One day while browsing on the Internet, I came across the following problem:

 A 7 is written at the end of a two-digit number, increasing its value by 700. Find the original number.

Well, as I’m always hunting for good problems about numbers, I feel that I have developed a sixth sense for such things. And this problem proved to be as good as any I’ve run across recently. It is the kind of problem that looks very easy, at least to understand what’s going on, while at the same time proves to be a distinct challenge to young students. Of course, being a septophile, I was intrigued by the use of the numbers 7 and 700. And my interest only but increased when I found the solution: 77!

I solved the problem rather quickly, because I used some basic algebra. Then I set myself to thinking: if I didn’t know how to do it by algebra (like the vast majority of the students I teach), how would I proceed? Naturally, by “trial and check”. I would choose some two-digit number, place the 7 at the end, thus forming a much larger three-digit number. To find out how much the resulting increase was, it was a simple matter: subtract the original two-digit number from the new three-digit number; if the difference was 700, my search was through. If not, I would adjust my original trial appropriately, either up or down as needed.

That’s just what my students did, once they understood what the problem was talking about. Most of them had never been presented with a problem of this sort, especially in a math class. They often had difficulty knowing just which way to go. So I decided to devote a little more time to it. This sort of problem is easy to create variations for, such as:

1. If 6 is written at the end of the two-digit number 53, the new number formed
in this way would represent an increase over the original. How much was the increase?
2. If a digit is written at the end of the two-digit number 48, the original
number (48) would be increased by 434. What is the single digit?

I hear you saying about now, “Those are certainly easy problems.” And you are correct. After all, all these problems follow this basic pattern:

XY + increase = XYZ

where the capital letters represent the various digits, and Z being the special single digit part of the problem. Let’s analyze how this structure helps us to attack the three forms of the problem.

In form #1 above, we are given the X, the Y, and the Z. So it’s a simple matter to find the increase.

XYZ – XY = increase

Form #2 is easier still. Here we have the XY number and the “increase”. So it’s a trivial matter to add the two. As long as the sum starts out with “XY-“, we should have achieved our goal. The Z-digit is there for the picking.

Therefore, the tougher nut to crack is the original problem. In that form we only know the Z-digit and the increase. But as any good algebra student should be able to show, the positive difference of the single digit and the increase is divisible by 9, and further, the quotient of that difference divided by 9 is our desired two-digit number!! The proof might go something as follows:

Let n = the original two-digit number, d = the single digit, and i = the increase.

If we write a digit behind a number, the place value of each of its digits is multiplied by 10. [In effect, each of its digits is “shifted” to the left.] This means our three-digit number has this structure: 10n + d. Using the fact explained above for form #1, we can state

(10n + d) – n = i

Solving this equation gives us

n = (i – d)/9

Now solving problems like the original 7’s problem becomes a mere two-step process:

1. Subtract the given digit from the increase.
2. Divide that result by 9.

In the case of the 7’s problem, this means 700 – 7 = 693, and 693 ÷ 9 = 77.

While working on the various extra homework exercises that I gave my students, Cristina found an interesting fact that had escaped my attention. And she did it without using algebra! She told all of us in her class the next day that if you add the digits of the “increase” number, and continue to add the digits if any sum is greater than 9 until a single digit is obtained, then that digit is the single digit in the problem statement. Using an example from above (form #2) we see that

434 + 48 = 482, so 2 was the so-called Z-digit.

But notice this:

4 + 3 + 4 = 11 and 1 + 1 = 2.

This is no freak accident. This is true all the time. Just get out your algebra and prove it; I challenge you to do this.

Next the two-step solution procedure explained above works equally well if the original number has three places (or even more). Try this problem and see.

 A 3 is written at the end of a three-place number, increasing it by 5367. Find the original number.

At first glance it looks a little scary. But using the steps on it produces the desired result rather quickly.

Finally we “up the ante” just a bit here with this extension.

 The number 17 is written at the end of a three-place number, increasing it by 16847. Find the original number.

(Put that one in your algebra “pipe” and smoke/prove/solve it!!!)