Challenging Math Problems

Terry Stickels

• Publisher: Dover Publications
• Year: 2015
• ISBN:
  9780486808574
• Edition: 1
• Topic: Mathematics
• Subtopic: Mathematics in Education
• Index: Mathematics

`Fun and highly formidable math problems and puzzles from noted puzzle creator Terry Stickels.` — Window on Resources
Two friends wish to meet for breakfast twice a month throughout the year. In how many ways can they choose those two days so that they never meet on consecutive days?
You want to measure 30 seconds and you have two pieces of string, each of which burns for 40 seconds. How can you accomplish this without bending, folding, or cutting the strings?
A positive whole number is divisible by 3 and also by 5. When the number is divided by 7, the remainder is 5. What is the smallest number that could work?
These are but a few of this book's assembly of the most challenging puzzles imaginable ― and they require no background in higher math, just good thinking skills. Terry Stickels, a well-known puzzle-maker, has compiled 101 of some of the best and most entertaining problems ever published. All of the challenges, which range from probability puzzles to dice games, have two things in common: each offers the `Aha!` moment of discovery that puzzle-solvers love, and they're all fun. Complete solutions for all puzzles explain every detail.

SOLUTIONS

1

Remove one student from each bench, including the one with four students to make a new bench of eight students. There must be seven full benches in order to do this. Therefore, there were seven benches of nine students plus the four for 67 students.

Following that there are some other ways to this. First:

Let	n	= Number of benches needed when seating eight per bench
	m	= Number of benches needed when seating nine per bench
	s	= Number of students

Writing the two equations for the numbers of students in terms of the seating on the benches, we have

s = 8n + 3

and s = 9m + 4.

Recognizing that the number of students is the same, we set the quantities equal:

8n + 3 = 9m + 4.

Then, noticing that n must be greater than m, we condense, to write

8n – 9m = 1.

Pulling out a common factor, we have

8(n – m) – m = 1.

Then, noticing that the quantity (n – m) is actually the difference in the number of benches used to seat the students (between the 8-per-bench and 9-per-bench approaches), we define the difference as

d = n – m.

Then, substituting the difference d and rearranging, we have

m = 8d – 1.

Substituting back in one of the original equations yields

s = 9(8d – 1) + 4, or

s = 72d – 5.

Since the number of students in a class must be a positive number, we see by inspection that d must be greater than 0. So, setting d to 1, to get the minimum number of students satisfying the equation, we have

s = 72(1) – 5 = 67

Minimum number of students = 67

Here is another way to solve this:

We know that 3 + 8x = 4 + 9y

subtract 8y from both sides

3 + 8(x – y) = 4 + y

The smallest value of (x – y) is 1

(since x is the = number of benches with eight students per bench and y is the number with nine per bench, so x > y because it will take fewer benches to seat more students)

So the smallest value is when (x – y) = 1 plug in the equation

3 + 8 = 4 + y

so y = 7.

That’s it!

x + 3 = y + 4

or x – y = 1

which, according to what I said above, is the smallest positive difference in the number of benches!

2

5/8. When you switch the marbles to their new boxes, you have the probability of having

W and BW or

B and WW

In the four cases,

1/4 the time you get a white marble by choosing A. 1/4 the time you get a white marble from choosing D. In the case of choosing B, you get a white marble 1/2 of 1/4 or 1/8 the time.

Total probability of a white marble: 1/4 + 1/4 + 1/8 = 5/8.

3

The watermelons weigh 100 kg at the end of the day. In the morning, the 200 kg of watermelons are 99% water. So the non-water part of the watermelons has a mass of 2 kg. At the end of the day, the melons are 98% water. The remaining 2% is still the 2 kg of non-water material (which does not change when the water evaporates). If 2% equals 2 kg, then 100% equals 100 kg. So, Molly has 100 kg of watermelons left at the end of the day.

4

From lowest to highest value:

5

–5. One way to solve this is to examine the difference between the numbers to see if a pattern develops.

Once you see a pattern, then work your way back to the top to find the first number. Keep in mind when finding the differences between numbe...