math-eval/TAL-SCQ5K huggingface dataset:
Problem: Ms. Osborne asks each student in her class to draw a rectangle with integral side lengths and a perimeter of $$50$$ units. All of her students calculate the area of the rectangle they draw. What is the difference between the largest and smallest possible areas of the rectangles? 
Options:
aoVal: A
content: $$76$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$128$$ 
aoVal: D
content: $$132$$ 
aoVal: E
content: $$136$$ 
Answer: D
Analysis: As we know, the sum of the length and width is $$25$$. The largest area is $$13\times12=156$$ and the smallest area is $$24\times1=24$$, so the difference is $$156-24=132$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Extreme Value with Fixed Sums
Problem: Sandy, Sam, and Steve are observing a cat in the distance. The cat is eating something but it\textquotesingle s hard to tell what it is.  Sandy says:"The cat is eating fish."  Sam says:"I agree with Sandy!"  Steve says: "The cat is eating meat."  Then they come closer and realize only one of them guessed right.  The cat is eating . 
Options:
aoVal: A
content: 　meat 
aoVal: B
content: 　fish 
aoVal: C
content: 　Not certain 
Answer: A
Analysis: We can directly find that Sandy\textquotesingle s point and Sam\textquotesingle s point are identical,so both of them tell lies.Therefore, Steve tells the truth. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Today is Tuesday, and Cindy eats cakes today. She will eat cakes again after $5$ days. What day will it be? 
Options:
aoVal: A
content: Sunday 
aoVal: B
content: Saturday 
aoVal: C
content: Friday 
aoVal: D
content: Thursday 
aoVal: E
content: Wednesday 
Answer: A
Analysis: Today is Tuesday, and $5$ days later will be Sunday. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Counting the Number of Figures
Source: 其它
Problem: One day, Pip asks his parents: "What day is it today?"  His mother says: "Today is Monday."  His father says: "Today is Tuesday."  From the options below, which one do you agree with? 
Options:
aoVal: A
content: One of these two sentences is definitely wrong and the other one is correct. 
aoVal: B
content: It is possible that both of Pip\textquotesingle s parents are wrong. 
aoVal: C
content: It is possible that both of Pip\textquotesingle s parents are right. 
aoVal: D
content: If Pip\textquotesingle s mother is wrong, then his father must be right 
Answer: B
Analysis: "Today is Monday" is not the opposite of "Today is Tuesday".i.e. they can both be false.  "Today is Monday" is the direct opposite of "Today is not Monday". One must be true and the other must be false. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: How many different three-digit numbers can we make using the digits $$2$$, $$4$$, and $$6$$? (Each digit can be used only once.) 
Options:
aoVal: A
content: $$3$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
Answer: B
Analysis: $246, 264, 426, 462, 624, 642$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations->Basic Operations of Combinations
Problem: What is $$542$$ hundredths? 
Options:
aoVal: A
content: $$0.0542$$ 
aoVal: B
content: $$0.542$$ 
aoVal: C
content: $$5.42$$ 
aoVal: D
content: $$54.02$$ 
aoVal: E
content: $$54.2$$ 
Answer: C
Analysis: $$542$$ hundredths $=5$ ones, $4$ tenths and $2$ hundredths $=5.42$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: How many two-digit numbers are there where the ones digit is greater than the tens digit?． 
Options:
aoVal: A
content: $$26$$ 
aoVal: B
content: $$18$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$30$$ 
aoVal: E
content: $$36$$ 
Answer: E
Analysis: $$8+7+6+5+4+3+2+1=36$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many different four-digit numbers can be formed by rearranging the four digits in $2004$?~ 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$16$$ 
aoVal: D
content: $$24$$ 
aoVal: E
content: $$81$$ 
Answer: B
Analysis: 0 shouldn\textquotesingle t be put in the highest digit. There are two 0s  The solutions are: 2400, 2040, 2004, 4200, 4020, 4002. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: When Coco the Caterpillar is not sleeping, he eats $$5$$ grams of leaves per hour. Yesterday he slept $$20$$ hours. How many grams of leaves did he eat yesterday? 
