math-eval/TAL-SCQ5K huggingface dataset:
Problem: Four friends, Edwin, Fred, Gary and Howard, were playing together when one of them broke a vase. The teacher asked,~"Who is the culprit?"  \textbf{Both Edwin and Howard said, "Not me."}  \textbf{Fred said, "Howard broke the vase."}  \textbf{Gary said, "Fred is the culprit."}  If only one of four boys was lying, who broke the vase? 
Options:
aoVal: A
content: Edwin 
aoVal: B
content: Howard 
aoVal: C
content: Fred 
aoVal: D
content: Gary 
Answer: C
Analysis: Either Fred or Howard must be lying since what they said is conflicting. Since only one person was lying, Gary was telling the truth, which means Fred broke the vase. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Teacher Anderson has pens with three different colours, $$12$$ black, $$8$$ red and $$10$$ green. He put all of these pens in his pencil case. If he close his eyes and randomly pick a pen, at least how many pen he must pick to guarantee he get a red pen?． 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$22$$ 
aoVal: E
content: $$23$$ 
Answer: E
Analysis: To guarantee he get a red pen, the worse cases scenario is he picked all the green and black pens. Which is $$12+10=22$$. Then, the next pen must be the red. Thus, $$22+1=23$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Sophia's average score on six tests is $$82$$. Her average score on the $$7^{}\text{th}$$ and $$8^{}\text{th}$$ test is $$98$$. What is her average score on all eight tests? 
Options:
aoVal: A
content: $$86$$ 
aoVal: B
content: $$88$$ 
aoVal: C
content: $$90$$ 
aoVal: D
content: $$94$$ 
Answer: A
Analysis: Sophia\textquotesingle s total score on the first six tests is $$6\times82=492$$. Her total score on all eight tests is $$492+2\times98=688$$, and her average score is $$688\div8=86$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Problem: In my suitcase I have $$5$$ sweaters and $$6$$ pairs of pants. If I make an outfit of a sweater and a pair of pants, how many different outfits can I select? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$22$$ 
aoVal: C
content: $$25$$ 
aoVal: D
content: $$30$$ 
Answer: D
Analysis: I have $$5$$ sweaters and $$6$$ pairs of pants. For each sweater, there are $$6$$ pairs of pants with which that sweater can be paired. There are $$5$$ sweaters, so there are $$5\times6=30$$ different possible outfits. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The time on a $$12$$-hour circular clock is $$11:00$$ A.M. When the \emph{minute} hand goes around $$3$$ times, the time will be． 
Options:
aoVal: A
content: $$11:03$$ A.M. 
aoVal: B
content: $$11:30$$ A.M. 
aoVal: C
content: $$1:00$$ P.M. 
aoVal: D
content: $$2:00$$ P.M. 
Answer: D
Analysis: It takes $$3$$ hours for the \emph{minute} hand to go around $$3$$ times, so the time will be $$2$$ P.M. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: How many ways are there to rearrange the letters in the word \textquotesingle BEAUTY\textquotesingle{} if the vowels are never together? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$48$$ 
aoVal: C
content: $$144$$ 
aoVal: D
content: $$480$$ 
aoVal: E
content: $$720$$ 
Answer: C
Analysis: $3\times2\times1=6$  $4\times3\times2=24$  $24\times6=144$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Four friends, Edwin, Fred, Gary and Howard, were playing together  when one of them broke a vase.  The teacher asked: "Who is the culprit?"  Both Edwin and Howard said, "Not me."  Fred said, "Howard broke the vase."  Gary said, "Fred is the culprit."  If only one of four boys was lying, who broke the vase? 
Options:
aoVal: A
content: Edwin 
aoVal: B
content: Fred 
aoVal: C
content: Gary 
aoVal: D
content: Howard 
Answer: B
Analysis: Either Fred or Howard must be lying since what they said did not tally.  Since only one person was lying, Gary was telling the truth i.e, Fred broke the vase. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Logical Reasoning->Reasoning using Hypothesis
Problem: Eight pupils stand in a row to take a photo. Four of them insist on standing together. How many different ways are there to arrange them? 
