Algebra
The top of one Christmas tree is 16 feet higher than the top of another Christmas tree. The heights of the smaller tree to the taller tree are in the ratio 3:4. In feet, how tall is the taller tree?
64 feet
Solution: Set the height of the taller tree as h and the smaller tree as h-16. We know the ratio between the small tree and tall tree is 3:4. We can rewrite that as (h-16)/h = 3/4. Solving for h gives us h=64.
Santa’s reindeer have several calves that are kept in various pens with integer side lengths (feet) and an area of 60 square feet. How many unique (uncongruent) pen shapes can there be at once?
6 configurations
Solution: 1x60, 2x30, 3x20, 4x15, 5x12, 6x10, 6 total possible configurations.
Santa asks each of his reindeers to draw a rectangle with integer sides and a perimeter of 50. The reindeers then calculate the area of the rectangle. What is the difference between the smallest and largest areas of the rectangles drawn?
132
Solution: The areas of the rectangles are larger when the side lengths have a smaller difference, in this case the largest area can be 12·13=156. The smallest area occurs when the side lengths have the greatest difference possible, in this case the smallest area can be 24·1 = 24. So, the difference is 156 - 24 = 132.
How many ways are there to give 7 candy canes to 3 kids if all the candy canes are the same and each kid must get at least one candy cane?
15 ways
Solution: Stars and bars: 6 choose 2 ways to put two dividers between 7. 6 choose 2 = 15 ways.
Santa Claus and Mrs. Claus are sitting on two couches facing each other and trying to get some sleep while elves play outside. Santa is getting some Σleep, or discrete sleep, in which Santa only gets moments of sleep before waking up again. Mrs. Claus is getting ∫leep, or continuous sleep, in which she gets intervals of sleep before waking up again. Over five hours, Mrs. Claus slept during t∈[1,3)∪[4,4.5). If Mrs. Claus thinks Santa never slept, at which of the following 4 times couldn’t Santa have been asleep, t=1, t=2, t=3, or t=4?
t=4
Solution: [1,3) does not include 3, but it does include 1 and 2. [4,4.5) includes 4.
Santa and his elves make presents in all sorts of colors. This year, in a certain pile, he is trying out the following strategy: Three eighths are green, 0.3 are red, 12.5% are blue, twelve percent are purple, twenty one out of every three hundred fifty are yellow, and four are orange. How many presents are in that pile?
200 presents
Solution: (3/8+3/10+1/8+3/25+3/50)· n+4=(75/200+60/200+25/200+24/200+12/200)· n+4=(196/200)· n+4=n. 4=n/50. n=200
Santa’s reindeer are racing for fun, although some are running rather slow due to being tired. Dasher, Dancer, Prancer, Vixen, Comet, Cupid, Donner, and Blitzen complete 1, 2, 3, 4, 5, 6, 7, and 8 laps per minute, respectively. If currently, they are all at a certain point in the lap at the same time, after how many minutes will they all be there again?
1 minute
Solution: They all finish integer laps in a minute, some of which are coprime, so 1 minute is all that is needed.
Santa’s reindeer racetrack is a perfect circle with an inside radius of 99 meters and outside radius of 101 meters. Dunder (original name of Donner) runs on the inside of the track and Blixem (original name of Blitzen) runs on the outside of the track. In one lap, to the nearest integer, how much farther does Blixem run?
13 meters
Solution: 101·2pi-99·2pi=4pi which is roughly 13.
Originally, Derek couldn’t stop writing hard questions. After writing his nth hard question, Zion told him to write easy questions n times. How many hard questions did he write for this jeopardy game before Zion told him to stop writing hard questions 50 times?
10 questions
Solution: It is well known that the sum of the first ten positive integers is 55, which is less than ten greater than 50. The answer is thus 10.
2 of 5 elves, Alice, Bob, Charlie, David, and Emily, each holding cards numbered 1, 2, 3, 4, and 5, respectively, have been naughty. Only they are lying. Alice says David has been nice. Bob says Emily has been nice. Charlie says Alice has been nice. David says Charlie has been nice. Emily says Bob has been nice. What is the sum of the numbers on the naughty elves’ cards?
