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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Greatest Positive Integer | AIME I, 1996 | Question 2

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 1996 based on Greatest Positive Integer.

Positive Integer – AIME I, 1996


For each real number x, Let [x] denote the greatest integer that does not exceed x,find number of positive integers n is it true that \(n \lt 1000\) and that \([log_{2}n]\) is a positive even integer.

  • is 107
  • is 340
  • is 840
  • cannot be determined from the given information

Key Concepts


Inequality

Greatest integer

Integers

Check the Answer


Answer: is 340.

AIME I, 1996, Question 2

Elementary Number Theory by Sierpinsky

Try with Hints


here Let \([log_{2}n]\)=2k for k is an integer

\(\Rightarrow 2k \leq log_{2}n \lt 2k+1\)

\(\Rightarrow 2^{2k} \leq n \lt 2^{2k+1}\) and \(n \lt 1000\)

\(\Rightarrow 4 \leq n \lt 8\)

\(16 \leq n \lt 32\)

\(64 \leq n \lt 128\)

\(256 \leq n \lt 512\)

\(\Rightarrow 4+16+64+256\)=340.

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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Consecutive positive Integers | AIME I, 1990| Question 11

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1990 based on consecutive positive integers.

Consecutive positive integer – AIME I, 1990


Someone observed that 6!=(8)(9)(10). Find the largest positive integer n for which n! can be expressed as the product of n-3 consecutive positive integers.

  • is 107
  • is 23
  • is 634
  • cannot be determined from the given information

Key Concepts


Integers

Inequality

Algebra

Check the Answer


Answer: is 23.

AIME I, 1990, Question 11

Elementary Number Theory by David Burton

Try with Hints


The product of (n-3) consecutive integers=\(\frac{(n-3+a)!}{a!}\) for a is an integer

\(n!=\frac{(n-3+a)!}{a!}\) for \(a \geq 3\) \((n-3+a)! \geq n!\)

or, \(n!=\frac{n!(n+1)(n+2)….(n-3+a)}{a!}\)

for a=4, n+1=4! or, n=23 which is greatest here

n=23.

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AMC 10 AMC 12 AMC 8 India Math Olympiad Math Olympiad PRMO Regional Mathematics Olympiads USA Math Olympiad

Can we prove that the length of any side of a triangle is not more than half of its perimeter?

Can we Prove that ……..


The length of any side of a triangle is not more than half of its perimeter

Key Concepts


Triangle Inequality

Perimeter

Geometry

Check the Answer


Answer: Yes we can definitely prove that by Triangle Inequality

Mathematical Circles – Chapter 6 – Inequalities Problem 3

Mathematical Circles by Dmitri Fomin , Sergey Genkin , Llia Itenberg

Try with Hints


We can start this sum by using this picture below

The length of the three sides of this triangle are a,b and c. So if we apply triangle inequality which implies that the length of one side of a triangle is less than the sum of the lengths of the two sides of that triangle. In reference to the theorem

b + c > a

Proof based on triangle

So can you try to do the rest of the sum ????????

According to the question we have to find the perimeter at first

Perimeter is the sum of the length of all sides of the triangle = a + b + c

And the length of each side is a or b or c.

We have to prove : a + b + c > length of any one side

This can be one of the most important hint for this problem. Try to do the rest of the sum …………………………..

Here is the rest of the sum :

As stated above if we use triangle inequality :

b + c > a

Lets add a to both the sides

a + b + c > a + a

a + b + c > 2 a

The left hand side of the above inequality is the perimeter of this triangle.

perimeter > 2 a

So , \(\frac {perimeter}{2} > a \)

\(\frac {perimeter}{2} \) = semi perimeter

Hence this is proved that the length of one side of a triangle is less than half of its perimeter.

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AMC 10 PRMO USA Math Olympiad

Inequality (Forerunner Problem 2)

Problem – Inequality (Forerunner Problem 2)


The maximum number of passengers a bus can hold is 40. Which of the following is correct?

  • Passengers < 40
  • Passengers ≤ 40
  • Passengers > 40
  • Passengers ≥ 40

Key Concepts


Inequality

Mathematical Analysis

Number

Check the Answer


Answer: Passengers ≤ 40

Forerunner Problem Number 2 (Math is Fun)

Secrets in Inequalities.

Try with Hints


This problem is very easy to solve.We can start with a basic chart

Inequality- chart

I think now its quite clear from the above chart that what we have to do

If the maximum number of passengers a bus can hold is 40 then the expression will be like Passenger \(\leq 40 \) .So the number of passenger it van hold is either 40 or less than 40. So the second option is the correct option.

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