Curious Identities Involving Integer Products

This is a continuation of the page where I started to collect identities with expressions that include repeated identical digits. To make describing such numbers a manageable undertaking, I'll use subscripts to denote the number of repetitions of a digit, e.g., 1₅3₂ will denote 1111133, i.e., five 1s followed by 2 3s. In general d_n means a sequence of n d's.

The identities evolved from CSIRO newsLetter, MathByEmail. An early morning exchange on twitter.com with Gary Davis and subsequent comments by James Tanton, who has observed that the repunits (the integers 111..11) with an even number of 1s are represented by the difference of squares of which one is in the form 555...56 in two ways.

3_k · 3_k-14 = 1_k2_k, e.g., 3333 · 3334=11112222. (Proof)
3_k · 3_k-15 = 1_k5_k, e.g., 3333 · 3335=11115555. (Follows from the previous one)
3_k · 3_k-16 = 1_k8_k, e.g., 3333 · 3336=11118888. (Follows from the previous one)
6_k · 6_k-17 = 4_k2_k, e.g., 6666 · 6667 = 44442222. (Proof)
6_k · 6_k-18 = 4_k8_k, e.g., 6666 · 6668 = 44448888. (Follows from the previous one)
9_k · 10_k-11 = 9_2k, e.g., 9999 · 10001 = 99999999.
5_k-16² - 4_k-15² = 1_2k, e.g., 5556² - 4445² = 11111111. (Proof)
5_k-16² - 5_k² = 1_k+1, e.g., 5556² - 5555² = 11111.
6_k · 6_k-19 = 4_k-15_k4, e.g., 6666 · 6669 = 44455554.

Subscripts can be easily converted to decimal notations. Observe that

1_t = 9_t / 9 = (10^t - 1) / 9.

So, for example,

3_k5_m6_n7 = 7 + 10 · 6 · (10ⁿ - 1) / 9 + 10^{n + 1} · 5 · (10^m - 1) / 9 + 10^{m + n + 1} · 3 · (10^k - 1) / 9.

Number Curiosities

Number 8 Is Interesting
Curious Identities Involving Integer Squares
Curious Identities Involving Integer Products
Decimal Sums of Successive Integers
Curious Identities In Pythagorean Triangles
Hardy's Example of Non-Serious Theorems

_k-1

3_k · 3_k-14 = 3 · (10^k - 1)/9 × [3 · (10^k - 1)/9 + 1]

= (10^k - 1)/3 × [(10^k - 1)/3 + 1]

= (10^2k - 2 · 10^k + 1)/9 + (10^k - 1)/3

= (10^2k + 10^k - 2)/9

= (10^2k - 1)/9 + (10^k - 1)/9

= 1_2k + 1_k

= 1_k2_k.

_k-1

6_k · 6_k-17 = 6(10^k - 1)/9 × (6(10^k - 1)/9 + 1)

= 2(10^k - 1)/3 × (2(10^k - 1)/3 + 1)

= 4(10^2k - 2 · 10^k + 1)/9 + 2(10^k - 1)/3

= 4(10^2k - 2 · 10^k + 1)/9 + 6(10^k - 1)/9

= 4 · 10^2k - 2 · 10^k - 2)/9

= 4 · (10^2k - 1)/9 - 2 · (10^k - 1)/9

= 4_2k - 2_k

= 4_k2_k

Of course it is alos possible to derive this identity from 3_k · 3_k-14 = 1_k2_k. Indeed, let's multiply the latter by 4, splitting 4 into two factors of 2 on the left:

6_k · 6_k-18 = 4_k8_k,

which is the next identity. Subtracting 6_k from both sides gives

6_k · 6_k-17 = 4_k2_k,

To remind:

More Curious Identities Involving Integer Squares

71547973

3_k · 3_k-14	= 3 · (10^k - 1)/9 × [3 · (10^k - 1)/9 + 1]
	= (10^k - 1)/3 × [(10^k - 1)/3 + 1]
	= (10^2k - 2 · 10^k + 1)/9 + (10^k - 1)/3
	= (10^2k + 10^k - 2)/9
	= (10^2k - 1)/9 + (10^k - 1)/9
	= 1_2k + 1_k
	= 1_k2_k.

6_k · 6_k-17	= 6(10^k - 1)/9 × (6(10^k - 1)/9 + 1)
	= 2(10^k - 1)/3 × (2(10^k - 1)/3 + 1)
	= 4(10^2k - 2 · 10^k + 1)/9 + 2(10^k - 1)/3
	= 4(10^2k - 2 · 10^k + 1)/9 + 6(10^k - 1)/9
	= 4 · 10^2k - 2 · 10^k - 2)/9
	= 4 · (10^2k - 1)/9 - 2 · (10^k - 1)/9
	= 4_2k - 2_k
	= 4_k2_k