MaxWong

Similar problem here: https://web2.0calc.com/questions/modular-number-theory_1#r1

Try to solve this problem using the method explained in that thread.

So, suppose the digits were \(a_1, a_2, \cdots , a_{10}\) and a_1 is the first digit, a_2 is the second digit, etc.

Note that the first digit can't be 0. So the required number is 

\((\#\text{ of cases where }a_1 + a_2+ \cdots + a_{10} = 15) - (\#\text{ of cases where }a_2 + a_3 + \cdots + a_{10} = 15)\)

Suppose you mean the exponents form a geometric sequence.

Just combine the exponents using the formula \(a^m a^n = a^{m+n}\).

\(2^{1/3} \cdot 2^{1/9} \cdot 2^{1/27} \cdots = 2^{1/3 + 1/9 + 1/27 + \cdots}\)

Note that the exponent is now the sum of a geometric sequence. Can you complete it now?

Suppose that m is the smallest positive integer such that m! is a multiple of 4125.

First we need to take a look at what 4125 really is. 

\(4125 = 3 \times 5^3 \times 11\)

So, is 11! a multiple of 4125? No, because if you expand 11! = 11 * 10 * 9 * ... * 1, you will notice only 5 and 10 has a factor of 5, so the exponent of 5 in the prime factorization of 11! is 2, not 3. We need another multiple of 5 in the factorial so to make it 3. Therefore, m = 15.

Can you try to find the value of n on your own?

Simplify the polynomial first.

\(\quad 9x^2 + 6x + 8x - 5x^2 + \underline{\phantom{\text{aaaaa}}}\\ = 4x^2 + 14x + \underline{\phantom{\text{aaaaa}}}\\\)

Note that (a + b)^2 = a^2 + 2ab + b^2. What happens if we take a = 2x?

(2x + b)^2 = 4x^2 + 4bx + b^2

There we matched the first term of the polynomial. The rest is just finding the value of b that makes the second term match, and then the answer is just b^2.

Hint: Consider the coefficient. What b would make 4b = 14?

Simplify the function first.

\(g(x) = \dfrac2{2 + 4x^2 + 5 + 9x^2} = \dfrac2{13x^2 + 7}\)

Note that 13x^2 + 7 is always positive. So g(x) is always positive.

Also note that the minimum of 13x^2 + 7 is attained at x = 0, with a minimum value of 7.

Therefore, \(\displaystyle\max_{x\in\mathbb R} g(x) = \dfrac2{\displaystyle\min_{x\in\mathbb R}(13x^2 + 7)} = \dfrac27\).

Also, note that \(\displaystyle\lim_{x\to +\infty} g(x) = \displaystyle\lim_{x\to -\infty} g(x) = 0\).

Therefore, the range is \(\left(0, \dfrac27\right]\).

Note that \(13 + 2(17) = 47\).

Multiply both sides by 29 gives 

\(13 \times 29 + 2 \times 17 \times 29 = 47 \times 29 \equiv 0 \pmod{47}\\ 1 + 17 \times (2 \times 29) \equiv 0 \pmod {47}\\ 17 \times (-2\times 29) \equiv 1\pmod{47}\\ 17 \times (-58) \equiv 1 \pmod{47}\\ 17 \times 36 \equiv 1 \pmod{47} \)

Therefore, by definition of modular inverse, \(17^{-1} \equiv 36 \pmod{47}\).

Consider the prime factorization of the answer. The factorization cannot contain primes smaller than 20.

That means all prime factors must be greater than or equal to 20.

The smallest prime greater than or equal to 20 is 23.

Since it cannot be just 23 (it would be a prime, not a composite number), we arrive at the conclusion that \(23^2 = 529\) is the smallest composite number that has no prime factors less than 20.

So, 

\(\begin{cases}x + y = 3\\xy = -18\end{cases}\)

You can check by calculating the sum and product that x = 6, y = -3 is a solution.

Alternatively, if you know how to solve a quadratic equation, by Vieta's formula, x and y are the roots of \(t^2 - 3t - 18 = 0\).

So \(f(x) \neq 0\) and \(x \geq 0\) since \(x^{1/2} = \sqrt x\).

\(f(x) \neq 0\\ 2 - x^{1/2} \neq 0\\ x \neq 4\)

Combine \(x \geq 0\) and \(x \neq 4\) to get \(x \in [0, 4) \cup (4, +\infty)\).

Therefore, the domain of g(x)/f(x) is \([0, 4) \cup (4, +\infty)\), or written in inequality form, \(\{x \in \mathbb R : 0 \leq x < 4 \text{ or }x > 4\}\).