Categories

## Arithmetic Sequence Problem | AIME I, 2012 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2012 based on Arithmetic Sequence.

## Arithmetic Sequence Problem – AIME 2012

The terms of an arithmetic sequence add to $715$. The first term of the sequence is increased by $1$, the second term is increased by $3$, the third term is increased by $5$, and in general, the $k$th term is increased by the $k$th odd positive integer. The terms of the new sequence add to $836$. Find the sum of the first, last, and middle terms of the original sequence.

• is 107
• is 195
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Number Theory

Algebra

AIME, 2012, Question 2.

Elementary Number Theory by David Burton .

## Try with Hints

After the adding of the odd numbers, the total of the sequence increases by $836 – 715 = 121 = 11^2$.

Since the sum of the first $n$ positive odd numbers is $n^2$, there must be $11$ terms in the sequence, so the mean of the sequence is $\frac{715}{11} = 65$.

Since the first, last, and middle terms are centered around the mean, then $65 \times 3 = 195$

Hence option B correct.

Categories

## Theory of Equations | AIME I, 2015 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2015 based on Theory of Equations.

## Theory of Equations – AIME I, 2015

The expressions A=$1\times2+3\times4+5\times6+…+37\times38+39$and B=$1+2\times3+4\times5+…+36\times37+38\times39$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers.Find the positive difference between integers A and B.

• is 722
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Equations

Number Theory

AIME I, 2015, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

A = $(1\times2)+(3\times4)$

$+(5\times6)+…+(35\times36)+(37\times38)+39$

B=$1+(2\times3)+(4\times5)$

$+(6\times7)+…+(36\times37)+(38\times39)$

B-A=$-38+(2\times2)+(2\times4)$

$+(2\times6)+…+(2\times36)+(2\times38)$

=722.

Categories

## Equations with number of variables | AIME I, 2009 | Question 14

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2009 based on Equations with a number of variables.

## Equations with number of variables – AIME 2009

For t=1,2,3,4, define $S^{t}=a^{t}_1+a^{t}_2+…+a^{t}_{350}$, where $a_{i}\in${1,2,3,4}. If $S_{1}=513, S_{4}=4745$, find the minimum possible value for $S_{2}$.

• is 905
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2009, Question 14

Polynomials by Barbeau

## Try with Hints

j=1,2,3,4, let $m_{j}$ number of $a_{i}$ s = j then $m_{1}+m{2}+m{3}+m{4}=350$, $S_{1}=m_{1}+2m_{2}+3m_{3}+4m_{4}=513$ $S_{4}=m_{1}+2^{4}m_{2}+3^{4}m_{3}+4^{4}m_{4}=4745$

Subtracting first from second, then first from third yields $m_{2}+2m_{3}+3m_{4}=163,$ and $15m_{2}+80m_{3}+255m_{4}=4395$ Now subtracting 15 times first from second gives $50m_{3}+210m_{4}=1950$ or $5m_{3}+21m_{4}=195$ Then $m_{4}$ multiple of 5, $m_{4}$ either 0 or 5

If $m_{4}=0$ then $m_{j}$ s (226,85,39,0) and if $m_{4}$=5 then $m_{j}$ s (215,112,18,5) Then $S_{2}=1^{2}(226)+2^{2}(85)+3^{2}(39)+4^{2}(0)=917$ and $S_{2}=1^{2}(215)+2^{2}(112)+3^{2}(18)+4^{2}(5)=905$ Then min 905.

Categories

## Probability of divisors | AIME I, 2010 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability of divisors.

## Probability of divisors – AIME I, 2010

Ramesh lists all the positive divisors of $2010^{2}$, she then randomly selects two distinct divisors from this list. Let p be the probability that exactly one of the selected divisors is a perfect square. The probability p can be expressed in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME I, 2010, Question 1

Elementary Number Theory by Sierpinsky

## Try with Hints

$2010^{2}=2^{2}3^{2}5^{2}67^{2}$

$(2+1)^{4}$ divisors, $2^{4}$ are squares

probability is $\frac{2.2^{4}.(3^{4}-2^{4})}{3^{4}(3^{4}-1)}=\frac{26}{81}$ implies m+n=107

Categories

## Geometric Sequence Problem | AIME I, 2009 | Question 1

Try this beautiful problem from American Invitational Mathematics Examination I, AIME I, 2009 based on geometric sequence.

