### Converging Product

In the limit you get the sum of an infinite geometric series. What about an infinite product (1+x)(1+x^2)(1+x^4)... ?

### Circles Ad Infinitum

A circle is inscribed in an equilateral triangle. Smaller circles touch it and the sides of the triangle, the process continuing indefinitely. What is the sum of the areas of all the circles?

### Binary Squares

If a number N is expressed in binary by using only 'ones,' what can you say about its square (in binary)?

# Sixty-seven Squared

##### Stage: 5 Challenge Level:

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 666667² 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.

 67 2 = 4489 667 2 = 444889 6667 2 = 44448889 66667 2 = 4444488889

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 $9$).

 666 ... 666667 666 ... 666667 4666 ... 666669 (x7) 40000 ... 000020 (x60) 400000 ... 000200 (x600) . . . 40 ... 000020 ... 000000 (x 6 x 10 m -1 ) 400 ... 000200 ... 000000 (x 6 x 10 m ) 444 ... 444888 ... 888889 Total

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.