Sixty-Seven Squared

Markland from The John Roan School, Gareth; Euen and Alex from Madras College, Scotland; and Chin Siang from Tao Nan School, Singapore; all sent in good solutions .

Jack from The Ridings High School described the pattern:

I noticed that the number of 4s and 8s each increased by 1 for each extra 6 and that the last digit was always a 9. I then predicted that 666667 ² would equal 444444888889 and I was correct. Therefore according to this pattern: (1 million 6s followed by a 7) ² would be written 1000001 4s followed by 1000000 8s followed by a nine.

Doing these four calculations by long multiplication shows how this pattern works. If $m$ is the number of sixes in the number that is squared, the pattern is:

( $m$ sixes followed by $7$)$^2 =$ ($(m+1)$ $4$'s followed by $m$ $8$'s followed by a $9$).

 

So

(one million sixes followed by a $7$)$^2 =$ (one million and one $4$'s followed by a million $8$'s followed by a $9$).

To prove the result using the sums of series, evaluate

$$[6(1 + 10 + 10^2 + \dots + 10^m) + 1]^2$$

to get

$$\left[6\left(\frac{10^{m+1} - 1}{9}\right) + 1\right]^2$$

Multiplying out and simplifying this gives

$$\frac{1}{9}\left(4 \times 10^{2(m+1)} + 4 \times10^{m+1} + 1\right)$$

Using

$$\frac{10^{p+1}}{9} = 1 + 10 + \dots + 10^p + \frac{1}{9}$$

for $p = 2m+1$ and $p = m$, we get

$$4\left(1 + 10 + 10^2 + \dots + 10^{2m+1}\right) + 4\left(1 + 10 + 10^2 + \dots + 10^{m}\right) + 1$$

which is written $444\dots888\dots9$ that is as $(m+1)$ $4$'s followed by $m$ $8$'s followed by a 9.