## Geometric Sequence Problem – AIME 2009

Call a 3-digit number geometric if it has 3 distinct digits which, when read from left to right, form a geometric sequence. Find the difference between the largest and smallest geometric numbers.

• is 500
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Sequence

Series

Real Analysis

AIME, 2009

Introduction to Real Analysis, 4th Edition  by Robert G. Bartle, Donald R. Sherbert

## Try with Hints

3-digit sequence a, ar, $ar^{2}$. The largest geometric number must have a<=9.

ar $ar^{2}$ less than 9 r fraction less than 1 For a=9 is $\frac{2}{3}$ then number 964.

a>=1 ar and $ar^{2}$ greater than 1 r is 2 and number is 124. Then difference 964-124=840.

Categories

## Coordinate Geometry Problem | AIME I, 2009 Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2009 based on Coordinate Geometry.

## Coordinate Geometry Problem – AIME 2009

Consider the set of all triangles OPQ where O  is the origin and P and Q are distinct points in the plane with non negative integer coordinates (x,y) such that 41x+y=2009 . Find the number of such distinct triangles whose area is a positive integer.

• is 107
• is 600
• is 840
• cannot be determined from the given information

### Key Concepts

Algebra

Equations

Geometry

AIME, 2009, Question 11

Geometry Revisited by Coxeter

## Try with Hints

let P and Q be defined with coordinates; P=($x_1,y_1)$ and Q($x_2,y_2)$. Let the line 41x+y=2009 intersect the x-axis at X and the y-axis at Y . X (49,0) , and Y(0,2009). such that there are 50 points.

here [OPQ]=[OYX]-[OXQ] OY=2009 OX=49 such that [OYX]=$\frac{1}{2}$OY.OX=$\frac{1}{2}$2009.49 And [OYP]=$\frac{1}{2}$$2009x_1$  and [OXQ]=$\frac{1}{2}$(49)$y_2$.

2009.49 is odd, area OYX not integer of form k+$\frac{1}{2}$ where k is an integer

41x+y=2009 taking both 25  $\frac{25!}{2!23!}+\frac{25!}{2!23!}$=300+300=600.

.

Categories

## Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences.

## Problem on Series and Sequences (SMO Test)

For each positive integer $n \geq 1$ , we define the recursive relation given by $a_{n+1} = \frac {1}{1+a_{n}}$.

Suppose that $a_{1} = a_{2012}$.Find the sum of the squares of all

possible values of $a_{1}$.

• 2
• 3
• 6
• 12

### Key Concepts

Series and Sequence

Functional Equation

Recursive Relation

Challenges and Thrills – Pre – College Mathematics

## Try with Hints

If you got stuck start from here :

At first we have to understand the sequence it is following

Given that : $a_{n+1} = \frac {1}{1+a_{n}}$

Let $a_{1} = a$

so , $a_{2} = \frac {1}{1 + a_{1}}$ = $\frac {1}{1 + a}$

Again, $a_{3} = \frac {1}{1+a_{2}} = \frac {1+a}{2+a}$

For , $a_{4} = \frac {1}{1+a_{3}} = \frac {2+a}{3+2a}$

And , $a_{5} = \frac {1}{1+a_{4}} = \frac {3+2a}{5+3a}$

and so on……..

Try to do the rest …………………………….

Looking at the previous hint ………………

In general we can say ……………..

$a_{n} = \frac {F_{n} + F_{n-1}a}{F_{n+1} +F_{n}a}$

Where $F_{1} = 0 , F_{ 2} = 1$ and $F_{n+1} = F_{n}$ for all value of $n\geq 1$

Try to do the rest …….

