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AMC 10 Combinatorics Math Olympiad Number Theory Probability USA Math Olympiad

Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability in Marbles based on smallest value AMC 10 A, 2010. You may use sequential hints to solve the problem.

Probability in Marbles – AMC-10A, 2010- Problem 23


Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles. What is the smallest value of $n$ for which $P(n)<\frac{1}{2010}$ ?

,

  • $20$
  • $22$
  • $44$
  • $45$
  • $46$

Key Concepts


Probability

Combination

Marbles

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-23

Check the answer here, but try the problem first

$45$

Try with Hints


First Hint

Given that Each of 2010 boxes in a line contains a single red marble, and for $1 \leq k \leq 2010$, the box in the $k$ th position also contains $k$ white marbles..

Therefore The probability of drawing a white marble from box $k$ is $\frac{k}{k+1}$ and the probability of drawing a red marble from box $k$ is $\frac{1}{k+1}$

Now can you finish the problem?

Second Hint

Also given that She stops when she first draws a red marble. Let $P(n)$ be the probability that Isabella stops after drawing exactly $n$ marbles.

Therefore we can say $P(n)=\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \cdots \frac{n-1}{n}\right) \cdot \frac{1}{n+1}=\frac{1}{n(n+1)}$

Now Can you finish the Problem?

Third Hint

Therefore the probability $\frac{1}{n(n+1)}<\frac{1}{2010}$ or $n(n+1)>2010$

Now $ n^2+n-2010>0$

Now to find out the factorization we see that $45 \times 46=2070$ and $ 44 \times 45 =1980$

As $n$ is smallest so $n=45$

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AMC 10 Math Olympiad Number Theory USA Math Olympiad

Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Points on a circle from AMC 10 A, 2010. You may use sequential hints to solve the problem.

Points on a circle – AMC-10A, 2010- Problem 22


Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle are created?

,

  • $20$
  • $22$
  • $12$
  • $25$
  • $28$

Key Concepts


Number theory

Combination

Circle

Suggested Book | Source | Answer


Suggested Reading

Pre College Mathematics

Source of the problem

AMC-10A, 2010 Problem-22

Check the answer here, but try the problem first

$28$

Try with Hints


First Hint

To create a chord we have to nedd two points. Threfore three chords to create a triangle and not intersect at a single point , we have to choose six points.

Now can you finish the problem?

Second Hint

Now the condition is No three chords intersect in a single point inside the circle.Now we know that for any six points we pick, there is only 1 way to connect the points such that a triangle is formed in the circle’s interior

Now Can you finish the Problem?

Third Hint

Therefore the required answer is $ 8 \choose 6$=$28$

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India Math Olympiad Math Olympiad Number Theory PRMO USA Math Olympiad

Sum of digits Problem | PRMO 2016 | Question 6

Try this beautiful problem from Number system based on sum of digits.

Sum of digits | PRMO | Problem 6


Find the sum of digits in decimal form of the number \((9999….9)^3\) (There are 12 nines)

  • $200$
  • $216$
  • $230$

Key Concepts


Number system

Digits

counting

Check the Answer


Answer:$216$

PRMO-2016, Problem 6

Pre College Mathematics

Try with Hints


we don’t know what will be the expression of \((9999….9)^3\). so we observe….

\(9^3\)=\(729\)

\((99)^3\)=\(970299\)

\((999)^3\)=\(997002999\)

……………..

……………

we observe that,There is a pattern such that…

In \((99)^3\)=\(970299\) there are 1-nine,1-seven,1-zero,1-two,2-nines & \((999)^3\)=\(997002999\) there are 2- nines,1-seven,2-zeros,1-two,3-nines….so in this way…..\((999….9)^3\) will be 11-nines,1-seven,11-zeros,1-two,12-nines……….

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)

can you finish the problem?

