Mr. Richards has a number of pretzel sticks.
He eats 3 of the pretzel sticks before somebody comes and joins him.
When the guest arrives he eats three more sticks and then divides the rest equally between the two of them.
The piles come out equally, but just as Mr. Richards is done dividing up the piles between the other person and himself, another
person joins them.
He combines the piles and divide them up for 3 people, again eating 3 pretzel sticks before doing so.
The piles come out equally again.
But before anyone has a chance to eat any of them, another person comes.
So Mr. Richards combines all the piles, eats 3 pretzel sticks, and divides them up among the 4 of them.
People keep coming in this fashion until there are 10 piles (including Mr. Richards'), and then nobody comes once there are 10 piles, so they all eat their pretzel sticks.
Mr. Richards is
happy that the piles came out evenly each time, and notes that he has the least amount of pretzel sticks with which this would happen.
How many pretzel sticks did each person who joined Mr. Richards have, and how many pretzel sticks would Mr. Richards have had if he didn't have to share them with
the guests?
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.
Correctly solved by:
| 1. Davit Banana | Istanbul, Turkey |
| 2. Colin (Yowie) Bowey | Beechworth, Victoria, Australia |
| 3. Kelly Stubblefield | Mobile, Alabama, USA |
| 4. Kamal Lohia |
Holy Angel School, Hisar, Haryana, India |
| 5. Ivy Joseph | Pune, Maharashtra, India |
| 6. Austin Hale | Grand Junction, Colorado, USA |