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Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 195.

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.

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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

Check the Answer


Answer: is 722.

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.

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Algebra Arithmetic Functional Equations Math Olympiad Math Olympiad Videos USA Math Olympiad

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

Check the Answer


Answer: is 905.

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.

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AIME I Algebra Arithmetic Combinatorics Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 107.

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

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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

Check the Answer


Answer: is 840.

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.

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Coordinate Geometry Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 600.

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.

.

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AMC 10 USA Math Olympiad

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

Check the Answer


Answer: 3

Singapore Mathematics Olympiad

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)

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 72.

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})

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AIME I Algebra Arithmetic Calculus Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 648.

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.

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AIME I Arithmetic Combinatorics Math Olympiad USA Math Olympiad

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

Check the Answer


Answer: is 107.

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.

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