MaxWong

Since \(x \equiv 3 \pmod{20}\), let x = 20k + 3 for some integer k.

\(20k + 3 \equiv 13\pmod{45}\\ 20k \equiv 10 \pmod{45}\\ 4k \equiv 2 \pmod{9}\\ 4k \equiv 20 \pmod{9}\\ k \equiv 5 \pmod 9\)

Then we can let k = 9n + 5 for some integer n. Substituting this into x = 20k + 3 gives x = 180n + 103.

Therefore, taking (mod 6), we have:

\(x \equiv 180n + 103\pmod 6\\ x \equiv 103 \pmod 6\\ x \equiv 1 \pmod 6\)

It turns out the system has a solution if and only if \(r \equiv 1 \pmod 6\), i.e., \(r = 6p + 1\) for some integer p.

Note that he can choose (i) 4 green peppers and 2 red peppers, or (ii) 5 green peppers and 1 red pepper. Then,

\(\begin{array}{rcl} \text{Probability} &=&\dfrac{\displaystyle \binom{5}4\binom{10}2+\binom{5}5\binom{10}1}{\displaystyle \binom{10+5}6}\\ &=& \dfrac{47}{1001} \end{array}\)

It is quite complicated to consider those numbers which have a digit sum that is at least 6. 

We can consider those numbers which have a digit sum less than 6 instead.

Numbers with digit sum 4: 1111

Numbers with digit sum 5: 2111, 1211, 1121, 1112

Therefore, there are only 5 numbers with a digit sum less than 6. All other numbers have a digit sum of at least 6.

Note that there are 3^4 = 81 numbers which can be written with only 1, 2, 3.

\(\text{Required number} = 3^4 - 5 = 76\).

According to Wolfram Alpha, the only real value that satisfies this equation is \(x = \dfrac13 \left(4 - \dfrac{\sqrt[3]{4}}{\sqrt[3]{113+3\sqrt{1419}}} + \sqrt[3]{2\left(113+3\sqrt{1419}\right)}\right)\).

Graph: https://www.desmos.com/calculator/t08bdptj6k

I forgot to attach the graph when I sent the response.

Edit: Updated the graph to the new one.

This problem is equivalent to the probability of choosing two real numbers x, y with 0 < x < y < 5 such that x < 3, y - x < 3, and 5 - y < 3.

We can view this geometrically.

Let \(D\) be the square with vertices (0, 0), (5, 0), (0, 5), (5, 5).

Let \(\Omega\) be the region bounded by 0 < x < 3, y - x < 3, 5 - y < 3, x < y and \(D\).

Basically, the probability is equal to \(\dfrac{\text{area}(\Omega)}{\text{area}(D)} = \dfrac{9 - 2 - \dfrac12}{25} = \dfrac{13}{50}\).

I believe that is the answer.

Edit: Edited. This should be correct now.

Note that by property of centroid, AG : GD = 2 : 1.

Since MN is parallel to BC, MG is parallel to BD and GN is parallel to DC.

Since MG is parallel to BD, \(\triangle AMG \sim \triangle ABD\) by AA postulate.

Since GN is parallel to DC, \(\triangle AGN \sim \triangle ADC\) by AA postulate.

Then, \(MG = \dfrac23 BD = \dfrac23 DC = GN\).

Hence, G divides MN internally in the ratio 1 : 1.

Therefore, area of triangle AGN = area of triangle AGM = 144.

Nice to see you too, CPhill :)

It has been a long time since I was last active on this site. It feels great to come back to this site and see that it is still a very active forum.

Nice!