Here is the rest of the solution,

If $a_{2012} = \frac {F_{2012}+F_{2011}a}{F_{2013} + F{2012}a} = a$

Then $(a^2+a-1 )F_{2012} = 0$

Since $F_{2012}>0$ we have $a^2 +a -1 = 0$ ……………………….(1)

Assume x and y are the two roots of the $eq^n (1)$, then

$x^2 + y^2 = (x+y)^2 -2xy = (-1)^2 – 2(-1) = 3$ (Answer)

Categories

## Theory of Equations | AIME I, 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Theory of Equations.

## Theory of Equations – AIME 2015

Let (f(x)) be a third-degree polynomial with real coefficients satisfying[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.]Find (|f(0)|).

• is 107
• is 72
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Algebra

AIME, 2015, Question 10.

Polynomials by Barbeau.

## Try with Hints

Let (f(x)) = (ax^3+bx^2+cx+d). Since (f(x)) is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up. By drawing a coordinate axis, and two lines representing (12) and (-12), it is easy to see that (f(1)=f(5)=f(6)), and (f(2)=f(3)=f(7)); otherwise more bends would be required in the graph.

Since only the absolute value of f(0) is required, there is no loss of generalization by stating that (f(1)=12), and (f(2)=-12). This provides the following system of equations.[a + b + c + d = 12][8a + 4b + 2c + d = -12][27a + 9b + 3c + d = -12][125a + 25b + 5c + d = 12][216a + 36b + 6c + d = 12][343a + 49b + 7c + d = -12]

Using any four of these functions as a system of equations yields (|f(0)| = \boxed{072})

Categories

## Number and Series | Number Theory | AIME I, 2015

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2015 based on Number Theory and Series.

## Number Theory and Series – AIME 2015

The expressions A = $(1 \times 2)+(3 \times 4)+….+(35 \times 36)+37$ and B = $1+(2 \times 3)+(4 \times 5)+….+(36 \times 37)$ are obtained by writing multiplication and addition operators in an alternating pattern between successive integers. Find the positive difference between integers A and B.

• is 107
• is 648
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Theory of Equations

Number Theory

AIME, 2015, Question 1

Elementary Number Theory by David Burton

## Try with Hints

B-A=$-36+(2 \times 3)+….+(2 \times 36)$

=$-36+4 \times (1+2+3+….+18)$

=$-36+(4 \times \frac{18 \times 19}{2})$=648.

Categories

## Probability Biased and Unbiased | AIME I, 2010 Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME 2010 based on Probability Biased and Unbiased.

## Probability Biased and Unbiased – AIME 2010

Ramesh and Suresh have two fair coins and a third coin that comes up heads with probability $\frac{4}{7}$,Ramesh flips the three coins, and then Suresh flips the three coins, let $\frac{m}{n}$ be the probability that Ramesh gets the same number of heads as Suresh, where m and n are relatively prime positive integers. Find m+n.

• is 107
• is 250
• is 840
• cannot be determined from the given information

### Key Concepts

Series

Probability

Number Theory

AIME, 2010, Question 4

Combinatorics by Brualdi

## Try with Hints

No heads TTT is $\frac{1.1.1}{2.2.7}=\frac{3}{28}$and $(\frac{3}{28})^{2}=\frac{9}{784}$

One Head HTT THT TTH with $\frac{3}{28}$ $\frac {3}{28}$ and $\frac{4}{28}$ then probability is $\frac{4(3.3)+4(3.4)+1(4.4)}{28^{2}}$=$\frac{100}{784}$

Two heads HHT $\frac{4}{28}$ HTH $\frac{4}{28}$ THH $\frac{3}{28}$ then probability is $\frac{1(3.3)+4(3.4)+4(4.4)}{28^{2}}$=$\frac{121}{784}$.

Three heads HHH is $\frac{4}{28}$ then probability $\frac{16}{784}$

Then sum is $\frac{9+100+121+16}{784}=\frac{123}{392}$ then 123+392=515.