1222...² Problem

We define $(1)^{(n)}$ , with parenthesis at the power position, to be $n$ repetitions of digit $1$ , in decimal representation. For instance $(1)^{(2)}=11$ and $(1)^{(3)}=111$ .

The numbers of the form $1222...$ , with $n$ digits in total, can be written as $10^n+2\times (1)^{(n-1)}$ . Take it to the power two to have

$10^{2n}+4\times \big( (1)^{(n-1)} \big)^2+4\times10^n\times (1)^{(n-1)} \ (*)$ .

The swapped version of $1222...$ can be written as $10\times 2 \times (1)^{(n-1)}+1$ . Its square is

$10^2\times 4\times \big( (1)^{(n-1)} \big)^2+1+40 \times (1)^{(n-1)} \ (**)$

In order to have the property true, we need the following to be true (the LHS is from ( $*$ ), when the digit "1", from the beginning is dropped. the RHS is from ( $**$ ), when digit "1", from the end, is subtracted and the remaining is divided by $10$ ).

$4\times \big( (1)^{(n-1)} \big)^2+4\times10^n\times (1)^{(n-1)}=(\frac{1}{10})\times \{10^2\times 4\times \big( (1)^{(n-1)} \big)^2+40 \times (1)^{(n-1)}\}$

the equation, above, can be simplified to

$(10^n-1)\times (1)^{(n-1)}=9\times ((1)^{(n-1)})^2$

or

$(10^{n-1}+\dots+1)\times (1)^{(n-1)}=((1)^{(n-1)})^2$

which can be proven, knowing the definition of multiplication.

Thing to prove:

1222…² = 1ABC… (ABC… represents a sequence of digits that are the same in both numbersl)

222…1² = ABC..1

Proof:

1222… (222.. shall represent n number of twos)

= 10ⁿ + 222…

= 10ⁿ + 2/9 * 999… (999.. also represents n number of nines)

= 10ⁿ + 2/9 * (10ⁿ - 1)

= 11/9 * 10ⁿ - 2/9

222…1

= 1 + 222…0

= 1 + 10 * 222…

= 1 + 10 * 2/9 * 999…

= 1 + 10 * 2/9 * (10ⁿ - 1)

= 20/9 * 10ⁿ - 11/9

Now we must find out how many digits the number 1222…² has.

10ⁿ < 1222… < 10ⁿ * √10

(10ⁿ)² < (1222…)² < (10ⁿ * √10)²

log10(10ⁿ)² < log10(1222…)² < log10(10ⁿ * √10)²

2n < x < 2n + 1

So the number 1222…² must have 2n digits after the 1.

For this next step we’ll call 1ABC… the variable name a.

a = 1ABC…

a = 10²ⁿ + ABC… |-10²ⁿ

a - 10²ⁿ = ABC… | *10

10 * (a - 10²ⁿ) = ABC…0 | +1

10 * (a - 10²ⁿ) + 1 = ABC…1

Now we can replace the a with 1ABC… .

10 * (1ABC… - 10²ⁿ) + 1 = ABC…1

We can now replace 1ABC… with 1222…² and replace ABC…1 with 222…1².

10 * (1222…² - 10²ⁿ) + 1 = 222…1²

For our last step we can subtitute the numbers 1222… and 222…1 with their respective definitions.

10 * ((11/9 * 10ⁿ - 2/9)² - 10²ⁿ) + 1 = (20/9 * 10ⁿ - 11/9)²

10 * (121/81 * 10²ⁿ - 44/81 * 10ⁿ + 4/81 - 10²ⁿ) + 1 = 400/81 * 10²ⁿ - 440/81 * 10ⁿ + 121/81 | *81

10 * (121 * 10²ⁿ - 44 * 10ⁿ + 4 - 81 * 10²ⁿ) + 81 = 400 * 10²ⁿ - 440 * 10ⁿ + 121

1210 * 10²ⁿ - 440 * 10ⁿ + 40 - 810 * 10²ⁿ + 81 = 400 * 10²ⁿ - 440 * 10ⁿ + 121

400 * 10²ⁿ - 440 * 10ⁿ + 121 = 400 * 10²ⁿ - 440 * 10ⁿ + 121 | -400 * 10²ⁿ + 440 * 10ⁿ - 121

0 = 0

This equation is true so our original assumtion must have also been true.