O God, I could be bounded in a nutshell and count myself a
king of infinite space, were it not that I have bad dreams.
                                                                              - Hamlet II, ii, 240

When I was eleven, it was discovered that I could repeat a thirteen digit random string of numbers backwards after hearing it once.

Seventeen digits if presented visually.

The numbers seemed to form a pattern in my brain^[1] and from childhood such patterns, their beauty and their anomalies, have fascinated me. They have guided my memory and, in a way, realizing those patterns has guided my life.

The Early Numbers:

Consider the set of the squares of integer (whole) numbers in the first line, which I found to be, frankly, boring.

1  4  9  16  25  36  49  64....

                        3  5   7   9    11   13   15....

But the second line -- showing the difference between the successive squares -- briefly held me by its lure of orderly progression until it undid its promise by the triviality of its proof^[2] .

Not only did the patterns have to be beautiful, so did the proofs.

12, 69, 324, 1245, 2768931

324 and 2768931

The numbers above have no discernible pattern, but what grabbed my young imagination was not just being told or recognizing that a similar set of numbers were all divisible by 3 and those in the second line also by 9 -- but being able to go beyond the "baby" proofs and prove the theorem in full ^[3].

The Binomial Theorem and Pascal's Triangle:

Ah, the Queen of Theorems, such elegance in its sinuous form, such depth of refinement in its inductive proof -- a thing of beauty and a joy forever.

(x+y)ⁿ = b₀xⁿ + b₁x^n-1y + b₂x^n-2y² ....+ b_n-1xy^n-1 + b_nyⁿ

But as much as the theorem itself, it was the binomial coefficients (b₀, b₁, b₂.......b_n  above) arranged geometrically for ascending powers of n as shown in Pascal's Triangle below which captivated the young pup.

How did I love thee? Let me count the ways: the horizontal symmetry, the second diagonal with the natural numbers in order, the third with the triangular numbers, next the tetrahedrals and the pentatopes und so weiter. (I didn't fully understand these then, but later they unfolded as a class of d-simplex numbers leading to combinatorics and... aah, you don't want to go there.)

But see for yourselves in all the glory of 3-D animation (whereas all we had was imagination) a tetrahedral in action. If you could build it with marbles, a three-sided pyramid on a triangular base with side length 5 would contain 35 spheres. And each layer  would comprise the preceding triangular numbers: 1, 3, 6, 10, 15.  All Blaise Pascal!

Source: Wikimedia Commons

And if that were not enough, depict the same numbers in a right triangle:

The sum of the numbers in each color-coded diagonal gives us the series 1, 1, 2, 3, 5, 8, 13 .....as presented by the world famous, the one and only Leonardo Fibonacci of Pisa in 1202!

Feast upon the exquisite shape of the Nautilus in the head-piece diagram  constructed from the Fibonacci series and prepare yourselves for Part II: Fibonacci Numbers, the Golden Ratio and Infinite Continued Fractions in both Canonical and Square Root Forms. I kid you not.

WOOF

Notes: 

[1]  See Matthew Decoursey's excellent posts on different ways of thinking: I do not think in language and  Similes are cool; metaphors are hot

[2] Difference of successive squares proof:

Let n, n+1 be successive integers.

Then x, the difference between their squares can be written as

x = (n+1)² - n²= (n² + 2n + 1) - n² = 2n + 1

Example: If the successive integers are 6 and 7, x = 2*6 + 1 = 13, which is indeed the difference between their squares 36 and 49. 

[4] The "rule" is that if the sum of the digits is divisible by 3 or 9, the entire number is divisible by 3 or 9 respectively.

The "baby" proof goes something like this. Take a number abc .

abc = 100*a + 10*b + c
= (100*a - a) + (10*b - b) + (c - c) + (a + b + c)
= (99*a + 9*b) + (a + b + c)

Since the expression (99*a + 9*b) is divisible by 3 or 9, if (a + b + c) is also divisible by 3 or 9, the entire right hand side is divisible by 3 or 9. Therefore the left hand side i.e. the original number abc is divisible by 3 or 9. And (take it on faith) this can be extended to "n" digits. QED

The proof in full for an eleven year old disbeliever:

Let x be a positive integer with n+1 digits i.e. a_n..... a₂ a₁ a₀

Then x = a₀ + a₁*10 + a₂*10²+...... a_n10ⁿ
= (a₁*10 - a₁₎ + (a₂*10² - a₂₎ + .... (a_n*10ⁿ - a_n) + (a₀ + a₁ + ... a_n)
= a₁(10 - 1) + a₂(10² - 1) + a_n (10ⁿ -1) + (a₀ +a₁ + a₂ + ..... a_n)

Now, IF both (10ⁿ - 1) AND (a₀ + a₁ + a₂ + ..... a_n) are divisible by 3 and 9, we have our proof.

Using the Binomial Theorem, where b0, b1, b2 etc. are the binomial coefficients:

10ⁿ - 1 = (9+1)ⁿ -1 = (b₀*9ⁿ + b₁*9^n-1 + b₂*9^n-2 .... + 1) -1
= b₀*9ⁿ + b₁*9^n-1 + b₂*9^n-2 +.......

Since, the right hand side is divisible by 3 or 9, the left hand side (10ⁿ - 1) is also divisible by 3 or 9.

Therefore, if the remaining condition is met i.e. if the sum of the digits (a₀ + a₁ + a₂ + ..... a_n) is divisible by 3 or 9, the original number is divisible by 3 or 9 respectively. QED