One familiar name is a Math
Puzzler winner in June

Courtney Lauer of St. Louis is getting to be a familiar name in the winner's circle of Young Saint Louis.com's Math Puzzler competition. But, she was the only one who answered all six questions correctly last month.

Entries in the June competition fell off some as most of you headed off to summer vacations. But, Mr. Math Puzzler reminds you to continue to think about having some fun with math in the summer. It will help you keep sharp for next fall's math classes.

In order to enter the contest this month, just click here. You fill out the attached entry blank and figure your answers. Then, mail your entry or entries to the address on the entry form. (Remember, your entry must be postmarked by the 15th of the month to be eligible to win.)

Answers for June, 2002, Math Puzzler Contest:

1. John wants to ship a baseball bat to his sister. The bat is 4 feet, 11 inches long. He places it in a rectangular box that is 5 feet long. When he takes it to the shipper, they can't send the package because it is too long. All dimensions of the package must be 4 feet or less in order to be shipped. How can he ship the bat with this shipper? (A hint: If you want help, maybe you should ask Mr. Pythagorus.)

Answer: Bat will fit in 4x4 box if put in diagonally.

The explanation: You can use Pythagorus' Theorem to check the answer. The side of the box (Side A) is 4' and the bottom of the box (Side b) is 4'. Then, consider the bat put in diagonally as the third side of a triangle. Pythagorus' formula says Side A squared plus Side B squared = Six C squared.

• 16 (4 squared) + 16 (4 squared) = 32
• The square root of 32 is approximately more than 5.6.
• That answer of 5.6' is more than long enough to handle the 4'11' bat.

 

2. There's a tile below that doesn't fit with the other three in the group. Which one doesn't belong? (Hint: Think about a math product.)

Answer: The 54/22 tile

The explanation: The common relationship of the tile numbers, except for Tile C, is that if the top number is separated and then multiplied, it equals the bottom number. For example: 8 x 6 = 48; 3 x 5 = 15 and 7 x 4 = 28. But, in Tile 3, 5 x 4 = 20, not 22.

 

3. Can you exchange one card from each pile to form three piles with equal sums? (Aces count as one.)

Answer: Numbers need to be rearranged
so each column totals to 15

The explanation: To find what common number is needed, you add the total of the three groupings (No. 1 = 13, No. 2 = 19, No. 3 = 13). That total is 45. Then divide 45 by 3 to see what common per-column total is; the common number is 15. Then just shift one card from each line until each column adds up to 15. There are several different combinations you could use to get the 15 for each column.

 

4. If warrior princess Mistar were to ascertain how many men and how many horses she has under her command by counting both legs and heads, she was could 45 heads and 120 legs. How many horses are under Mistar's command? (Hint: Think about setting up two different equations for this answer.)

Answer: 15 horses

The explanation: This involves two equations. In the first one, each man and each horse has 1 head so m + h = 45. In the second one, each man has 2 legs and each horse has 4 legs or 2m + 4h = 120. Now, we need to combine the equations and get rid of the m (for man) variable to focus on the horses. If you multiple the top formula by minus 2, you can do that.

-2(m + h = 45) or -2m - 2h = -90

2m + 4h = 120 or 2m + 4h =120 Then, subtract, leaving as m's cancel out

2h   30
-- = --
 2    2

or h = 15 horses

 

5. At a royal banquet, there are 15 knights around the table. Each knight clinked his mug with knights on his immediate left and right. How many times did mugs clink? (Hint: Mr. Math Puzzler asked a similar knight question in January, 2002. The answer was in the February, 2002, edition.)

Answer: 15 clinks

The explanation: If everyone clinks to the right, there are 15 clinks. If everyone clinks to the left, there are 15 clinks. But, the second round of clinks is redundant because everyone has already clinked with each other on the first round.

 

6. Distal had sufficient hay and corn to feed his six horses for only 30 more days of harsh winter, not enough for the remaining 75 days before Spring arrived. On the seventh day, before feeding time, Distal sold four of his horses. Will he be able to feed his remaining two horses for the rest of the Winter?

Answer: Yes

The explanation: You can prove this answer with straight math, no algebra. First you need to find out how many horse-days of feed Distal has. (Just multiply 6 (number of horses) by 30 (number of days) to get 180 horse-days of feed. Therefore, he uses up 36 horse days of feed to feed his six horses for six days. That means he's got 144 horse-days of feed left at the time he sells four of the horses. But, by that time, there are only 69 days left before winter is over. Since he now has only two horses to feed, he'll only need 138 horse-days of feed to get to the end of winter. He'll have 6 horse-days of feed left or enough to feed his two horses an additional 3 days.