 # Convergence to the Number 2997 from any Positive Number

Pure mathematics consists entirely of such asseverations as that, if such and such a proposition is true of anything, then such and such another proposition is true of that thing. It is essential not to discuss whether the first proposition is really true, and not to mention what the anything is of which it is supposed to be true …. If our hypothesis is about anything and not about some one or more particular things, then our deductions constitute mathematics. Thus mathematics may be defined as the subject in which we never know what we are talking about, nor whether what we say is true.

Bertrand Russell

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Mathematics is all about numbers, and one mystical number is 2997. In this blog post, let’s play with the number 2997. Also, at the end let us consider various representations of this number.

The number 2997 is indeed interesting. From any positive integer (one, two, three, four or more digits), following certain rules, we can always get 2997! What are the rules? They are given below.

Rules for obtaining the number 2997

(a) Choose a number that has any number of digits. Let the number be xyz

(b) Extract the digits of the number and multiply each digit by 111, that is, x × 111, y × 111, z × 111, ….

(c) Take the sum of these product terms, that is, consider the sum

x × 111 +  y × 111 + z × 111 + ⋅⋅⋅ = (xy + z + ⋅⋅⋅) × 111.

(d) If this yields 2997, stop; otherwise, continue.

(e) Repeat the procedure.

Let’s see the rule in action.

Example 1. Consider the sixdigit number 459801.

The digits of this number are 4, 5, 9, 8, 0 and 1. Multiplying each of these digits by 111 and then taking the sum yields

4 × 111 + 5 × 111 + 9 × 111 + 8 × 111 + 0 × 111 + 1 × 111

= 444 + 555 + 999 + 888 + 111

= 2997, as expected.

Only one step is required here to reach 2997.

Note. Here the sum of the digits of the number is 27 and 27 × 111 = 2997.

Example 2. Consider the fourdigit number 4761.

The digits of this number are 4, 7, 6 and 1. Multiplying each of these digits by 111 and then taking the sum yields

4 × 111 + 7 × 111 + 6 × 111 + 1 × 111

= 444 + 777 + 666 + 111

= 1998.

Clearly, we need to move for the next step. The digits of this number are 1, 9, 9 and 8. Multiplying each of these digits by 111 and then taking the sum yields

1 × 111 + 9 × 111 + 9 × 111 + 8 × 111

= 111 + 999 + 999 + 888

= 2997, as required.

Only two steps are required here to reach 2997.

Note. The sum of the digits of the given number is 18 and that of the number taken in the next step is 27.

Example 3. Consider the fivedigit number 12345.

The digits of this number are 1, 2, 3, 4, and 5. Multiplying each of these digits by 111 and then taking the sum yields

1 × 111 + 2 × 111 + 3 × 111 + 4 × 111 + 5 × 111

= 111 + 222 + 333 + 444 + 555

= 1665.

Clearly, we need to move for the next step. The digits of this number are 1, 6, 6 and 5. Multiplying each of these digits by 111 and then taking the sum yields

1 × 111 + 6 × 111 + 6 × 111 + 5 × 111

= 111 + 666 + 666 + 555

= 1998.

For the next step, consider the digits 1, 9, 9 and 8. Multiplying each of these digits by 111 and then taking the sum yields

1 × 111 + 9 × 111 + 9 × 111 + 8 × 111

= 111 + 999 + 999 + 888

= 2997, as needed.

Only three steps are required here to reach 2997.

Example 4. Consider the threedigit number 526.

Following the similar lines of argument as above, we have the first step as

5 × 111 + 2 × 111 + 6 × 111

= 555 + 222 + 666

= 1443.

The second step will be

1 × 111 + 4 × 111 + 4 × 111 + 3 × 111

= 111 + 444 + 444 + 333

= 1332.

In the third step, we have

1 × 111 + 3 × 111 + 3 × 111 + 2 × 111

= 111 + 333 + 333 + 222

= 999.

The fourth step will be

9 × 111 + 9 × 111 + 9 × 111

= 999 + 999 + 999

= 2997, as expected.

Only four steps are required here to reach 2997.

For any number, repeating these steps of multiplying and adding will guarantee that the answer becomes 2997 in a maximum of four repetitions.

An obvious question: how the numbers 111 and 2997 are related?

Clearly, 2997 = 111 × 27 and this can be achieved when (xy + z + ⋅⋅⋅) = 27 in step (c).

Your suggestions are eagerly and respectfully welcome! See you soon with a new mathematics blog that you and I call Math1089 – Mathematics for All!“.