calculation puzzle - The Magic Stick

Here's a relatively easy puzzle that's actually entirely my own creation.

An old guru in an Asian monastery is showing his beloved student his gallery of items that he has either collected or kept from previous generations.

One item that he is particularly fond of is an odd-looking walking stick. He shows it to his student, waving his hands over the evenly spaced notches in the stick. His student understands without talking that the stick served two purposes – as a walking stick and as a measuring device.

He then proceeds to pop the handle off the stick. The student almost gasps but utters no sound. The guru continues to break the stick into pieces, while the student watches in silent horror at how inconsiderate the guru is being of an old relic.

The student regains his calm, though, after noticing that the guru has not destroyed the stick after all, but is snapping pieces of the stick back together again, rearranging them into two smaller sticks. He then lines the two sticks on the ground, and the student sees that one piece is one notch longer than the other.

He then proceeds to break the sticks into pieces again. The student watches curiously as he then rearranges the pieces into three sticks, each of them one notch longer than the next. The guru continues to do this for five sticks and seven sticks, and in each of the arrangements, each stick is one notch longer than the next.

Finally, the guru snaps off all the pieces that he had been using to make the smaller sticks, and lines them up individually. The student knows what to expect. Each individual piece, ten of them in total, is one notch longer than the next.

How long is this stick, in notches?

Answer

Part 1

Xynariz's answer seems to be correct. We can note all the infinite solutions if the stick has more than 10 breaks as follows: $$5(2j+9)=3*5*7*k \quad \text{for some integers \(j\) and \(k\)} $$

If $j = 21x+6$ and $k=2x+1$ for any integer $x$, then these equations are true.

This means the number of sticks is

$210x+105$ for any integer $x$. This number is odd and divisible by 3,5, and 7. Also if you subtract 45, the answer is divisible by 10.

Part 2

If there are only 9 breaks, each stick is $L+s$ long where $s$ is the stick number $0,1,2,3...$ while L is $21x+6$.

If we try to make 3 sticks out of the 10 parts, the minimum length of the stick that contains 4 parts (as 10 is not divisible by 3) is $(21x+6)*4+0+1+2+3=84x+30$. The largest stick that can be made out of 3 small sticks is $(21x+6)*3+9+8+7 = 63x+42$. The smalles difference between the 4 and 3 stick lengths is $21x-12$ while the guru's game requires at least one combination that differ in length by no more than $1$. As the 4-stick is more than $1$ greater than the 3-stick for any $x>0$, the only solution is $x=0$. This means that the only solution is

$105$.