Parade of Math Puzzlers winners
continued in December

There were nine more winners in the December Math Puzzler competition. And that's even though Mr. Math Puzzler threw in a somewhat unusual question about pies.

Actually, the total number of winning entries could have been higher. Two former Math Puzzlers winners also answered all six questions correctly but their entries were postmarked after the December 15 deadline.

Remember to get your entries mailed well ahead of the deadline. You can't be sure when people at the Post Office are going to get around to registering your mailing.

Of those who got all the puzzlers correct, one didn't want her name listed among the winners. The eight others who got all six questions correct were: Ben Harman, Nina Zanaboni, Amy Lange, Lauren Kloeppinger, Marc Painter, Jim Hakenewerth, Charlie Lowell and Leah Vandiver.

Those winners who get the extra bonus of a $10 Borders gift certificate were Nina Zanaboni, Jim Hakenewerth and Leah Vandiver.

The winners all stuck to their guns on the answer to Question 2 even though it was somewhat confusing. The answer was 10 2/3 pies. But, Mr. Math Puzzler didn't tell who ate that one big piece out of the last pie before they were distributed.

Our contest entrants understand that Mr. Math Puzzler sometimes comes up with some unusual questions....and answers.

With the nine winners in December, that means there have been 30 winners in the last three months of 2002. There were nine winners in October and 12 in November.

Young Saint Louis.com is always looking for new contestants for the Math Puzzlers. While your home on holiday break, why not suggest to your neighborhood friends that they join in the fun.

Before anyone enters the contest for the first time, they might like to check out some past questions and answers to learn how Mr. Math Puzzler thinks.

YSL.com started the Math Puzzlers nearly 1 1/2 years ago. To check past questions, go to the Past Stories tab at the top of the home page. Click on any month since September, 2001, and you can check both questions and answers. (The answers to one month's questions are available in the next month's edition.)

When you're ready, just click here for the January entry blank and questions.

Here are answers to the December, 2002, questions, including that funny Question No. 2.

December Math Puzzler Answers

1. Ryan is trying to arrange his toy soldiers in even rows with no individual soldiers left over. At five across, he had four left over; at six across, he had one left over; at eight across, he had one left over, and at 12 across, he had one left over. What is the smallest number that would give him even rows and how many in each row.

Answer: 49 and 7 rows

The explanation: Educated guessing will lead to this answer. We need to figure out what number will end up with rows where there are no soldiers left over. To start the comparison, lets start with the largest number, 12. One row of 12 with one left over is 13. But, 13 doesn't work for the other rows. Two rows of 12 plus one is 25. That doesn't work for the others either. Three rows plus one is 37. That doesn't work either. But, four rows of 12 plus one is 49. That would result in seven rows of seven each. And, if you divide 49 by the other rows, there's the right number left over.

 

2. Simon bought a number of pies. He gave one-eighth to Tweedledum and one-fourth to the Black Knight. He then gave four pies to his brother and was left with one-fourth of the original number. How many pies did he start with?

Answer: 10 2/3 pies.

The explanation: Here's a way to build a formula.
Tweedledum gets 1/8 of the total (X). Black Knight gets 1/4. Simon's brother gets 4 pies. And another 1/4 is left over. Add all those together and you get:

5/8X + 4 = X

then subtract 5/8th from each side and you get:

4 = 3/8X

then, we use a reciprocal to simplify the problem:

8/3 (4) = (3/8X) 8/3

that leaves:

32/3 = X or 10 2/3 pies

 

3. Following the logic used in the pattern below, fill in the missing number in the third group.

Answer: 24

The explanation: Using educated guessing, you try different mathematical maneuvers to see what works. The logic that works for all three involves multiplying the two numbers in the top squares and then subtracting the number in the third square. That leads to the number in the bottom triangle. Thus, 4 times 3 minus 6 equals 6 in the left-hand illustration. Then, 7 times 5 minus 5 equals 30 in the middle illustration. Using the same pattern in the right-hand illustration, 9 times 3 minus 3 equals 24.

 

4. There's a local candy store that charges some unusual prices. A chocolate bar costs $.50, a soda costs $.16, and a peppermint costs $.44. According to the same system, how much will a lollipop cost? (Hint: Sweets are made up of more than their ingredients. Their words are made up of different kinds of letters.)

Answer: 34 cents

The explanation: Using the hint, we know that words are made up of vowels and consonants. We need to figure out the types of letters in the name of each sweet. Then, assign a value for each vowel and consonant that results in the price of each sweet. In trial and error, you find that a value of 3 cents for each vowel and 5 cents for each consonant will result in the prices of the three sweets that we know. Then, figuring 3 vowels and five consonants for lollipop, the price would be 34 cents.

 

5. "I want two gallons of water for my horse," said Mongol to the stable owner.
     The stable owner replied, "I have a three-gallon bucket and a four-gallon bucket. I cannot measure out exactly two gallons."
     "I can," said Mongol.
     How can he measure exactly two gallons using a three-gallon and a four-gallon bucket.

Answer and explanation: Fill 3-gallon bucket and pour it into 4-gallon bucket. Then, fill 3-gallon bucket again and put in the 1 gallon needed to fill the 4-gallon bucket. What's left in the 3-gallon bucket is 2 gallons.

 

6. An outdoor amphitheater holds 120 adults or 144 children. If 90 adults are already inside, how many children can also be admitted?

Answer: 36 kids

Explanation: The 90 adults already inside make up 3/4ths of the amphitheater's capacity. Therefore, there's 1/4 capacity left to be filled with kids. One-fourth of the 144 kids that would fill the amphitheater is 36.