Therefore \((999….9)^3\)=\((99999999999) 7 (00000000000) 2(999999999999)\)…….

total numbers of Nines are (11+12) and (7+2)=9(another one) …..so total (11+12+1)=24 nines and the sum be \((24\times 9)\)=\(216\)

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Math Olympiad Number Theory PRMO USA Math Olympiad

Integer based Problem | PRMO-2018 | Question 20

Try this beautiful Integer based Problem from Algebra, from PRMO 2018.

Integer based Problem – PRMO 2018, Question 20


Determine the sum of all possible positive integers n, the product of whose digits equals \(n^2 -15n – 27 \)

  • $9$
  • $17$
  • $34$

Key Concepts


Algebra

Integer

multiplication

Check the Answer


Answer:$17$

PRMO-2018, Problem 17

Pre College Mathematics

Try with Hints


Product of digits = \(n^2 – 15n – 27 = n(n – 15) – 27\)

so at first we observe when n=one digit ,two digit and 3 digit numbers…..

If n is a more than 2-digit number, say 3-digit number, then product has to be\(\leq 9 × 9 × 9 = 729\) but \((n(n – 15) – 27)\) is more than 729 (in fact it a more than 3-digit numbers for any 3-digit n). Hence, n can be either one-digit or 2-digit.

If n is 1-digit then \(n^2 – 15n – 27 = n\) \(\Rightarrow n\)= not an integer , so n is a two digit number

now we will observe for 2-digit numbers…..

Can you now finish the problem ……….

For Two-digit numbers:

As product is positive so n(n-15)-27>0\(\Rightarrow n\geq 17\)

Now two digit product is less than equal to 81

so \(n(n-15)-27\leq 1\)\(\Rightarrow n(n-15)\leq 108\) \(\Rightarrow n\leq 20\)

Therefore n can be \(17\),\(18\),\(19\) or \(20\)

Can you finish the problem……..

For \(n\)= \(17\),\(18\),\(19\) or \(20\)

when n=17,then \(n(n-15)-27=7=1 \times 7\)

when n=18,then \(n(n-15)-27=27\neq 1\times 8\)

when n=19,then \(n(n-15)-27=49=1 \neq 9\)

when n=20,then \(n(n-15)-27=73=1 \neq 0\)

Therefore \(n\)=17

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AMC 8 Math Olympiad Number Theory USA Math Olympiad

Prime numbers | AMC 8, 2006| Problem 25

Try this beautiful problem from Algebra based on Prime numbers.

Algebra based on Number theory – AMC-8, 2009 – Problem 23


Barry wrote 6 different numbers, one on each side of 3 cards, and laid the cards on a table, as shown. The sums of the two numbers on each of the three cards are equal. The three numbers on the hidden sides are prime numbers. What is the average of the hidden prime numbers?

  • $14$
  • $12$
  • $16$

Key Concepts


Algebra

Number theory

card number

Check the Answer


Answer:$14$

AMC-8 (2006) Problem 25

Pre College Mathematics

Try with Hints


Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained.

Can you now finish the problem ……….

Obtain this even number would be to add another even number to 44

Can you finish the problem……..

Notice that 44 and 38 are both even, while 59 is odd. If any odd prime is added to 59, an even number will be obtained. However, the only way to obtain this even number would be to add another even number to 44 , and a different one to 38. Since there is only one even prime ( 2 ), the middle card’s hidden number cannot be an odd prime, and so must be even. Therefore, the middle card’s hidden number must be 2, so the constant sum is 59+2=61. Thus, the first card’s hidden number is 61-44=17, and the last card’s hidden number is 61-38=23

Since the sum of the hidden primes is 2+17+23=42, the average of the primes is \(\frac{42}{3}=14\)

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AMC 8 Math Olympiad Number Theory

Largest and smallest numbers | AMC 8, 2006 | Problem 22

Try this beautiful problem from Algebra about Largest and smallest numbers from AMC-8, 2006.

Largest and smallest numbers | AMC-8, 2006|Problem 22


Three different one-digit positive integers are placed in the bottom row of cells. Numbers in adjacent cells are added and the sum is placed in the cell above them. In the second row, continue the same process to obtain a number in the top cell. What is the difference between the largest and smallest numbers possible in the top cell?

