Categories

## 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 $P(n)<\frac{1}{2010}$ ?

,

• $20$
• $22$
• $44$
• $45$
• $46$

Probability

Combination

Marbles

## Suggested Book | Source | Answer

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 $n=45$

Categories

## 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?

,

• $20$
• $22$
• $12$
• $25$
• $28$

Number theory

Combination

Circle

## Suggested Book | Source | Answer

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 $8 \choose 6$=$28$

Categories

## 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

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 $(24\times 9)$=$216$

Categories

## 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

Answer:$17$

PRMO-2018, Problem 17

Pre College Mathematics

## Try with Hints

Product of digits = $n^2 â€“ 15n â€“ 27 = n(n â€“ 15) â€“ 27$

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$\leq 9 Ã— 9 Ã— 9 = 729$ but $(n(n â€“ 15) â€“ 27)$ 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 $n^2 â€“ 15n â€“ 27 = n$ $\Rightarrow n$= 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

Categories

## 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

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 $\frac{42}{3}=14$

Categories

## 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?

• 34
• 12
• 26

### Key Concepts

Number theory

Number counting

integer

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……..

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

Categories

## 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

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, $\frac{2x}{5} – 34 = 62$

$\Rightarrow 2x-170=310$

$\Rightarrow 2x=480$

$\Rightarrow x=240$

Categories

## 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

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 .