Options:
aoVal: A
content: $$0$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$10$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$40$$ 
Answer: D
Analysis: Yesterday he did not sleep for $$24 - 20 = 4$$ hours, so he ate $$4 \times 5 = 20$$ grams of leaves. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: Eight pupils from Victory Primary School take a Mathematics test, but none of the pupils wrote his/her name on the test. The tests are therefore handed back to the pupils at random. In how many ways can exactly $5$ of the $8$ pupils get the correct test back? 
Options:
aoVal: A
content: $$30$$ 
aoVal: B
content: $$36$$ 
aoVal: C
content: $$56$$ 
aoVal: D
content: $$112$$ 
aoVal: E
content: None of the above 
Answer: D
Analysis: If exactly $5$ pupils get the correct test, then exactly $3$ pupils must get the wrong test.  No. of ways to choose $5$ pupils to get the correct test is $$\frac{8 \times 7 \times 6 \times 5 \times 4}{5 \times 4 \times 3 \times 2 \times 1}-56.$$  To make sure that the other $3$ pupils get the wrong tests, the correct number is $2$.  Hence, the total no, of ways $=56 \times2=112$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: In how many ways can the letters in $BEEKBBPER$ be rearranged so that two or more $E$s do not appear together? 
Options:
aoVal: A
content: $$4200$$ 
aoVal: B
content: $$900$$ 
aoVal: C
content: $$800$$ 
aoVal: D
content: $$720$$ 
aoVal: E
content: $$700$$ 
Answer: A
Analysis: There are $3$ $E$s in total now with other $6$ letters remaining. But pay attention to $B$: there are $3$ $B$s here.  There are $\_6P\_3$ ways for us to arrange the $6$ letters\textquotesingle{} positions. Then, we can put the $3$ $E$s in the $7$ intervals.  So the answer is $\_6P\_3 \times \_7C\_3=4200$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Pip multiplies all the digits in the number $$145$$ to get $$20$$. How many three digit numbers are there whose digits multiply to give $$20$$? 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$7$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
Answer: D
Analysis: $$145, 154, 415, 451, 514, 541, 225, 252, 522$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations->Basic Operations of Combinations
Problem: Ranson holds $3$ white and $9$ purple jelly beans in his hand. Nancy holds $2$ white, $5$ yellow, and $2$ purple jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match? (adapted from $2013$ AMC $8$ Problem, Question \# $14$) 
Options:
aoVal: A
content: $\dfrac{1}{6}$ 
aoVal: B
content: $\dfrac{1}{3}$ 
aoVal: C
content: $\dfrac{2}{9}$ 
aoVal: D
content: $\dfrac{1}{2}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: The probability that both show a white bean is $\dfrac{3}{12}\times \dfrac{2}{9}=\dfrac{1}{18}$. The probability that both show a purple bean is $\dfrac{9}{12}\times \dfrac{2}{9}=\dfrac{1}{6}$. Therefore, the probability is $\dfrac{1}{18}+\dfrac{1}{6}=\frac29$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
Problem: Linda picks $3$ different numbers from $$1-15$$. To make the sum of the three numbers divisible by $3$, how many different groups are there for Linda to pick? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$910$$ 
aoVal: C
content: $$91$$ 
aoVal: D
content: $$155$$ 
aoVal: E
content: $$30$$ 
Answer: D
Analysis: Group $A$: $$1$$, $$4$$, $$7$$, $$10$$, $13$;  Group $B$: $$2$$, $$5$$, $$8$$, $$11$$, $$14$$;  Group $C$: $$3$$, $$6$$, $$9$$, $$12$$, $$15$$.  Linda can choose three numbers from the same group, or choose each number from a different group to get the sum she needs.  There are $3\times\_5C\_3+5\times5\times5=155$ groups. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Problem: There are $$5$$ dancers. Every $$2$$ dancers will have a dance. How many dances would they have in total? 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$25$$ 
aoVal: E
content: $$8$$ 
Answer: B
Analysis: $4+3+2+1=10$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Problem: If your average score on your first six mathematics tests was $84$ and your average score on your first seven mathematics tests was $85$, then your score on the seventh test was ． 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$91$$ 
aoVal: E
content: $$92$$ 
Answer: D
Analysis: The total score of the first six tests was $84\times6=504$, and the total score of the first seven tests was $85\times7=595$. Therefore, the score of the seventh test equals to the difference: $595-504=91$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Questions Involving Average->Questions Involving Average (ordinary type)
