DICE
225
No, no, way too high.
BIG clue;
. o .
. . .
. . . can only be a 6
Mark:
ELVES
Cher - 6 trains - yellow
Jane - 5 balls - red
Johnny - 3 cars - green
Marcia - 7 sleds - orange
Sue - 9 tops - blue
Solution:
Each elf made at least three toys (line 4). One elf made three times as many as another (13-14), and three elves made consecutive amounts (5-8). If you eliminate duplicate amounts (3), two possible number combinations total 30: 12/6/5/4/3 or 9/7/6/5/3.
With either solution, Sue made the most (13-14). And we know that she made tops 17-18). She was not dressed in green (13-14), orange (15), red (5-6), or yellow (11-12), so she must be the elf in blue.
The two elves who made the least must be dressed in green (13-14) and red (5-6). Therefore the elf who made the least can not be Marcia (15). It also can not be Cher (5-8), or Jane (10), so Johnny must have made the least. And he made cars (9).
The elf who made the second most was not Jane (10)or Cher (5-8), so it must have been Marcia.And Marcia dressed in orange (15).
Since the two elves who made the least were in green and red, the elf in yellow must have made the third-smallest amount of toys. And that elf made trains (11-12).
Since the elf making sleds could not have made the second-smallest amount of toys (7-8), Marcia must have made the sleds. Cher is therefore the elf in yellow, and the person who made the second-smallest amount of toys is in red (5-8).
By elimination, Johnny is in green, and Jane is in red and made the balls. Since the elf in green made three toys, Sue must have made nine toys (13-14), and our amounts of toys must be: 9/7/6/5/3.
Sue (blue) made 9 tops.
Marcia (orange) made 7 sleds.
Cher (yellow) made 6 trains.
Jane (red) made 5 balls.
Johnny (green) made 3 cars.
"Let N = 111...1222...2"
N is even so it is a multiple of 2, but the last two digits are not divisible by 4, so N is not a multiple of 4. The sum of its digits is 1999 ⋅ 1 + 1999 ⋅ 2 = 1999 ⋅ 3, so it is a multiple of 3 but not 9. 5 is not a factor because N does not end with 0 or 5.
Potential third factors can be tested by a technique that I devised based on remainder repetition intervals. I will use the division of 7 into a short series of ones and twos as an example:
1 5 8 7 3 0 3 1 7 4 6 7 ) 1/1
14 16 15 12 10 22 21 25 23 24 20
As you can see, it divides evenly into 6 ones, and divides evenly into 6 twos. Because the remainder at the end of each section is zero, it can loop around as many times as desired. In addition, at equal distances from the change from 1 to 2 are two remainders of 2, so the cycle can be interrupted at that point. Therefore, 7 is a factor of any number like N with 6x ones and 6x twos, or a number with 6x+5 ones and 6x+5 twos, with x as an integer greater than or equal to 0. I write this cycle structure as 6x and 6x+5. Since 1999 does not equal 6x or 6x+5 for any integer x, 1999 does not fit seven's cycle, and 7 is not a factor.
Now, this method of testing primes until a factor is found can be used. The results starting after 7
are:
Prime Cycle structure
11 2x
13 6x
17 16x, 16x+2
19 18x, 18x+8
23 22x, 22x+19
1999 = 22 ⋅ 90 19 + =22 ⋅ x 19 + , so the third factor of N is 23. The results of division cycle with the remainder cycles, so N/23 can be expressed as follows:
(Notation: X indicates that the previous series of numbers should be repeated the indicated number of times, ... indicates that the next number should be concatenated to the previous one)
N/23 = 0048309178743961352657X90...
0048309178743961352661837.
00966178357487922705314X.
Each section alone is not a multiple of 3, but the three sections together are. In addition, the middle section (put together) is also a multiple of 3. So
N
-------------=
323
⋅
001610305958132045088566827697262479871175523349436392914653784219X30...
0016130595813204508872785829307568438... 003220611916264090177133655394524959742351046698872785829307568438X30
Only the first section is odd, but the remainder can be contained in the first digit of the next repetition.
Talking of Santa's little helper:
Santa needs some help from you this year - he'll supply the laughs (HO HO) if you do the wrapping. In each of the words and phrases below, Santa's filled in HO twice, but the rest of the letters are missing.
Can you reconstruct each word or phrase from the clues and blanks given?
1. Hershey addict: _ H O _ _ H O _ _ _
2. Listing of famous and influential people:
_ H O ' _ _ H O
3. Word pronounced like another, as "heir" or "air"
H O _ _ _ H O _ _
4. Garage where stolen cars are dismantled for parts:
_ H O _ _ H O _
5. Be Patient!:
H O _ _ _ _ _ _ H O _ _ _ _ !