Options:
aoVal: A
content: $$24$$ 
aoVal: B
content: $$120$$ 
aoVal: C
content: $$576$$ 
aoVal: D
content: $$2880$$ 
aoVal: E
content: $$14400$$ 
Answer: D
Analysis: $8-4+1=5$  $5\times4\times3\times2\times1=120$  $4\times3\times2\times1=24$  $120\times24=2880$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Two different numbers are randomly selected from the set $-2, -1, 0, 3, 4, 5$~and multiplied together. What is the probability that the product is $0$? ($2016$ AMC $8$ Problem, Question \#$13$) 
Options:
aoVal: A
content: $\frac{1}{6}$ 
aoVal: B
content: $\frac{1}{5}$ 
aoVal: C
content: $\frac{1}{4}$ 
aoVal: D
content: $\frac{1}{3}$ 
aoVal: E
content: $\frac{1}{2}$ 
Answer: D
Analysis: The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are $15$ ways of choosing $2$ numbers randomly. Thus $\frac{5}{15} = \frac{1}{3}$. The answer is $D$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability
Source: 其它
Problem: Four students asked their teacher, Mr. Carter, to line up with them to take a picture.  ①If Mr. Carter does not want to stand on either ends, how many different ways can they line up for the picture?  ②If Mr. Carter insists on standing on one of the $2$ ends, how many different ways can they line up for the picture? 
Options:
aoVal: A
content: $72$ , $24$ 
aoVal: B
content: $96$ , $24$ 
aoVal: C
content: $72$ , $48$ 
aoVal: D
content: $96$ , $48$ 
Answer: C
Analysis: ①$$3\times 4\times 3\times 2\times 1=72$$ ,  ②$$2\times 4\times 3\times 2\times 1=48$$ . 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Queuing Problems
Problem: Molly, Dolly, Sally, Elly and Kelly are sitting on a park bench. Molly is not sitting on the far right and Dolly is not sitting on the far left. Sally is not sitting at either end. Kelly is not sitting next to Sally and Sally is not sitting next to Dolly. Elly is sitting to the right of Dolly but not necessarily next to her. Who is sitting at the far right end? 
Options:
aoVal: A
content: Molly　 
aoVal: B
content: Dolly　 
aoVal: C
content: Sally　 
aoVal: D
content: Kelly　 
aoVal: E
content: Elly　 
Answer: E
Analysis: The question tells us that Sally is not sitting at either end. This leaves three possible positions for Sally, which we will call positions $$2$$, $$3$$ and $$4$$ from the left-hand end. Were Sally to sit in place $$2$$, neither Dolly nor Kelly could sit in places $$1$$ or $$3$$ as they cannot sit next to Sally and, since Elly must sit to the right of Dolly, there would be three people to fit into places $$4$$ and $$5$$ which is impossible. Similarly, were Sally to sit in place $$3$$, Dolly could not sit in place $$2$$ or $$4$$ and the question also tells us she cannot sit in place $$1$$ so Dolly would have to sit in place $$5$$ making it impossible for Elly to sit to the right of Dolly. However, were Sally to sit in place $$4$$, Dolly could sit in place $$2$$, Kelly in place $$1$$, Molly (who cannot sit in place $$5$$) in place $$3$$ leaving Elly to sit in place $$5$$ at the right-hand end. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: The faces of each of two fair dice are numbered $1$, $2$, $3$, $5$, $7$, and $8$. When the two dice are tossed, what is the probability that their sum will be an even number? ． 
Options:
aoVal: A
content: $\dfrac{4}{9}$ 
aoVal: B
content: $\dfrac{1}{2}$ 
aoVal: C
content: $\dfrac{5}{9}$ 
aoVal: D
content: $\dfrac{3}{5}$ 
aoVal: E
content: $\dfrac{2}{3}$ 
Answer: C
Analysis: There are two cases in which the sum can be an even number: both numbers are even and both numbers are odd. This results in only one case where the sum of the numbers are odd (one odd and one even in any order). We can solve for how many ways the $2$ numbers add up to an odd number and subtract the answer from $1$.  How to solve the problem: The probability of getting an odd number first is $\dfrac{4}{6}=\dfrac{2}{3}$. In order to make the sum odd, we must select an even number next. The probability of getting an even number is $\dfrac{2}{6}=\dfrac{1}{3}$. Now we multiply the two fractions: $\dfrac{2}{3}\times\dfrac{1}{3}=\dfrac{2}{9}$. However, this is not the answer because we could pick an even number first then an odd number. The equation is the same except switched, and by the Communitive Property of Multiplication, it does not matter if the equations are switched. Thus we do $\dfrac{2}{9}\times2=\dfrac{4}{9}$. This is the probability of getting an odd-number sum. In order to get the probability of getting an even number we do $1-\dfrac{4}{9}=\left (\text{C}\right )\dfrac{5}{9}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Statistics and Probability->Questions Involving Probability->Basic Concepts of Probability
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: $$112$$ 
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: What is the angle between the hour hand and the minute hand at seven o\textquotesingle clock?~ ~ ． 
Options:
aoVal: A
content: $50^{}\circ $ 
aoVal: B
content: $120^{}\circ $ 
aoVal: C
content: $135^{}\circ $ 
aoVal: D
content: $150^{}\circ $ 
aoVal: E
content: $165^{}\circ $ 
Answer: D
Analysis: The smaller angle is $\frac 5{12}$ of a full circle.  A full circle has $360$ degrees, so the angle is $\frac 5{12}\times 360^{}\circ =150^{}\circ $.  So, the answer is $\rm D$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Reading the Clock
Problem: A fair $6-$sided die is rolled twice. What is the probability that the sum of the two rolls\textquotesingle{} outcomes is a perfect square? 
