Solution:
Each person who joined Mr. Richards ate 249 pretzels.
Mr. Richards would have had 2,520 pretzels had he not shared them with his guests.
Here is what happened:
There are 2,520 pretzels to start with.
Mr. Richards eats 3 pretzels, so now there are 2,517.
Mr. Richards eats 3 more pretzels, so now there are 2,514, which is divisible by 2.
Mr. Richards eats 3 more pretzels, so now there are 2,511, which is divisible by3.
Mr. Richards eats 3 more pretzels, so now there are 2,508, which is divisible by 4.
Mr. Richards eats 3 more pretzels, so now there are 2,505, which is divisible by 5.
Mr. Richards eats 3 more pretzels, so now there are 2,502, which is divisible by 6.
Mr. Richards eats 3 more pretzels, so now there are 2,499, which is divisible by 7.
Mr. Richards eats 3 more pretzels, so now there are 2,496, which is divisible by 8.
Mr. Richards eats 3 more pretzels, so now there are 2,493, which is divisible by 9.
Mr. Richards eats 3 more pretzels, so now there are 2,490, which is divisible by 10.
Here is K. Sengupta's solution:
Let the original number of pretzel sticks possessed by Mr. Richards be s.
Now, Mr. Richards has eaten 6 pretzels by the time he divides them evenly by 2.
He has eaten 9 pretzels by the time he divides the rest by 3, etc.
Accordingly, we must have:
(s - 3k) divisible by k, whenever k=2,3,4,5,6,7,8,9
=> s must be divisible by k for k = 2 to 9 inclusively.
Therefore, the least value of s must correspond to LCM (2,3,4,5,6,7,8,9) = 2,520
By this time Mr. Richards has eaten 3*10=30 pretzels, so that:
The remaining 2520 - 30 = 2490 pretzels are divided into 10 piles of 2490/10= 249 each, so the number of sticks possessed by each of the 10 individuals is 249.
Mr Richards ate 249 + 30 = 279 pretzels, after adding on the 3 that he ate each time.
Mr. Richards would have had all 2520 pretzels to himself if people did not come to join him.
Here is Kamal Lohia's solution:
Going reverse, let each of the 10 persons get x number of sticks in the end.
So, before the distribution in 10 piles, the total number of sticks must be 10x + 3 as Mr. Richards ate 3 sticks before this distribution.
Also, this 10x + 3 must be a multiple of 9 as it was compiled from the 9 persons earlier. So, x = 9a + 6 and hence the total becomes 90a + 63. Thus, befor the distribution in 9 piles, the total number of sticks must be 90a + 66 as Mr. Richards ate 3 sticks before this distribution.
Also, this 90a + 66 must be a multiple of 8 by the same earlier logic. So, a = 4b + 3 and total sticks before this distribution must be 360b + 339 as Mr. Richards ate 3 sticks before this distribution.
Also, this 360b + 339 must be a multiple of 7. So, b = 7c + 6 and total sticks before this distribution must be 2520c + 2502 as Mr. Richards ate 3 sticks before this distribution.
Also, this 2520c + 2502 must be a multiple of 6 which is always the case. Thus, the total number of sticks before this distribution must be 2520c + 2505 as Mr. Richards ate 3 sticks before this distribution.
Also, this 2520c + 2505 must be a multiple of 5 which is always the case. Thus, the total number of sticks before this distribution must be 2520c + 2508 as Mr. Richards ate 3 sticks before this distribution.
Also, this 2520c + 2508 must be a multiple of 4 which is always the case. Thus, the total number of sticks before this distribution must be 2520c + 2511 as Mr. Richards ate 3 sticks before this distribution.
Also, this 2520c + 2511 must be a multiple of 3 which is always the case. Thus, the total number of sticks before this distribution must be 2520c + 2514 as Mr. Richards ate 3 sticks before this distribution.
Also, this 2520c + 2514 must be a multiple of 2 which is always the case. Thus, the total number of sticks before this distribution must be 2520c + 2517 as Mr. Richards ate 3 sticks before this distribution.
Thus, initial number of sticks must be 2520c + 2520 as Mr. Richards ate 3 sticks before anyone came.
Taking c = 0, the least number of sticks initially must be 2520.
So, Mr. Richards ate 3 before anyone came and balance is 2517.
When 1st guest came, Mr. Richards ate 3 more and balance is 2514 which was evenly divisible among the two.
When 2nd guest came, Mr. Richards ate 3 more and balance is 2511 which was evenly divisible among the three.
When 3rd guest came, Mr. Richards ate 3 more and balance is 2508 which was evenly divisible among the four.
When 4th guest came, Mr. Richards ate 3 more and balance is 2505 which was evenly divisible among the five.
When 5th guest came, Mr. Richards ate 3 more and balance is 2502 which was evenly divisible among the six.
When 6th guest came, Mr. Richards ate 3 more and balance is 2499 which was evenly divisible among the seven.
When 7th guest came, Mr. Richards ate 3 more and balance is 2496 which was evenly divisible among the eight.
When 8th guest came, Mr. Richards ate 3 more and balance is 2493 which was evenly divisible among the nine.
When 9th guest came, Mr. Richards ate 3 more and balance is 2490 which was evenly divisible among the ten.
So, minimum number of sticks received by each guest is 249.