Problem: Take out $$4$$ digits from $101112131415$ to make the rest of the digits into a new $8-$digit number without changing the order of digits. The least possible value of the new number is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$10111111$$ 
aoVal: B
content: $$10111110$$ 
aoVal: C
content: $$10111100$$ 
aoVal: D
content: $$10111011$$ 
aoVal: E
content: $$10111112$$ 
Answer: A
Analysis: To make it the least, you should make the digit in the first place as small as possible. Thus the $$2$$, $$3$$, $4$ and $$5$$ should be taken out. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many two-digit numbers have digits whose sum is a perfect square? (2006 AMC 8 Problem, Question \#11) 
Options:
aoVal: A
content: $$13$$ 
aoVal: B
content: $$16$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$19$$ 
Answer: C
Analysis: There is 1 integer whose digits sum to $1: 10$. There are 4 integers whose digits sum to $4: 13,22,31$, and 40 . There are 9 integers whose digits sum to $9: 18,27,36,45,54,63,72,81$, and 90 . There are 3 integers whose digits sum to $16: 79,88$, and 97 . Two digits cannot sum to 25 or any greater square since the greatest sum of digits of a twodigit number is $9+9=18$ Thus, the answer is $1+4+9+3=$ (C)17. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Three students go apple-picking. Eddie picks $30$ apples, Avril picks $27$ apples, and Mike picks $33$ apples. How many apples does each of them pick on average? 
Options:
aoVal: A
content: $$90$$ 
aoVal: B
content: $$45$$ 
aoVal: C
content: $$31$$ 
aoVal: D
content: $$30$$ 
Answer: D
Analysis: They pick $30+27+33=90$ apples in total. Therefore, on average, each of them picks $90\div3=30$ apples. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: SASMO 2014 P2 Q1  What is 2014 + 2 x 0 x 1 x 4 equal to? 
Options:
aoVal: A
content: $$2014$$ 
aoVal: B
content: $$2016$$ 
aoVal: C
content: $$2021$$ 
aoVal: D
content: $$2022$$ 
aoVal: E
content: None of the above 
Answer: A
Analysis: anything x 0 = 0 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: There is a ball in a box and three kids are guessing what colour it is.  Val says: "The ball is~red."  John says: "The ball is~green."  Elvis says: "I agree with~John."  They open the box and find only one of them guessed right.  What colour is the ball? 
Options:
aoVal: A
content: red 
aoVal: B
content: green 
aoVal: C
content: Uncertain 
Answer: A
Analysis: We can spot that John\textquotesingle s guess and Elvis\textquotesingle{} guess are the same, so both of them must be wrong. Therefore, Val guessed it right. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Comparing
Problem: Each of the 20 balls is tossed independently and at random into one of the 5 bins. Let $p$ be the probability that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Let $q$ be the probability that every bin ends up with 4 balls. What is $\frac{p}{q}$ ? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$4$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$12$$ 
aoVal: E
content: $$16$$ 
Answer: E
Analysis: Solution 1: For simplicity purposes, we assume that the balls and the bins are both distinguishable. Recall that there are $5^{20}$ ways to distribute $20$ balls into $5$ bins. We have $$ p=\frac{5 \cdot 4 \cdot\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 3,5,4,4,4 \end{array}\right)}{5^{20}} \text { and } q=\frac{\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 4,4,4,4,4 \end{array}\right)}{5^{20}} \text {. } $$ Therefore, the answer is $$ \frac{p}{q}=\frac{5 \cdot 4 \cdot\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 3,5,4,4,4 \end{array}\right)}{\left(\begin{array}{c} 20 \textbackslash\textbackslash{} 4,4,4,4,4 \end{array}\right)}=\frac{5 \cdot 4 \cdot \frac{20 !}{3 ! \cdot 5 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}}{\frac{20 !}{4 ! \cdot 4 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}}=\frac{5 \cdot 4 \cdot(4 ! \cdot 4 ! \cdot 4 ! \cdot 4 ! \cdot 4 !)}{3 ! \cdot 5 ! \cdot 4 ! \cdot 4 ! \cdot 4 !}=\frac{5 \cdot 4 \cdot 4}{5}=(\mathbf{E}) 16 . $$  Solution 2:  For simplicity purposes, we assume that the balls and the bins are both distinguishable. Let $q=\frac{x}{a}$, where $a$ is the total number of combinations and $x$ is the number of cases where every bin ends up with 4 balls.  