Options:
aoVal: A
content: $\dfrac{1}{27}$ 
aoVal: B
content: $\dfrac{7}{9}$ 
aoVal: C
content: $\dfrac{7}{36}$ 
aoVal: D
content: $\dfrac{1}{9}$ 
aoVal: E
content: $\dfrac{5}{18}$ 
Answer: C
Analysis: There are $7$ out of $36$ outcomes are perfect squares: $1+3$, $2+2$, $3+1$, $3+6$, $4+5$, $5+4$, $6+3$. 
Knowledge points: Overseas Competition->Knowledge Point->Number Theory Modules->Perfect Square Numbers->Basic Applications of Square Numbers
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: $$30$$ 
aoVal: D
content: $$36$$ 
Answer: D
Analysis: $$8+7+6+5+4+3+2+1=36$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Problem: $$150$$ students went to a $$\text{CCA}$$ fair and all of them at least tried out one activity. $$85$$ students tried out Sports activity. $$77$$ students tried out Art activity. How many students tried out both Sport activity and Art activities? 
Options:
aoVal: A
content: $$11$$ 
aoVal: B
content: $$12$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$14$$ 
Answer: B
Analysis: $$85+77-150=12$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Maria went to the market at $$9:35$$ a.m. She spent $$1.5$$ hours at the market, and she took an hour to reach home. What time did she reach home? 
Options:
aoVal: A
content: $11:05$ am. 
aoVal: B
content: $12:05$ am. 
aoVal: C
content: $11:45$ am. 
aoVal: D
content: $11:55$ am. 
aoVal: E
content: $12:00$ am. 
Answer: B
Analysis: $9:35+1:30+1=12:05$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Time Problem->Time Calculation
Source: 其它
Problem: $$9+99+999+9999+99999=$$ 
Options:
aoVal: A
content: $$111109$$ 
aoVal: B
content: $$111119$$ 
aoVal: C
content: $$111100$$ 
aoVal: D
content: $$111105$$ 
Answer: D
Analysis: $9+99+999+9999+99999=10+100+1000+10000+100000-5=111105$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: What is the average of $$1$$, $$2$$, $$3$$, $$4$$, $$5$$, $$6$$, $$7$$, $$8$$, and $$9$$? 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$9$$ 
aoVal: D
content: $$45$$ 
Answer: A
Analysis: The average is the middle number, $$5$$. 
Knowledge points: Overseas Competition->Knowledge Point->Word Problem Modules->Average Problems ->Using Formulas
Problem: Pat and Lee counted leaves on two plants. Pat\textquotesingle s got a $$1$$-digit number. Lee got a $$3$$-digit number. If the difference between their numbers was $$91$$, what was the sum of their numbers? 
