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

## 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.

Categories

## 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 ## Subscribe to Cheenta at Youtube Categories ## 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 ## Subscribe to Cheenta at Youtube Categories ## 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 ## Subscribe to Cheenta at Youtube Categories ## 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: $|C|=0$.