The
CENTURY
Problem

     What student wouldn't like to "make a 100"? Now, I ask you, don't you just love seeing that number at the top of your quiz and test papers? Of course, you do. But unfortunately, it doesn't happen all that often to enough of us -- if at all :( . However, don't give up; here is a way we can all make a 100, numerically speaking, that is.

     Observe this interesting numerical expression:

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8×9

     That all the digits 1 to 9 are used once and only once is immediately obvious, and they are arranged in the regular counting order. But that's not the big idea here. Compute the value and what do you get? 100, that's right! We did it! We made a 100.

     Can we do it again, and again, etc.? That's the challenge before us now. Luckily for us, the answer is "YES". Here are some more expressions that "make a 100":

1 + 2 + 3 - 4 + 5 + 6 + 78 + 9

1 + 2×3 + 4×5 - 6 + 7 + 8×9

[(1 + 2) ÷ 3] + 4.5(-67 + 89)

12 + 34 - 5 + 6 + 7 + 8 - 9

     In those four examples we saw some new tricks to achieve our goal: use subtraction, division, decimal points, exponents, and grouping symbols. And later we will see a few more variations of this.

     Since the value of all the expressions equals 100, and there are 100 years in a century, this numerical pastime has come to be known in my classroom as The Century Problem. Back in August 1935 the magazine Sphinx, devoted to recreational mathematics, published a table by G. Fistie of more than 100 ways of writing 100 by using the nine digits 1, 2, 3 4, 5, 6, 7, 8, 9 and mathematical symbols and operations. Here are a couple of those expressions:

(7 - 5)2 + 96 + 8 - 4 - 3 - 1

32 + 91 + 7 + 8 - 6 - 5 - 4

     You have noticed, I'm sure, that the digits are not in their natural order. Although this obviously makes the task a little easier, there is still lot of challenge left in this style. Here is one of my own efforts along this line:

31 + 54 + 9 + 6 × (8-7)2

     By means of getting accustomed to the idea, I recommend that you start out with the easier style, then graduate to the "in order" style. In his Recreational Mathematics Magazine (Feb. 1961), Joseph S. Madachy presented more than 100 ways to make 100 keeping the digits in order. That's the source of the first five expressions.

     Later in his book Mathematics on Vacation (1966), Madachy gives the problem again, plus another important variation: the digits in REVERSE order. Here is the first way he gave:

98 - 76 + 54 + 3 + 21

     In my classes I have given the challenge to my students. Here are some of their results.

               Laura:

(98 - 76)(5) - 4 × 3 + 2 × 1

               Crystal:

(9 + 8) × 7 - 65 + 43 + 2 + 1

9 × 8 + 76 - 54 + 3 + 2 + 1

               Valeria:

[98 - (7 + 6) + 5 × 4 - 3 - 2] × 1

               Eduardo:

(9 + 8)(7) - 6 - 5 - 4 - 3 - 2 + 1

9 × 8 × 7 ÷ 6 × 5 ÷ 4 - 3 - 2 × 1

     I've always liked Eduardo's second one; just look at all those multiplication and division signs in it. Nice.

New Style Century

     A real surprise came to me one day when Karla brought in a new twist to the old problem. Her idea was to use the squares of the numbers 1 to 10, in descending order. Here is her expression:

102 - 92 + 82 + 72 - 62 + 52 - 42 - 32 + 22 × 12

     I was so impressed that I gave her a new challenge: to try the powers of 2 and the cubes of the numbers. Again she came through with flying colors.

210 ÷ 29 + 28 - 27 - 26 + 25 + 24 - 23 - 22 - 21

103 - 93 + 83 - 73 - 63 - 53 - 43 - 33 - 23 × 13

     Yes, Karla and I know that the expression of cubes does not make a 100, but rather a 0, but isn't that just the opposite end of the 100 to 0 grading scale for examinations in school? It's still a nice expression, and we liked it enough to show you.

"100 or Bust!"

     Below are some expressions that equal 100, but also some that do NOT equal 100. Your task is to find which ones do and which ones do not.