We can take 1 ball from one bin and place it in another bin so that some bin ends up with 3 balls, another with 5 balls, and the other three with 4 balls each. Note that one configuration of $4$-$4$-$4$-$4$-$4$ corresponds to $5 \cdot 4 \cdot 4=80$ configurations of $3$-$5$-$4$-$4$-$4$. On the other hand, one configuration of $3$-$5$-$4$-$4$-$4$ corresponds to 5 configurations of $4$-$4$-$4$-$4$-$4$. Therefore, we have $$ p=\frac{80}{5} \cdot \frac{x}{a}=16 \cdot \frac{x}{a}, $$ from which $\frac{p}{q}=$ (E) 16 . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Yuki and Claire are looking for a place to eat dinner. They know there are four Chinese restaurants, three French restaurants, and two Peruvian restaurants nearby. There are~\uline{~~~~~~~~~~}~different choices in total for them to eat one meal in one place. 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$24$$ 
aoVal: C
content: $$27$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$14$$ 
Answer: A
Analysis: They can only choose one place, so it can only be either Chinese, French, or Peruvian restaurants. Therefore, we can add each one up to get $$4+3+2 = 9$$.~ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: There are $12$ black balls, $27$ red balls, and $11$ blue balls in an opaque box. The balls are the same except for their colors. Bob adds several red balls in the box and mixes the balls. Now, if the probability of taking out a black ball is $\frac15$, how many red balls does Bob add? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$15$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$25$$ 
Answer: B
Analysis: $\frac15=\frac{12}{60}$, which means there are $60$ balls in total. Thus, Bob adds $60-12-27-11=10$ red balls. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: There are $$10$$ players in a chess tournament. If each game is played by $$2$$ players, and each player plays every other player exactly once, what is the total number of games played in the tournament? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$45$$ 
Answer: D
Analysis: There are $10$ players and each player plays $9$ games, so there are $10 \times 9 \div 2 = 45$ games in total. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Sports Competition
Problem: Among $$30$$ children, there are at least~\uline{~~~~~~~~~~}~children born in the same month. 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
Answer: C
Analysis: There are $$12$$ months. Thus, $$30\div 12=2R6$$ and hence $$2+1=3$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: If the sum of $$9$$ numbers is $$1998$$, then their average is ． 
Options:
aoVal: A
content: $$9+1998$$ 
aoVal: B
content: $$9\times 1998$$ 
aoVal: C
content: $$1998\div 9$$ 
aoVal: D
content: $$9\div 1998$$ 
Answer: C
Analysis: The average of any $$9$$ numbers is their sum divided by $$9$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: In the Coin Game, you toss three coins at the same time. You win only if the 3 coins are all showing heads, or if the 3 coins are all showing tails. If you play the game once only, what is the probability of winning? 
Options:
aoVal: A
content: $\frac{1}{6}$ 
aoVal: B
content: $\frac{1}{3}$ 
aoVal: C
content: $\frac{2}{27}$ 
aoVal: D
content: $\frac{2}{3}$ 
aoVal: E
content: $\frac{1}{4}$ 
Answer: E
Analysis: When tossing a single coin, there are two possible outcomes, a head (H) or a tail (T).  When tossing 2 coins, there are $$2 \times 2 = 4 $$possible outcomes.  These are HH, HT, TH, and TT.  When tossing 3 coins, there are $$2 \times 2 \times 2 = 8 $$possible outcomes. These are HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Of these 8 possible outcomes, there are 2 winning outcomes, HHH and TTT.  Thus, the probability of winning the Coin Game is $\frac{1}{4}$ .  Answer: E 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Typical Probability Problems->Tossing Coins
Source: 其它
Problem: How many different four-digit numbers can be formed by rearranging the four digits in $2021$?~(Adapted from $2004$ AMC $8$ Problem, Question \#2) 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$18$$ 
aoVal: E
content: $$24$$ 
Answer: C
Analysis: When the thousand digit is $2$, there are $\_3P\_3=6$ ways. When the thousand digit is $1$, there are $\_3C\_1=3$ ways. So the answer is $3+6=9$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Source: 其它
Problem: Paula shoots arrows at the following target. When she misses, she obtains zero points. Paula shoots two arrows and adds the number of points. Which of the following sums cannot be her score? 
