Mr. Math Puzzler was too tough;
No June winners

Every once in awhile, Mr. Math Puzzler comes up with a set of questions that are stump everyone. June was one of those times.

Actually, it was Puzzler No. 6 that was the one that stymied everyone. There were plenty of answers. But, none of them were as low as Mr. Math Puzzler. He found a way to reduce the number of weight pieces to just 5.

The lowest of your answers was 7.

One of the June entries was from past Puzzler winner Phillip Hsu. The 11-year-old figured out his answers while he was on vacation. He said he had to go to a library to print out his entry.

Why don't you bring Young Saint Louis.com along on your vacation?

That's one of the reasons Mr. Math Puzzler and Young Saint Louis.com are continuing the math competition through the summer. The Puzzlers are a good way for you to keep exercising your math skills while on vacation.

Working the Math Puzzlers is a good summer activity. You can have fun and stay sharp.

Why not get together with some of your friends. Each of you can print out the July entry form and answer the questions. Then you can mail in your entries and see who wins.

Remember, if you are a first-time entrant in the Puzzlers, you can review some past Puzzlers before entering.

YSL.com Puzzlers started in September, 2001. Using the Past Stories tab on the home page, pick a month and review past questions. Then, move to the next month for those answers.

By checking past questions and answers, you can see how Mr. Math Puzzler thinks. He is Wayne Hesse, a math teacher at Green Park Lutheran School in south St. Louis.

(When you're ready for the July Puzzlers, click here.)

Those who get all the July Puzzlers correct will be listed in the August edition. Also, the winners will have a chance to win a $10 Borders gift certificate. Up to three certificates are awarded to those who get all questions correct.

The Answers for June's Math Puzzlers

1. What is the next number in this logical sequence of numbers?

4   5   7   11   19   ___

Answer: 35

The Explanation: This answer involves finding a pattern to the gaps between numbers. The gaps can be explained by applying a power to the number 2. For instance, the gap between 4 and 5 is 2 to the power of zero or 1. The 5-7 gap is 2 to the power of 1. The 7-11 gap is 2 to the power of 2; the 11-19 gap is 2 to the power of 3. And finally, the gap between 19 and the answer is 2 to the power of 4, or 16. Then, 19 and 16 equal 35.

 

2. If you take one-fifth of the sum of 16 and a certain number, it will be twice as much as if you had subtracted 2 from that number.

Answer: 4

The Explanation: This answer could be arrived at by an educated guess. But, you could also set up a formula to get the answer.

16 + x
------ = 2(x-2)
   5

 

    16 + x
5 ( ------ )= 5(2x-4)
       5

 

16 + x = 10x - 20
    -x    -x
-----------------
16     =  9x - 20

 

16 + 20 = 10x - 20 + 20

36 = 9x

36/9 = 9x/9

4 = x

 

3. What 3-digit number is the square of a number, and when rotated 180 degrees (upside down), is also the square of another number?

Answer: 196

The Explanation: The five numbers that are still recognizable numbers when put upside down are 0, 1, 6, 8 and 9. From there, you go through the perfect squares until you get to the one that is a square of another. For instance, 1x2=2, 2x2=4 and so on until 10x2=100, the first 3-digit number. By keeping going, you get to 14x2=196.

 

4. What two numbers when added together equal 56 as a sum, and when multiplied together equal 768 as the product?

Answer: 24 and 32

The Explanation: You can set up two formulas and then find the answers by using a substitution method. One formula is x + y = 56 or another way would be x = y - 56. The formula for the larger number is xy = 768. Then, you do a series of substitution formulas to reduce those to y being 32 and x being 24. Added they equal 56 and multiplied they equal 768. Then, by using the substitution method and then solving the resulting quadratic equation, you will get the two numbers, 24 and 32.

 

5. Timothy and Urban play a game with two dice. But, they do not use the numbers. Some of the faces are painted red and the others blue. Each player throws the dice in turn. Timothy wins when the two top faces are the same color. Urban wins when the colors are different. Their chances are even. The first di has 5 red faces and 1 blue face. How many red and how many blue faces are there on the second di? (This is a probability question.)

Answer: 3 red and 3 blue

The Explanation: This is a probability question. With two dice, there are 36 possible outcomes, 18 for each di. To have equal chances, that means both need 18 possibilities. Timothy's di has 5 red faces and 1 blue face. If Urban's di had 1 red face and 5 blues, there would be just 10 chances. If Urban's di had 2 red and 4 blue faces, there would be 14 possibly outcomes. If Urban's di had 3 red and 3 blue faces, the possible chances reach 18, a standoff.

The formula:

Timothy: 5 red and 1 blue
Urban:   3 red and 3 blue
         ----------------
         15       + 3 = 18

 

6. Julie owns a good balance, but no weights. She decides to make her own weights by cutting a 121-gram bar of metal into smaller pieces. She has a system enabling her to weigh all whole numbers of grams from 1 through 121. How does Julie divide the bar? What is the least number of pieces possible?

Answer: 1, 3, 9, 27, and 81

The Explanation: Obviously, Julie could be able to weigh to each gram by cutting the bar of metal into 1-gram sizes. But, then, she'd have 121 pieces. The task is to find a series of pieces that can span every weight but be the smallest number of pieces. One way to do that is to again use a base number and then apply a power number. For instance, if you start with 2 and multiply by a power of zero, power of 1, power or 2, etc. This doesn't quite fit. But, if you apply those same powers to the number 3, it works out exactly. 3 to power of zero is 1; 3 to power of 1 is 3; 3 to power of 3 is 9; 3 to power of 4 is 27, and 3 to power of 5 is 81. With this combination, you can cover each number by adding or subtracting those five weights. Another final proof is that the five weights also add up exactly to 121.