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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

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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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

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 $x = \pm 1$.