Options:
aoVal: A
content: $$60$$ 
aoVal: B
content: $$70$$ 
aoVal: C
content: $$80$$ 
aoVal: D
content: $$90$$ 
aoVal: E
content: $$100$$ 
Answer: D
Analysis: A is possible as 30+30 = 60  B is possible as 70+0+0 = 70  C is possible as 30 + 50 = 80  E is ossible as 30 + 70 = 100. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: How many different natural numbers can be formed with the numbers $$1$$, $$2$$, $$3$$?（without using the same number two times like $$33$$） 
Options:
aoVal: A
content: $$6$$ 
aoVal: B
content: $$9$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$15$$ 
aoVal: E
content: $$18$$ 
Answer: D
Analysis: Classify the number first by the digits and then enumerate．  One-digit number：$$1$$、$$2$$、$$3$$，  Two-digit number：$$12$$、$$13$$、$$21$$、$$23$$、$$31$$、$$32$$；  Three-digit number：$$123$$、$$132$$、$$213$$、$$231$$、$$312$$、$$321$$；  So totally $$3+6+6=15$$ different natural numbers can be formed． 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Dictionary Ordering
Problem: Three kids $$A$$、$$B$$、$$C$$ are playing the game "pass the ball". If it starts with $$A$$, he can pass the ball to~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$A$$ or $$B$$ or $$C$$ 
aoVal: B
content: $$B$$ or $$C$$ 
aoVal: C
content: $$A$$ or $$C$$ 
Answer: B
Analysis: A cannot pass the ball back to himself. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Tree Diagrams
Problem: An acronym is a word formed from the first one or more letters of each word in a group of words. If "UFO"is an acronym for "unidentified flying object," then for how many of the following word groups could MATH be an acronym?  $$\text{I}$$. Multiply All Those Hundreds  $$\text{II}$$. MArtians Take Hostages  $$\text{III}$$. MATthew Hides  $$\text{IV}$$. Minutes After The Hour 
Options:
aoVal: A
content: one　 
aoVal: B
content: two　 
aoVal: C
content: three　 
aoVal: D
content: four　 
Answer: D
Analysis: For $$\text{I}$$ and $$\text{IV}$$, use first letter of each word; for $$\text{II}$$, use first two letters of first word and first letter of other words; and for $$\text{III}$$, use first three letters of first word and first letter of second word. MATH could be an acronym for all four word groups.  $$\text{I}$$. Multiply All Those Hundreds  $$\text{II}$$. MArtians Take Hostages  $$\text{III}$$. MATthew Hides  $$\text{IV}$$. Minutes After The Hour 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning by Conditions
Problem: There are $$10$$ players in a chess tournament. If each game is played by $$2$$ players, and each player plays every other player exactly once, what is the total number of games played in the tournament? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$90$$ 
aoVal: C
content: $$50$$ 
aoVal: D
content: $$45$$ 
Answer: D
Analysis: There are $10$ players and each player plays $9$ games, so there are $10 \times 9 \div 2 = 45$ games in total. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Gregor forms two numbers with digits $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, and $$6$$. Both numbers have three digits, and each digit is used only once. He adds these two numbers. What is the greatest sum Gregor can get? 
Options:
aoVal: A
content: $$975$$ 
aoVal: B
content: $$999$$ 
aoVal: C
content: $$1083$$ 
aoVal: D
content: $$1173$$ 
aoVal: E
content: $$1221$$ 
Answer: D
Analysis: If we want to get the sum as large as possible, the hundred digits for both numbers should be as large as possible. Therefore, they should be $$6$$ and $$5$$. The sum of the hundreds digits is $$11$$. For the same reason, the sum of the tens digits should be $$3+4=7$$ and the sum of the ones digits should be $$2 + 1 = 3$$. Therefore, $$\rm D$$ is the correct answer. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Complex Forming Numbers