Dragon's 100 heads

Notice, that in every case the "delta of heads" (heads cut - heads grown) is divisible by 3. That means, that the number of heads will always have the same rest dividing by 3 - beacause we subtract only multiples of 3. Since 100 has rest 1 when dividing by 3, number of heads will always give rest 1 when dividing by 3, and thus it will never be 0 (which is divisible by 3.)

It's not sufficient to check that 100 is not divisible by 3.

Let's take up a scenario: We have a blow which cuts 17 heads and re-grows 2 heads. If, instead we had 16 heads cut and 1 head regrown.

It would seem, that now it is still not possible to kill the dragon (16 - 1 = 15)

Observe 16 + (16 - 1) = 31

If, we can bring the head count to 31, then now it is reducable to 0.

• 31 heads, cut the 16 ... 31-16 = 15, 1 grows back : 15 +1 = 16
• next cut of 16, brings to 0 ... 16-16 = 0

Now, is it reducable to 31 heads.

Let us only use the blows which reduce the heads:

15k + 6l = 69

5k + 2l = 23

l = 4 k = 3

So, we have to check one thing in this question:

each Individual cuts are of the form 3k or 3k+2

and we start with the initial number of heads as 100, viz. 3k + 1

Now, we can definitely say, it is not reducable to 0

Question is not very simple as it seems at face value ... Double thumbs up for this beautiful question :)

$Here$ $we$ $have$ $to$ $understand$ $that$

$15$ $=$ $24$ $(mod 3)$

$17$ $=$ $2$ $(mod 3)$

$20$ $=$ $14$ $(mod 3)$

$5$ $=$ $17$ $(mod 3)$ . $is$ $invariant$

$So$ $the$ $Dragon$ $will$ $never$ $die$

There is no possible last move since any of the 4 possible moves leaves a positive number of heads on the dragon.

Thus no strategy can reduce the number of heads to zero.

Note that the loss (or gain) of heads at each stroke is divisible by 3. But the total number of heads isn't.