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AIME I Algebra Arithmetic Math Olympiad USA Math Olympiad

Problem on Rational Numbers | AIME I, 1992 | Question 1

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1992 based on Rational Numbers.

Problem on Rational Numbers – AIME I, 1992


Find the sum of all positive rational numbers that are less than 10 and that have denominator 30 when written in lowest terms.

  • is 107
  • is 400
  • is 840
  • cannot be determined from the given information

Key Concepts


Integers

Rational Numbers

Euler’s Totient Function

Check the Answer


Answer: is 400.

AIME I, 1992, Question 1

Elementary Number Theory by David Burton

Try with Hints


For Euler’s Totient function, there exists 8 numbers that are relatively prime to 30, less than 30.

Here they are in (m,30-m) which in the form of sums of 1

\(\Rightarrow\) sum of smallest eight rational numbers=4

there are eight terms between 0 and 1 and there are eight terms between 1 and 2 where these we get as adding 1 to each of first eight terms

\(\Rightarrow\) 4(10)+8(1+2+3+…+9)=400.

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AMC 10 Singapore Math Olympiad

Functional Equation Problem from SMO, 2013 – Senior Section

Try this beautiful Functional Equation Problem from SMO, Singapore Mathematics Olympiad, 2013.

Problem – Functional Equation (SMO Test)

Let M be a positive integer .It is known that whenever \(|ax^2 + bx +c|\leq 1\) for all

\(|x|\leq 1\) then \(|2ax + b |\leq M \) for all \(|x|\leq 1\). Find the smallest possible value of M.


  • 4
  • 5
  • 6
  • 10

Key Concepts


Functional Equation

Function

Check the Answer


Answer: 4

Singapore Mathematics Olympiad

Challenges and Thrills – Pre – College Mathematics

Try with Hints


We cant this sum by assuming a,b,c as fixed quantity.

Let \( f(x) = ax^2 + bx + c \).

Then \( f(-1) = a – b + c \) ; \( f(0) = c \) ; \( f(1) = a + b + c\) ;

Try to do the rest of the sum …………………………..

Suppose \( |f(x)|\leq 1\) for all \(|x|\leq 1 \) . Then

\( |2ax + b| = | (x – \frac {1}{2} ) f(-1) – 2 f(0) x + (x+\frac {1}{2} f(1) |\)

\(\leq |x – \frac {1}{2}| + 2 |x| + |x + \frac {1}{2}|\)

\(\leq |x – \frac {1}{2} | + |x+\frac {1}{2}| + 2 \)

\(\leq 4 \)

Now I guess you have already got the answer but if not ………….

From the last step we can conclude ,

\(|2 x^2 – 1|\leq 1 \) whenever \(|x|\leq 4\) and \(|2x| = 4 \)

is achieved at \(x = \pm 1\).

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