7
Solution: Notice there is a loop of saying each other is nice from Alice to David to Charlie and between Bob and Emily. Everyone in one “loop” must be all naughty or all nice. Since there are only two naughty elves, the “loop” of 2 must be naughty. Bob is 2 and Emily is 5. 2+5=7.
Given C=1, H=2, R=3, I=4, S=1, T=4, M=3, A=2, and S=1, compute (C-H)(C-R)(C-I)(C-S)(C-T)(C-M)(C-A)(C-S)+(H-C)(H-R)(H-I)(H-S)(H-T)(H-M)(H-A)(H-S)+...+(S-H)(S-R)(S-I)(S-T)(S-A).
Note: The last product is different slightly due to S appearing twice. This is essentially (A-A)(A-B)(A-C)... with the (A-A) taken out, as it is a factor of zero.
0
Solution: Notice that C-S=0, H-A=0,…, I-T=0. Every product is equal to 0.
Santa is testing out whether he can deliver presents while asleep! He has 100 houses numbered 1 to 100. He first drops presents in every other house’s chimney (2nd, 4th, 6th, etc.). He realizes he missed some and goes back to drop it in every third house’s chimney (3rd, 6th, 9th, etc.). He then for the same reason drops in every fourth, fifth, and so on until he drops it into every hundredth. By this point, what is the highest number of a house that has an even number of presents?
House 100
Solution: If a present in round n-1 lands in a house, that house is a multiple of n. This is simply counting the number of factors of a number-1! To have an even number of presents, the number must have an odd number of factors, and thus must be a square. 100 is the largest square under or equal to 100.
Free 300 points! (Not because I was too lazy to write a question here)
TX has a sack of 50 red ornaments, 30 green ornaments, and 25 white ornaments. He selects random ornaments for each of 10 layers of his Christmas tree, with 1 beneath the star, 2 beneath that, all the way to 10 at the base. Given that the sum of the number of ornaments in the top four layers is not equal to the number in the bottom, what is the probability that the upper 4 layers have more red ornaments in total compared to the bottom layer?
50%
Solution: 1+2+3+4=10. 50% by symmetry.
Santa’s elves have recently discovered the wonderful magic of spinning chairs, as well as an elevated rate of vomiting. There are 2024 elves in a ring around Santa facing towards him in their chairs right now. Every second, they may spin around, although they gradually stop spinning. During the nth second, every elf has a 1/n chance of spinning around. After 2024 seconds, how many elves on average are facing Santa?
1/2
Solution: After the second second, half the elves are facing inward. For every second after that, on average, the same number of elves switch to facing inward as switch to facing outward.
The intersections of the circles (a+4)^+(b+3)^2=36 and (a+1)^2+(b+5)^2=25 all lie on a line. If the line’s equation is pa+qb=1, what is 24pq?
-4
Solution: a^2+b^2+8a+6b-11=0, a^2+b^2+2a+10b+1=0, 6a-4b=12, 1/2a-1/3b=1. 24pq=24/(-6)=-4.
Derek is having trouble writing good, original, and fun problems. He numbers every question he writes starting at 1, then going to 2 and so on. His good questions have numbers equal to 2 mod 3. His original questions have numbers equal to 3 mod 5. His fun problems have numbers equal to 0 mod 2. Given he wrote 25, how many questions did he write that are good, original, and fun?
0 questions
Solution: In every lcm(2,3,5)=30 questions, there is at most only 1. This can be found by finding the number. From 1 mod 3 and 3 mod 5, we know it is 13 mod 15, which does not work, so it must be 13+15=28. This is not in the first 25, so Derek wrote 0 questions that are good, original, and fun.
Frosty the snowman is made of three spheres of diameter 3, 4, and 5 inches. Frostly the iceman is made up of three cubes of side length 3, 4, and 5 inches. Ice weighs 4 times snow per cubic centimeter. To the nearest integer, how many times heavier is Frostly than Frosty?
