| |
The solution to this relatively
complex problem
bothered me for decades, but lately I solved it.
Fact and Observation One:
The rule of the game states "eat number one,
skip number two, eat number three, etc."
This tells us that all the raisins situated at odd
numbered positions will be eaten. So Fact and
Observation One is "The remaining one must be
at an even numbered position."
Fact and Observation Two:
The rule "eat number one, skip number two,
etc."
tells us that the binary number system is
somewhat involved.
Let's find out the remaining raisin number R for
the cases n = 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15,
and 17 for the number of raisins in the circles:
Let's express all the n values that we picked in the
binary number system mode:
n
(Decimal) |
Binary mode for n |
Note |
R
(Decimal) |
Binary mode for R |
| |
24 |
23 |
22 |
21 |
20 |
|
|
24 |
23 |
22 |
21 |
20 |
| 16 |
8 |
4 |
2 |
1 |
16 |
8 |
4 |
2 |
1 |
| 3 |
|
|
|
1 |
1 |
|
2 |
|
|
|
1 |
0 |
| 4 |
|
|
1 |
0 |
0 |
See * |
4 |
|
|
1 |
0 |
0 |
| 5 |
|
|
1 |
0 |
1 |
|
2 |
|
|
|
1 |
0 |
| 6 |
|
|
1 |
1 |
0 |
|
4 |
|
|
1 |
0 |
0 |
| 7 |
|
|
1 |
1 |
1 |
|
6 |
|
|
1 |
1 |
0 |
| 8 |
|
1 |
0 |
0 |
0 |
See * |
8 |
|
1 |
0 |
0 |
0 |
| 9 |
|
1 |
0 |
0 |
1 |
|
2 |
|
|
|
1 |
0 |
| 10 |
|
1 |
0 |
1 |
0 |
|
4 |
|
|
1 |
0 |
0 |
| 11 |
|
1 |
0 |
1 |
1 |
|
6 |
|
|
1 |
1 |
0 |
| 12 |
|
1 |
1 |
0 |
0 |
|
8 |
|
1 |
0 |
0 |
0 |
| 15 |
|
1 |
1 |
1 |
1 |
|
14 |
|
1 |
1 |
1 |
0 |
| 17 |
1 |
0 |
0 |
0 |
1 |
|
2 |
|
|
|
1 |
0 |
* When n is 2K, it is a special case, and it will be
discussed later.
Except for the case of n = 2K, which will be
discussed in details later, the algorithm to take us
from n to R could be looked at in two different
ways that results in the same solution.
| Outlook 1: |
|
Transfer the 1 at the utmost
left to the
utmost right and then change it to zero.
Of course, the R result must be
converted to an integer |
| |
|
|
| Outlook 2: |
|
Move the binary n one
column to the left,
put a zero on the 20 spot, and delete the
leftmost column. |
Let's do a few n cases:
| n=5 is | 1 | 0 | 1 |
|
|
Move it one column to the
left
and put zero on the 20 spot.
We get |1|0|1|0|, then delete the
first column. We get |0|1|0| = 2. |
| |
|
|
| n=17 is |1|0|0|0|1| |
|
Move it one column to the
left
and put zero on the 20 spot.
We get |1|0|0|0|1|0|, then delete
the first column. We get |1|0| = 2. |
Let's practice for n=24, 125, 27,
22:
n=24 in binary=11000. Algorithm gives 110000=16.
n=125 in binary = 1111101. R = 11111010 = 122.
n=27 in binary = 11011. R = 110110 = 22.
n=22 in binary = 10110. R = 101100 = 12.
Let's deal now when n = 2K, like 8 = 23:
Expressing
8 in binary we get 1000. Using our algorithm we get
R = 10000 = 0000, and converting it to an
integer,
we get a big fat ZERO! This result kept me puzzled
for years, and I questioned the validiy of the
algorithm. But here is the interpretation of a zero
for R: A zero is an integer which is always prior to
one, so when we construct the raisins circle, the
integer prior to one is the number of raisins in this
circle before the eating game starts.
So for n = 2K like 8, 16, 32, 64, 128, ..., 1024, ...,
32768, ..., R would be 2K; e.g., for a circle of 2048
raisins, the last remaining raisin would be raisin
number 2048, which is just before number one
because 2048 = 211.
|
|
Gem 4: My Grandfather's
Clever Raisins Game
