*There are three ruling ideas, three so to say, spheres of thought, which pervade the whole body of mathematical science, to some one or other of which, or to two or all three of them combined, every mathematical truth admits of being referred; these are the three cardinal notions, of Number, Space and Order. Arithmetic has for its object the properties of number in the abstract. In algebra, viewed as a science of operations, order is the predominating idea. The business of geometry is with the evolution of the properties of space, or of bodies viewed as existing in space. *

**J. J. Sylvester**

Welcome to the blog **Math1089 – Mathematics for All**.

Glad you came by. I wanted to let you know I appreciate your spending time here at the blog very much. I do appreciate your taking time out of your busy schedule to check out **Math1089**!

Mathematics is all about numbers, and one important number is 153. 153 is the natural number following 152 and preceding 154. In this blog post, let’s play with the number 153. Yes, we’ll play a game with the number 153 too.

The divisors of 153 are 1, 3, 9, 17, 51 and 153. The sum of divisors of 153 is 1 + 3 + 9 + 17 + 51 + 153 = 234 and the sum of *aliquot divisors* (any divisor excluding the number itself) of 153 is 1 + 3 + 9 + 17 + 51 = 81 = 9^{2}, a perfect square.

153 is a triangular number and it is the sum of the first 17 integers. That is, 153 = 1 + 2 + 3 + ⋅⋅⋅ + 17. Also, 153 is the sum of the first five positive factorials, i.e., 153 = 1! + 2! + 3! + 4! + 5!.

Since 153 = 1^{3} + 5^{3} + 3^{3}, it is a 3-Narcissistic number. It is the smallest three-digit number which can be expressed as the sum of cubes of its digits.

Sum of the digits of 153 is = 1 + 5 + 3 = 9 and it is divisible by the sum of its digits: 153/(1 + 5 + 3) = 17.

**A Game with 153**

The number 153 is indeed interesting. From any positive integer (one, two, three, four or more digits), divisible by 3, following certain rules, we can always get 153. What are the rules? They are given below.

**Rules for obtaining the number 153**

- (
**a**) Choose a number that is divisible by 3. Let the number be*xyz*…; - (
**b**) Extract the digits (in base 10) of the number. The digits are*x*,*y*,*z*, . . .; - (
**c**) Take the sum of their cubes. That is, find*x*^{3}+*y*^{3}+*z*^{3}+ ⋅⋅⋅; - (
**d**) If this yields 153, stop; - (
**e**) Otherwise, repeat the procedure.

Let’s see the rule in action.

**Example 1**. Consider the *single*–*digit* number 3.

Clearly, the only digit of this number is 3 and 3^{3} = 27.

Next, the digits of the last number are 2, 7 and 2^{3} + 7^{3} = 351.

Finally, the digits of the last number are 3, 5, 1 and 3^{3} + 5^{3} + 1^{3} = 153.

Here, ** three steps** are required to reach 153.

** Note**. In this process,

*if we get any three-digit number comprising the digits*1, 3

*and*5

*only*,

*we are done*. For example, if we proceed one step further, then we have 1

^{3}+ 5

^{3}+ 3

^{3}= 153.

**Example 2**. Consider the *two*–*digit* number 84.

The digits of this number are 8 and 4. Now,

8

^{3}+ 4^{3}= 512 + 64

= 576.

The digits of the last number are 5, 7 and 6. We have,

5

^{3}+ 7^{3}+ 6^{3}= 125 + 343 + 216

= 684.

Again, the digits of the last number are 6, 8 and 4. We have,

6

^{3}+ 8^{3}+ 4^{3}= 216 + 512 + 64

= 792.

Now, the digits of the last number are 7, 9 and 2. We have,

7

^{3}+ 9^{3}+ 2^{3}= 343 + 729 + 8

= 1080.

The digits of the last number are 1, 0, 8 and 0. We have,

1

^{3}+ 0^{3}+ 8^{3}+ 0^{3}= 1 + 0 + 512 + 0

= 513.

Finally, the digits of the last number are 5, 1 and 3. We have,

5

^{3}+ 1^{3}+ 3^{3}= 125 + 1 + 27

= 153.

Only ** six steps** are required here to reach 153.

**Example 3**. Consider the *three*–*digit* number 177.

Following the similar lines of argument as above, we obtain

*Step* 1 : 1^{3} + 7^{3} + 7^{3} = 687

*Step* 2 : 6^{3} + 8^{3} + 7^{3} = 1071

*Step* 3 : 1^{3} + 0^{3} + 7^{3} + 1^{3} = 345

*Step* 4 : 3^{3} + 4^{3} + 5^{3} = 216

*Step* 5 : 2^{3} + 1^{3} + 6^{3} = 225

*Step* 6 : 2^{3} + 2^{3} + 5^{3} = 141

*Step* 7 : 1^{3} + 4^{3} + 1^{3} = 66

*Step* 8 : 6^{3} + 6^{3} = 432

*Step* 9 : 4^{3} + 3^{3} + 2^{3} = 99

*Step* 10 : 9^{3} + 9^{3} = 1458

*Step* 11 : 1^{3} + 5^{3} + 5^{3} + 8^{3} = 702

*Step* 12 : 7^{3} + 0^{3} + 2^{3} = 351

*Step* 13 : 3^{3} + 5^{3} + 1^{3} = 153

Here, ** thirteen steps** are required to reach 153.

How this works? Can you think about a formal proof?

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