Fun with digits

Start with the sequence of non-zero digits 123456789. The problem is to place plus or minus signs between them so that the result of thus described arithmetic operation will be 100.

We got one answer

12 + 3 - 4 + 5 + 67 + 8 + 9 = 100

and suggested there existed at least one more. I do not claim to have made an exhaustive search but there appear to be more than just two answers. One of this is

123 + 4 - 5 + 67 - 89 = 100

I am sure there it at least another one. Want to find it?

There is a keen observation that in the two examples above at least one of the operations is subtraction. And this is also true for all additive (the ones in which the only allowed operations are addition and subtraction) examples below. In fact, it is impossible to avoid subtraction even if the digits come in an arbitrary order. To figure out why, it may be helpful to recollect the notion of digital roots.

You may allow for operations other than addition and subtraction. This leads to a completely new set of problems with numbers having fractional parts. Variations include setting targets other than 100. Here, for example, a representation of one that uses all ten digits:

1 = 148/296 + 35/70

There are many ways to merrily spend time solving arithmetic problems. One way is to attempt representing numbers with limited means. For example, I can represent 100 with five threes as 100 = 33×3 + 3/3. It's surprising how many numbers could be represented this way.

In the 1960s another kind of number puzzles have become very popular. Cryptarithms are brain teasers obtained when digits in numerical calculations have been replaced by letters. Customarily, distinct letters stand for different digits. Stars substitute for any digit and are not related to each other.

I received the following letter from Belgium:

From: Gui et Nicole RULMONT
Date: Tue, 22 Apr 1997 17:02:44 +0200

Dear Cut-the-Knot,

First please excuse my English. I am Belgian and I am very interested by your site!

You wrote in "Fun with digits": Start with the sequence of non-zero digits 123456789. The problem is to place plus or minus signs between them so that the result of thus described arithmetic operation will be 100.

Some years ago, il found in the french magazine Science et Vie the 11 solutions:

1 + 2 + 34 - 5 + 67 - 8 + 9 = 100
12 + 3 - 4 + 5 + 67 + 8 + 9 = 100
123 - 4 - 5 - 6 - 7 + 8 - 9 = 100
123 + 4 - 5 + 67 - 89 = 100
123 + 45 - 67 + 8 - 9 = 100
123 - 45 - 67 + 89 = 100
12 - 3 - 4 + 5 - 6 + 7 + 89 = 100
12 + 3 + 4 + 5 - 6 - 7 + 89 = 100
1 + 23 - 4 + 5 + 6 + 78 - 9 = 100
1 + 23 - 4 + 56 + 7 + 8 + 9 = 100
1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100

If we put a "-" before 1, we have one more solution:

-1 + 2-3 + 4 + 5 + 6 + 78 + 9 = 100

Using the "." decimal separation I found another solution:

1 + 2.3 - 4 + 5 + 6.7 + 89 = 100 (solution of my own)

What about 987654321 ? There are 15 solutions, said Science et Vie:

98 - 76 + 54 + 3 + 21 = 100
9 - 8 + 76 + 54 - 32 + 1 = 100
98 + 7 + 6 - 5 - 4 - 3 + 2 - 1 = 100
98 - 7 - 6 - 5 - 4 + 3 + 21 = 100
9 - 8 + 76 - 5 + 4 + 3 + 21 = 100
98 - 7 + 6 + 5 + 4 - 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 + 3 - 2 - 1 = 100
98 + 7 - 6 + 5 - 4 - 3 + 2 + 1 = 100
98 - 7 + 6 + 5 - 4 + 3 - 2 + 1 = 100
98 - 7 + 6 - 5 + 4 + 3 + 2 - 1 = 100
98 + 7 - 6 - 5 + 4 + 3 - 2 + 1 = 100
98 - 7 - 6 + 5 + 4 + 3 + 2 + 1 = 100
9 + 8 + 76 + 5 + 4 - 3 + 2 - 1 = 100
9 + 8 + 76 + 5 - 4 + 3 + 2 + 1 = 100
9 - 8 + 7 + 65 - 4 + 32 - 1 = 100

Write the sign "-", three solutions:

-9 + 8 + 76 + 5-4 + 3 + 21 = 100
-9 + 8 + 7 + 65 - 4 + 32 + 1 = 100
-9-8 + 76 - 5 + 43 + 2 + 1 = 100

With the decimal point:

9 + 87.6 + 5.4 - 3 + 2 - 1 = 100 (solution of my own)

If I "shuffle" the digits there are many solutions. I found some when I was young, for example:

91 + 7.68 + 5.32 - 4 = 100
98.3 + 6.4 - 5.7 + 2 - 1 = 100
538 + 7 - 429 - 13 = 100
(8×9.125) + 37 - 6 - 4 = 100 etc etc etc ....

very interested by cryptarithms and I collect them. Do you want to receive french cryptarithms ? Do you know non-english cryptarithms ? Thanks!

Gui et Nicole Rulmont

Anthony Lesar notes that the solution 1 + 2 + 3 - 4 + 5 + 6 + 78 + 9 = 100 could be a little modified without changing the result: 1! + 2! + 3 - 4 + 5 + 6 + 78 + 9 = 100.

Note: There is a whole bunch of pages that offer practice problems of this kind. Also, Inder Jeet Taneja has gathered a fantastic collection of various sequential representations of numbers from 1 to 11111.


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  • A problem of representing 6 by three identical digits
  • A 9's Fan's Clock
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