Options:
aoVal: A
content: $$100$$ 
aoVal: B
content: $$109$$ 
aoVal: C
content: $$191$$ 
aoVal: D
content: $$200$$ 
Answer: B
Analysis: Pat and Lee counted leaves on two plants. Pat\textquotesingle s got a $$1$$-digit number. Lee got a $$3$$-digit number. If the dīfference of the numbers was $$91$$, the numbers were $$100$$ and $$9$$, and the sum is $$109$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Problem: How many whole numbers between $$1$$ and $$1000$$ do not contain the digit $$1$$? ($2009$ AMC $8$ Problems, Question \#$22$) 
Options:
aoVal: A
content: $$512$$ 
aoVal: B
content: $$648$$ 
aoVal: C
content: $$720$$ 
aoVal: D
content: $$728$$ 
aoVal: E
content: $$800$$ 
Answer: D
Analysis: One-digit number: $9-1=8$  Two-digit number: $8\times9=72$  Three-digit number: $8\times9\times9=648$  So, there are $8+72+648=728$ numbers that meet the requirements. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: The product of the formula "$486\times$~\uline{~~~~~~~~~~}~$5$ = $7$~\uline{~~~~~~~~~~}~$90$" is~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$7190$$ 
aoVal: B
content: $$7290$$ 
aoVal: C
content: $$7390$$ 
aoVal: D
content: $$7490$$ 
Answer: B
Analysis: $486\times 15 = 7290$ 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Number Puzzles->Number Puzzles (horizontal forms)
Problem: Three girls and two boys were dancing. They danced in pairs so that each girl danced with each boy for exactly one minute. At any time, there was only one pair on the dance floor. For how many minutes did they dance? (2021 Math Kangaroo Problem, Level 1-2, Question \#21) 
Options:
aoVal: A
content: $$5$$ 
aoVal: B
content: $$6$$ 
aoVal: C
content: $$8$$ 
aoVal: D
content: $$9$$ 
aoVal: E
content: $$10$$ 
Answer: B
Analysis: In total, there are $2+2+2=6$ different pairs. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication->Matching Objects
Source: 其它
Problem: ~$19+37+81=$~\uline{~~~~~~~~~~}~ 
Options:
aoVal: A
content: $$130$$ 
aoVal: B
content: $$137$$ 
aoVal: C
content: $$138$$ 
aoVal: D
content: $$129$$ 
Answer: B
Analysis: $19+37+81=19+81+37=100+37=137$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Law of Addition
Source: 其它
Problem: A cuboid has a length of $2\textasciitilde cm$, breadth of $4\textasciitilde cm$ and a height of $8\textasciitilde cm$. A cube has the same volume. Determine the side length of the cube. 
Options:
aoVal: A
content: $3\textasciitilde cm$ 
aoVal: B
content: $4\textasciitilde cm$ 
aoVal: C
content: $5\textasciitilde cm$ 
aoVal: D
content: $6\textasciitilde cm$ 
Answer: B
Analysis: Volume of cube $=2\times4\times8=64\textasciitilde cm^{3}$  Side length of cube $=$$\sqrt[3]{64}$$=4\textasciitilde cm$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: $$25\times 30\times 6=$$ 
Options:
aoVal: A
content: $$5800$$ 
aoVal: B
content: $$4500$$ 
aoVal: C
content: $$4300$$ 
aoVal: D
content: $$4750$$ 
Answer: B
Analysis: $$Nil$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication
Source: 其它
Problem: In Carl\textquotesingle s pencil case there are nine pencils. At least one of the pencils is blue. In any group of four pencils, at least two have the same colour. In any group of five pencils, at most three have the same colour. How many pencils are blue? 
Options:
aoVal: A
content: $$1$$ 
aoVal: B
content: $$2$$ 
aoVal: C
content: $$3$$ 
aoVal: D
content: $$4$$ 
aoVal: E
content: More information needed 
Answer: C
Analysis: The information that in any group of four pencils, at least two have the same colour, tells us that there at most three different coloured pencils in Carl\textquotesingle s pencil case. The information that in any group of five pencils, at most three have the same colour, tells us that there are at most three pencils of any single colour in the pencil case. Hence there are three pencils of each of the three different colours and so Carl\textquotesingle s pencil case contains three blue pencils. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations
Problem: Eddie is ordering lunch at a fast food restaurant that has sandwiches and burgers on the lunch menu, along with coffee, milk, and tea as drink options. If Eddie chooses one food and one drink from the lunch menu, he hasdifferent ways to order lunch. 
Options:
aoVal: A
content: $$4$$ 
aoVal: B
content: $$5$$ 
aoVal: C
content: $$6$$ 
aoVal: D
content: $$7$$ 
Answer: C
Analysis: $$2\times 3=6$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Law of Addition and Multiplication->Principle of Multiplication
Problem: Suzie flips a fair coin $6$ times. The probability that Suzie flips $3$ heads in a row but not $4$ heads in a row is given by $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. $m+n=$~\uline{~~~~~~~~~~}~.  $\textasciitilde$ 
Options:
aoVal: A
content: $$9$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$17$$ 
aoVal: D
content: $$19$$ 
aoVal: E
content: $$35$$ 
Answer: D
Analysis: Consider the distribution of the three consecutive heads. We have four cases:  1110xx, 01110x, x01110, xx0111.  Since case 1 and case4, case 2 and case 3 are the same, we only need to think about case 1 and case 2.  The probability of case 1: $(\dfrac{1}{2})^{4}=\frac{1}{16}$.  The probability of case 2: $(\dfrac{1}{2})^{5}=\frac{1}{32}$.  The sum of the probabilities of 4 cases is $2\times(\frac{1}{16}+\frac{1}{32})=\frac{3}{16}$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: Calculate:~$4+5+6+\cdots +19+20=$． 
Options:
aoVal: A
content: $$192$$ 
aoVal: B
content: $$196$$ 
aoVal: C
content: $$200$$ 
aoVal: D
content: $$204$$ 
aoVal: E
content: $$208$$ 
Answer: D
Analysis: $(4+20)\times17\div2=204$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules
Source: 其它
Problem: In Tim\textquotesingle s class, there are $$20$$ students who can swim, $$25$$ students who can play basketball, and $$10$$ students who can do both. If everyone in the class plays at least one sport, how many students are there in Tim\textquotesingle s class?~\uline{~~~~~~~~~~}~. 