1. 123 - 45 - 67 + 89 2. 1 + 2 + 34 - 5 + 67 - 8 + 9 3. 1 + 23 + 45 + (6 × 7) - 8 - 9 4. 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 5. 12 + (3 × 4 × 5) - (6 × 7) + (8 × 9) 6. 1 + 2 × 3 + 4 × 5 - 6 + 7 + 8 × 9 7. 1 × 234 - 56 - 78 + 9 8. 123 - 4 × 5 + 6 - 7 + 8 - 9 9. 1 + 23 - 4 + 56 + 7 + 8 + 9 10. 1 + 23 - 4 + 5 + 6 + 78 - 9 11. 1 + 2 + [3 × (4 + 5)] - 6 - 7 + 89 12. (12 - 3 - 4) ÷ 5 + 6(7 + 8) + 9 13. 12 + 3.4 + 5.6 + 7 + 8 × 9 14. 1 + (2 + 3) × 4 + 56 - 7 + 8 + 9 15. (1 × 2)(3 + 4.5) + 6 + 7 + 8 × 9 16. 1 - 2.3 + 45 - 6 + 7 × 8.9 17. 1.2 + 3 × 45 - 6 + 7.8 - 9 18. (1 + 2) ÷ 3 + 4.5(-67 + 89) 19. 1 + 23 + 45 - 6 + 7 × 8 + 9 20. -1 + 2 + 3 - 4[(-5)(6) + 7 + 8 - 9]

     Did you find all the 100s expressions? Add up all the non-100s answers now. You should have the square of 29. How'd you do?

     At the beginning of this page I promised you a few extra tricks for making expressions. Here we present the 100th expression that was given by Madachy in his magazine list. It uses the summation symbol (with the 3), the factorial symbol (with the 4), and the square root symbol (with the 9).

     As only the first symbol is not frequently covered in many math classes, I will explain that "summation 3" simply means to add up all the whole numbers from 1 to 3 -- (1 + 2 + 3), in this case. This means nothing more than "6" is to be used in that position of the expression.

     You have now seen a wide variety of ways to build expressions of any kind. There are other tricks left to learn but we'll save them for another day.

----------
Footnote: Throughout all this work, I have been assuming the use of the standard order of operations, just in case you didn't know.
---------



UPDATE: (7/26/01)

     Early this year (2001) I began an activity similar to the idea above that is currently viewable on the website
Math Forum. I say "similar" because there the use of all ten digits was required, i. e. 0, 1, 2, 3, ... , 9. Therefore, I gave it the title: All the King's Digits.

     If you go there, you will see the work of many, many students. And you'll note that the order of the digits wasn't a requirement.

     Today we received an email from a high school student from Nevada. He sent a regular solution (see it in AKD), and another rather irregular, but highly creative, one. We are presenting the latter work and the student's explanation below. You will soon see the reason why. Enjoy.

Here is my expression which uses each digit more than once but makes use of each symbol a great deal of times like requested. It is given in two parts, part B is subtracted from part A.

Part A: (= 102)

8 * sqrt(cbrt(Σ70 - 12^2 * cbrt(sqrt((6! - 5!)/Σ3) - 2)) - 4) + Σ12
8 * sqrt(cbrt(2485 - 144 * cbrt(sqrt(600/6) - 2)) - 4) + 78
8 * sqrt(cbrt(2485 - 144 * cbrt(10 - 2)) - 4) + 78
8 * sqrt(cbrt(2485 - 144 * 2) - 4) + 78
8 * sqrt(cbrt(2197) - 4) + 78
8 * sqrt(13 - 4) + 78
8 * sqrt(9) + 78
8 * 3 + 78
24 + 78
102



Part B: (= 2)

sqrt(cbrt(13^3 + 4*Σ4 + 13^2 * cbrt(sqrt(((7! - 6!)/Σ4)+2^3 + 17^2))) - Σ4)
sqrt(cbrt(2197+ 40 + 169 * cbrt(sqrt((4320/10) + 8 + 289))) - 10)
sqrt(cbrt(2197 + 40 + 169 * cbrt(sqrt(440 + 289))) - 10)
sqrt(cbrt(2197 + 40 + 169 * cbrt(27)) - 10)
sqrt(cbrt(2197 + 40 + 507) - 10)
sqrt(cbrt(2744) - 10)
sqrt(14 - 10)
sqrt(4)
2



Part A - Part B = 102 - 2 = 100


I was rather bored when I worked this out ;)

Rory



UPDATE: (6/24/02)

For a very definitive page on this topic, go to Mathews: Equations for Integers.


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