An Extraterrestrial Message

First, recall the divisibility rule for 11:

• add the digits in "odd" positions
• add the digits in "even" positions
• find the difference
• if this is a multiple of 11 then the original number is divisible by 11.

Next, observe that $43434343= 4343 \times 10001 = 43 \times 101 \times 10001$ .

As 101 is a factor of 43434343, it will also be a factor of 43434343 $^2$ .

By imagining the digits grouped in pairs (18 86 5a b1 51 84 16 49) we can consider it to be in base 100. This means we can utilise the same principle as the divisibility rule for 11 (working in base 10) to create a divisibility rule for 101 working, in effect, in base 100.

So:

• we add the "odd" position digit pairs: $49 + 84 + (10b+1) + 86 = 10b + 220$
• we add the "even" position digit pairs: $16 + 51 + (50 + a) + 18 = a + 135$
• we subtract these answers: $(10b + 220) - (a + 135) = 10b - a + 85$
• for the original number to be a multiple of 101, $10b - a + 85$ must be a multiple of 101.

For this to happen, $10b - a + 85$ must equal 101 (or else b will be greater than 9 or else negative).

This gives b = 2 and a = 4. So the answer to the problem is $2 \times 4 = 8$ .

$43434343=43(1+10^2+10^4+10^6)=43(1+10^4)(1+10^2)$

$43434343^2=43^2(1+10^4)^2 (1+10^2)^2 \\ =1849(10^8 +2.10^4+1)(10^4+2.10^2+1) \\ =1849(10^{12} +2.10^{10}+3.10^8+4.10^6+3.10^4+2.10^2+1) \\ =1849 \underbrace {000...00}_{\text{12 times}} + 3698 \underbrace {000...00}_{\text{10 times}} + 5547 \underbrace {000...00}_{\text{8 times}}+ 7396 \underbrace {000...00}_{\text{6 times}} + 5547 \underbrace {0..0}_{\text{4 times}} + 369800+1849$

Adding from backwards we get the result $1886542151841649$