MaxWong

Here we have the conditions \(\begin{cases}N\equiv 5 \pmod {10}\\N\equiv 10\pmod{13}\end{cases}\).

One way without using Chinese Remainder Theorem is to list all numbers with \(N \leq 130\) and \(N \equiv 10 \pmod{13}\).

You can start from 13 + 10 = 23, and add 13 each time you progress in the list.

The list is {23, 36, 49, 62, 75, ...}.

Now, note that 75 also satisfies \(N \equiv 5 \pmod{10}\). Then this is the answer we want.

Note that \(75 \operatorname{mod} 130 = \boxed{75}\). That's the answer.

Remark: This method is not rigorous at all. For a solution with more rigor, you would need Chinese Remainder Theorem.

Take two primes greater than or equal to 11, then multiply them together. It would result in a composite number satisfying the condition.

For example:

\(11 \times 13 = 143\) is a number with prime factors 11 and 13, which are both at least 11.

We can actually find \(f^{-1}(x)\) explicitly.

\(\begin{array}{rcl} f(x) &=& \dfrac{x - 3}{x - 4}\\ f\left(\color{red}f^{-1}(x)\right) &=& \dfrac{\color{red}f^{-1}(x)\color{black} - 3}{\color{red}f^{-1}(x)\color{black} - 4}\\ x &=& \dfrac{f^{-1}(x) - 3}{f^{-1}(x) - 4}\\ x(f^{-1}(x) - 4) &=& f^{-1}(x) - 3\\ xf^{-1}(x) - 4x &=& f^{-1}(x) - 3\\ (x - 1)f^{-1}(x) &=& 4x - 3\\ f^{-1}(x) &=& \dfrac{4x - 3}{x - 1} \end{array}\)

Therefore, \(f^{-1}(x)\) is undefined when the denominator is 0, i.e., x - 1 = 0. Solving gives x = 1 as the answer.

Sorry, I meant:

k = 71: 5041/k = 5041/71 = 71, which is an integer.

You can perform trial division with a calculator. Start from k = 2, use the calculator to compute 5041/k, and see if it is an integer. If not, make k the next prime. Here's a few lines as an example.

k = 2: 5041/k = 5041/2, not an integer.

k = 3: 5041/k = 5041/3 \(\approx\) 1680.33, not an integer

k = 5: 5041/k = 5041/5 = 1008.2, not an integer

...

k = 71: 5041/71 = 5041/71 = 71, which is an integer.

5041 is not prime.

We expand \(\sin(d\cdot x)\) using its Taylor series at x = 0.

\(\begin{array}{rcl} \sin(d\cdot x) &=& \displaystyle\sum_{k = 0}^\infty \dfrac{(d\cdot x)^{2k+1}(-1)^k}{(2k + 1)!}\\ &=&\displaystyle\sum_{k = 0}^\infty \dfrac{d^{2k+1}(-1)^k}{(2k + 1)!} x^{2k+1}\\ &=&d\cdot x - \dfrac{d^3}{6}x^3+\dfrac{d^5}{120}x^5-\mathcal O (x^7) \end{array}\)

For \(\displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5}\) to exist, the polynomial part must eliminate all terms with degrees lower than 5 of the Taylor expansion above.

Therefore, we have \(\begin{cases}a = -d\\b = \dfrac{d^3}6\\\end{cases}\). But we also know that \(b = \dfrac43\).

Comparing gives \(\dfrac{d^3}6 = \dfrac43\). Solving gives \(d = 2\).

Then since \(a = -d\), we have \(a = -2\).

Therefore, we have:

\(\begin{array}{rcl} \displaystyle\lim_{x\to0}\dfrac{\sin(d\cdot x) + cx^5 + bx^3 + ax}{x^5} &=& \displaystyle\lim_{x\to 0}\dfrac1{x^5}\left(d\cdot x - \dfrac{d^3}6x^3+\dfrac{d^5}{120}x^5-\mathcal O(x^7)+cx^5 + \dfrac{d^3}6x^3 - d\cdot x\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{d^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{2^5}{120} + c - \mathcal O(x^2)\right)\\ &=& \displaystyle\lim_{x\to0}\left(\dfrac{4}{15} + c - \mathcal O(x^2)\right)\\ &=& \dfrac{4}{15} + c \end{array}\)

But we also know that this limit is 0. Then:

\(\dfrac{4}{15} + c = 0\\ c = -\dfrac{4}{15} \)

Now we know the values of a and c, so 

\(\begin{array}{rcl} 15ac + 2 &=& 15 \cdot (-2) \cdot \dfrac{-4}{15} + 2\\ &=& 10 \end{array}\)

Remark: If you are not familiar with the Big \(\mathcal O\) notation, just treat it as \(\cdots\).

The number of intersections of the curve and the x-axis between -5 and 5, is the number of solutions.

i.e., there are two solutions on that interval.

You can see that it can be divided into two rectangular grids which only touch at 1 lattice point. Name that lattice point M.

Note that to walk from P to Q, your path must pass through M.

Number of ways to walk right/up from the bottom left to top right of a rectangular grid = \(\displaystyle\binom{\text{width} +\text{length}}{\text{width}}\).

So, the number of ways to walk right or up from P to M = \(\displaystyle\binom{2 + 3}{2} = 10\)

The number of ways to walk right or up from M to Q = \(\displaystyle\binom{3 + 4}{4} = 35\).

Now, total number of ways = \(10 \times 35 = 350\)