8 times
Solution: The volume of the top sphere is 4pi/3·3³, while the volume of the top cube is 3³. The ratio of Frostly’s volume to Frosty’s volume is 2³·3/4pi. The ratio of masses is 24/pi, which can be approximated as 8 because pi is roughly 3.
Mr. John is going to eat some candy canes in the next four minutes. After eating a candy cane the previous minute, he has a ½ chance of eating another one that minute. However, if he did not eat one in the previous minute, he will eat one that minute. If he eats one in the first minute, to the nearest candy cane, how many on average will he eat in the five total minutes?
3
Solution: He has a 1/2 of eating one the second minute, 1/2 (no eat second minute)+1/4 (eat second minute) of eating the third, and 1/4 (no eat third minute)+ 3/8 (eat third minute) of eating the last. 1+1/2+3/4+5/8=23/8, which is roughly 3.
Santa has a new toy making machine. Every day, it can either produce 5 presents or increase the current present count by a factor of 3. If Christmas has just passed and there are no presents now, how many days does it take the machine to make exactly 650 gifts?
9 days
Solution: Working backwards, the closest multiple of 3 is 645 which leads to 215. Then, the next is 210 which leads to 70. Continuing, 65, 60, 20, 15, 5, 0. Our sequence is 5, 15, 20, 65, 70, 210, 215, 645, 650, which is a sequence of length 9.
Given a quartic P(x) has P(0)=1, P(1)=1, P(2)=2, P(3)=3, and P(4)=5, find P(5).
11
Solution: Higher order differences: P: 1, 1, 2, 3, 5; Δ1: 0, 1, 1, 2; Δ2: 1, 0, 1; Δ3: -1, 1. Continuing the pattern, Δ3: -1, 1, 3; Δ2: 1, 0, 1, 4; Δ1: 0, 1, 1, 2, 6; P: 1, 1, 2, 3, 5, and finally our answer of 11.
There is a joke that 25 in base 10 is equal to 31 in base 8, so dec25=oct31, and thus Halloween is Christmas. Following this logic, which day in November is also Christmas? Nov is 9
The 27th
The largest multiple of 9 that is less than 25 is 2*9=18, with 7 left over.
Two presents sit on Santa's desk. They both are composed of a small gift in an abnormally large and massless box. One is a cylindrical gift of density 1 kg/in3, radius 1 inch, and height 2 inches in a spherical box of radius 3 inches. The other is an octahedral gift of density 1.5 kg/in3, with side length 2 inches in a cubic box of side length 10 inches. If the average density of the spherical box is n times higher than the cubic box and n can be expressed as a3√b/c in simplest form, what is √(a+b+c)?
5
Solution: The mass of the cylindrical gift is 1kg/in3·12in2·pi·2in=2pi kg. The density of the spherical box is 2pi/(4·33pi/3)=1/18 kg/in3. To derive the volume, recall that an octahedron with a side length of S is two pyramids of base with area S2. Due to symmetry, the heights of these pyramids are simply half the diagonal of the base. The mass of the octahedral gift is 1.5·2(23/√2)/3=4√2. The density of the cubic box is 4√2/1000=1/125√2 kg/in3. (1/18)/(1/125√2)=125√2/18. 5+2+18=25=52.
An elf is rolling down the stairs. She can either fall down one or two stairs. Is she rolls down 10 stairs, how many ways were there to roll down?
89 ways
Solution: Recursion: a1=1, a2=2, an=an-1+an-2. => a3=3, a4=5, a5=8, a6=13, a7=21, a8=34, a9=55, a10=89.
How many gifts were given in this carol? There are 12 days.
On the first day of Christmas my true love gave to me a partidge in a pear tree.
On the second day of Christmas my true love gave to me two turtle doves and a partidge in a pear tree.
On the third day of Christmas my true love gave to me three French hens, two turtle doves and a partidge in a pear tree.
No space for the rest of it…
364 gifts
Solution: 1+(1+2)+(1+2+3)+...+(1+2+...+12)=2· (12+22+30+36+40+42)=2· 182=364.