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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 Algebra Arithmetic Math Olympiad USA Math Olympiad

Number Theory of Primes | AIME I, 2015

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

Number Theory of Primes – AIME 2015


There is a prime number p such that 16p+1 is the cube of a positive integer. Find p.

  • is 307
  • is 250
  • is 840
  • cannot be determined from the given information

Key Concepts


Series

Theory of Equations

Number Theory

Check the Answer


Answer: is 307.

AIME, 2015

Elementary Number Theory by Sierpinsky

Try with Hints


Notice that 16p+1must be in the form \((a+1)^{3}=a^{3}+3a^{2}+3a\), or \(16p=a(a^{2}+3a+3)\). Since p must be prime, we either have p=a or a=16

p not equal to a then we have a=16,

p\(=16^{2}+3(16)+3=307

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AIME I Complex Numbers Math Olympiad Math Olympiad Videos USA Math Olympiad

Complex Numbers | AIME I, 2009 | Problem 2

Try this beautiful problem from AIME, 2009 based on complex numbers.

Complex Numbers – AIME, 2009


There is a complex number z with imaginary part 164 and a positive integer n such that $\frac{z}{z+n}=4i$, Find n.

  • 101
  • 201
  • 301
  • 697

Key Concepts


Complex Numbers

Theory of equations

Polynomials

Check the Answer


Answer: 697.

AIME, 2009, Problem 2

Complex Numbers from A to Z by Titu Andreescue .

Try with Hints


Taking z=a+bi

then a+bi=(z+n)4i=-4b+4i(a+n),gives a=-4b b=4(a+n)=4(n-4b)

then n=$\frac{b}{4}+4b=\frac{164}{4}+4.164=697$

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AIME I Algebra Combinatorics Math Olympiad Math Olympiad Videos USA Math Olympiad

Combinations | AIME I, 2009 |Problem 9

Try this problem from American Invitational Mathematics Examination, AIME, 2019 based on Combinations

Combinations- AIME, 2009


A game show offers a contestant three prizes A B and C each of which is worth a whole number of dollars from $1 to $9999 inclusive. The contestant wins the prizes by correctly guessing the price of each prize in the order A B and C. As a hint the digits of three prizes are given. On a particular day the digits given were 1,1,1,1,3,3,3. Find the total number of possible guesses for all three prizes consistent with the hint.

  • 110
  • 420
  • 430
  • 111

Key Concepts


Combinations

Theory of equations

Polynomials

Check the Answer


Answer: 420.

AIME I, 2009, Problem 9

Combinatorics by Brualdi .

Try with Hints


Number of possible ordering of seven digits is$\frac{7!}{4!3!}$=35

these 35 orderings correspond to 35 seven-digit numbers, and the digits of each number can be subdivided to represent a unique combination of guesses for A B and C. Thus, for a given ordering, the number of guesses it represents is the number of ways to subdivide the seven-digit number into three nonempty sequences, each with no more than four digits. These subdivisions have possible lengths 1/2/4,2/2/3,1/3/3, and their permutations. The first subdivision can be ordered in 6 ways and the second and third in three ways each, for a total of 12 possible subdivisions.

then total number of guesses is 35.12=420

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College Mathematics Corner USA Math Olympiad

Theory of Equation (TIFR 2014 problem 10)

Question:

Let \(C \subset \mathbb{ZxZ} \) be the set of integer pairs \((a,b)\) for which the three complex roots \(r_1,r_2,r_3\) of the polynomial \(p(x)=x^3-2x^2+ax-b\) satisfy \(r_1^3+r_2^3+r_3^3=0\). Then the cardinality of \(C\) is

A. \(\infty\)

B. 0

C. 1

D. \(1<|C|<\infty\)

Discussion:

We have \(r_1+r_2+r_3=-(-2)=2\)

\(r_1r_2+r_2r_3+r_3r_1=a\) and

\(r_1r_2r_3=-(-b)=b\)

Also, of the top of our head, we can think of one identity involving the quantities \(r_1^3+r_2^3+r_3^3=0\) and the three mentioned just above.

Let’s apply that:

\(r_1^3+r_2^3+r_3^3-3r_1r_2r_3=(r_1+r_2+r_3)(r_1^2+r_2^2+r_3^2-(r_1r_2+r_2r_3+r_3r_1))\) …\(…(1)\)

Also, note that \(r_1^2+r_2^2+r_3^2=(r_1+r_2+r_3)^2-2(r_1r_2+r_2r_3+r_3r_1) \)

So \(r_1^2+r_2^2+r_3^2=(2)^2-2(a)=4-2a \).

So by equation \((1)\),

\(0-3b=2(4-2a-a)\)

or, \(6a-3b=8\).

Now, we notice the strangest thing about the above equation: \(3|(6a-3b)\) but \(3\) does not divide \(8\).

So, we get a contradiction.

Why did this contradiction occur? We didn’t start off by saying “assume that … something …”. Well, even though we pretend to not assume anything, we did assume that this equation has a solution for some \(a,b\). So, the

ANSWER: \(|C|=0\).