Options:
aoVal: A
content: $$25$$ 
aoVal: B
content: $$35$$ 
aoVal: C
content: $$45$$ 
aoVal: D
content: $$55$$ 
Answer: B
Analysis: $$20+25-10=35$$. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle->Inclusion-Exclusion Principle for Two Sets
Problem: Dexter sprinted for the first $600\textasciitilde m$ of a race in $160\textasciitilde s$ and jogged the remaining $\frac{2}{5}$ of the race in $240\textasciitilde s$. What was his average speed? 
Options:
aoVal: A
content: $2.5\textasciitilde m/s$ 
aoVal: B
content: $5\textasciitilde m/s$ 
aoVal: C
content: $2\textasciitilde m/s$ 
aoVal: D
content: $0.4\textasciitilde m/s$ 
Answer: A
Analysis: Average speed $=\frac{Total distance}{Total Time}=\frac{1000\textasciitilde m}{400\textasciitilde s}=2.5\textasciitilde m/s$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Inclusion-Exclusion Principle
Source: 其它
Problem: April wants to split $$6$$ identical sweets into $$2$$ identical containers. How many different ways can she do so if a container can be empty? 
Options:
aoVal: A
content: $$2$$ 
aoVal: B
content: $$3$$ 
aoVal: C
content: $$4$$ 
aoVal: D
content: $$6$$ 
Answer: C
Analysis: $$6=0+6$$  $$6=1+5$$  $$6=2+4$$  $$6=3+3$$ 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Questions Involving Enumeration->Enumeration
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->Law of Addition and Multiplication->Law of Addition
Problem: Tiffany uses $3$ identical red gems, $6$ identical blue gems, and $9$ identical pearls to make a necklace. If any two gems cannot be adjacent, how many ways does she have to make the necklace? (After flipping or rotating, if two neckleces can be the same, then count them as $1$ way.) 
Options:
aoVal: A
content: $$7$$ 
aoVal: B
content: $$10$$ 
aoVal: C
content: $$12$$ 
aoVal: D
content: $$20$$ 
aoVal: E
content: $$36$$ 
Answer: A
Analysis: The $9$ pearls should be put in each of the two gems. Thus, just consider the combination of gems:  if all red gems are together, she has $1$ way;  if two red gems are together, she has $3$ ways;  if all the red gems are not together, the blue gems can be put $1-2-3$, $1-1-4$ or $2-2-2$, so there are $3$ ways.  There are $7$ ways. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations
Problem: The product of two whole numbers is $$30$$. What is the least possible value of their sum? 
Options:
aoVal: A
content: $$10$$ 
aoVal: B
content: $$11$$ 
aoVal: C
content: $$13$$ 
aoVal: D
content: $$31$$ 
Answer: B
Analysis: The product of two whole numbers is $$30$$. If the numbers are $$5$$ and $$6$$, their sum is $$5+6=11$$. 
Knowledge points: Overseas Competition->Knowledge Point->Combinatorics->Combinatorics Involving Extreme Values->Problems of Extreme Value with Fixed Products
Problem: Amos is taller than Eugene.  Leo is shorter than James but taller than Eugene.  James is shorter than Amos.  is the tallest andis the shortest. 
Options:
aoVal: A
content: Amos, James 
aoVal: B
content: James, Eugene 
aoVal: C
content: James, Leo 
aoVal: D
content: Amos, Eugene 
Answer: D
Analysis: From clue $$2$$, James is taller than Leo and Leo is taller than Eugene.  From clue $$3$$, Amos is taller than James.  Rank from tallest to shortest: \textbf{Amos}, James, Leo, \textbf{Eugene}. 
Knowledge points: Overseas Competition->Knowledge Point->Counting Modules->Permutations and Combinations->Combinations