Largest and smallest numbers

  • 34
  • 12
  • 26

Key Concepts


Number theory

Number counting

integer

Check the Answer


Answer:$26$

AMC-8, 2006 problem 22

Challenges and Thrills in Pre College Mathematics

Try with Hints


the lower cells contain A,B and C,

Can you now finish the problem ……….

the second row will contain A+B and B+C and the top cell will contain A+2B+C.

can you finish the problem……..

Largest and smallest numbers

If the lower cells contain A,B and C, then the second row will contain A+B and B+C and the top cell will contain A+2B+C. To obtain the smallest sum, place 1 in the center cell and 2 and 3 in the outer ones. The top number will be 7 . For the largest sum, place 9 in the center cell and 7 and 8 in the outer ones. This top number will be 33. The difference is 33-7=26

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AMC 8 Math Olympiad Number Theory

Page number counting |AMC 8- 2010 -|Problem 21

Try this beautiful problem from Algebra about Page number counting

Page number counting | AMC-8, 2010 |Problem 21


Hui is an avid reader. She bought a copy of the best seller Math is Beautiful. On the first day, Hui read \(\frac{1}{5}\) of the pages plus more, and on the second day she read  \(\frac{1}{4}\) of the remaining pages plus 15 pages. On the third day she read \(\frac{1}{3}\) of the remaining pages plus 18 pages. She then realized that there were only  62 pages left to read, which she read the next day. How many pages are in this book?

  • 320
  • 240
  • 200

Key Concepts


Algebra

Arithmetic

multiplication

Check the Answer


Answer:$240$

AMC-8, 2010 problem 21

Challenges and Thrills in Pre College Mathematics

Try with Hints


assume that the number of all pages be \(x\)

Can you now finish the problem ……….

count day by day

can you finish the problem……..

Let x be the number of pages in the book

First day ,Hui Read \(\frac{x}{5} + 12\) pages

After first day Remaining pages=\(\{x-(\frac{x}{5}+12)\}\)=\(\frac{4x}{5} -12\)

Second day ,Hui Read \(\frac{1}{4} (\frac{4x}{5} -12) +15=\frac{x}{5} +12\)

After Second day Remaining pages= \((\frac{4x}{5} -12) -(\frac{x}{5} +12)\)=\(\frac{4x}{5} -\frac{x}{5}-24\)=\(\frac{3x}{5} -24\)

Third day,Hui read \(\frac {1}{3} (\frac{3x}{5} -24) +18\) =\((\frac{x}{5} -8+18)\)=\(\frac{x}{5} +10\)

After Third day Remaining pages = \((\frac{3x}{5} -24) -(\frac{x}{5} +10)\) =\(\frac{2x}{5} – 34\)

Now by the condition, \(\frac{2x}{5} – 34 = 62\)

\(\Rightarrow 2x-170=310\)

\(\Rightarrow 2x=480\)

\(\Rightarrow x=240\)

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Algebra AMC 8 Math Olympiad Number Theory

Least common multiple | AMC 8, 2016 – Problem 20

LCM – AMC 8, 2016 – Problem 20


The least common multiple of a and b is 12 .and the lest common multiple of b and c is 15.what is the least possible value of the least common multiple of a and c?

  • 30
  • 60
  • 20

Key Concepts


Algebra

Division algorithm

Integer

Check the Answer


Answer:20

AMC-8, 2016 problem 20

Challenges and Thrills of Pre College Mathematics

Try with Hints


Find greatest common factors

Can you now finish the problem ……….

Find Least common multiple….

can you finish the problem……..

we wish to find possible values of a,b and c .By finding the greatest common factor 12 and 15, algebrically ,it’s some multiple of b and from looking at the numbers ,we are sure that it is 3.Moving on to a and c ,in order to minimize them,we wish to find the least such that the LCM of a and 3 is 12,$\to 4$.similarly with 3 and c,we obtain 5.the LCM of 4 and 5 is 20 .

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