multiple of 8, it's a number between 20 and 29.
c.) If the number of quarters I have isn't a multiple of 10, it's a number between 30 and 39.
What total number of quarters do I have?

Answer: 32 quarters

The explanation: This is a problem in logic. And it can be answered by trial and error. You want to figure out which number fits both parts of one of the questions. Start by laying out the numbers 1 through 39. Then try to find one that can be answered by one of the problems. The answer of 32 is one fits (c) because it isn't a multiple of 10 but its between 30-39.

 

4. Each morning a farm woman collects the eggs her hens have laid. One day, she stumbled as she left the coop and all the eggs were broken.

"How many eggs did you collect?" asked the daughter.

"I don't know," said the woman, "but I do remember that when I divided the number of eggs by 2, there was one egg left; when I divided the number by three, there were no eggs left, and when I divided by 5, there were three eggs left."

The woman had more than four eggs but fewer than 40. How many eggs were broken?

Answer: 33

The explanation: This is a problem of multiples and factors. You want to find one number that when divided by 2 leaves a remainder of 1 (x divided by 2 with r1); also when divided by 3 has no remainder (x divided by 3 with r0), and when divided by 5 has a remainder of 3 (x divided by 5, with r3). Using the last formula, you get 8, 13, 18, 23, 28 and 33. Then use those numbers with the other two formulas to see if they give you the right answers. The one that does is 33.

 

5. A man said to a friends, "I have three sons. They are all less than 10 years of age, but greater than one. The product of the ages of the two youngest equals the age of the oldest. The sum of their three ages equals a prime number. How old are my sons?

Answer: 2, 3 and 6

The explanation: Here's a problem where Mr. Math Puzzler likes to use a table to organize his calculations. The table has four columns, one for the youngest kid's age, one for the middle kid's age, one for the oldest kid's age and a fourth column so you can add the ages and see which series produces a primary number.

Youngest	Middle	Oldest	Sum
1	2	3	6 (all under 10 but not primary)
2	3	6	11 (all under 10 and primary)***
2	4	8	14(all under 10 but not primary)

 

6. On the Island of Odds, one third of the native people always lie, one third always tell the truth and one third are "normals" in that they sometimes lie and sometimes tell the truth. The chances of encountering any one of the three native people on the road on the island are the same. If a traveler meets a native on the road each of two successive days, what is the probability that at least one of the two native people is "normal?" (Hint: Make a diagram.)

Answer: 5/9ths

The explanation: This is a probability problem. Again, a chart helps to organize the material. T is for truth, L is for liar and N is for normal. The first day, a traveler has an equal chance of meeting one type of native. The second day, his has the same equal chance. If he met a truthteller the first day, the secnd day he'd have a one-third chance of meeting another truthteller, one-third a liar and one-third a normal. Then organize the chart this way:

1st meeting	2nd meeting
Truthteller ---	+-- Truthteller (T + T) | +-- Liar (T + L) | +-- Normal (T + N)
Liar ---	+-- Truthteller (L + T) | +-- Liar (L + L) | +-- Normal (L + N)
Normal ---	+-- Truthteller (N + T) | +-- Liar (N + L) | +-- Normal (N + N)