We have: 5 = 1 + 0 + 4 = 1 + 1 + 3 = 1 + 2 + 2 = 2 + 0 + 3 = 5 + 0 + 0

With 1, 0, 4, we can have 4 three-digit numbers including these 3 numbers

(104, 140, 401, 410)

With 1, 1, 3, we can have 3 three-digit numbers including these 3 numbers

(113, 131, 311)

With 1, 2, 2, we can have 3 three-digit numbers including these 3 numbers

(122, 212, 221)

With 2, 0, 3, we can have 4 three-digit numbers including these 3 numbers

(203, 230, 302, 320)

With 5, 0, 0, we can have 1 three-digit number including these 3 numbers

(500)

In total, we have 4 + 3 + 3 + 4 + 1 = 15 (numbers)

We have: 5 = 1 + 0 + 4 = 1 + 1 + 3 = 1 + 2 + 2 = 2 + 0 + 3 = 5 + 0 + 0

With 1, 0, 4, we can have 4 three-digit numbers including these 3 numbers

With 1, 1, 3, we can have 6 three-digit numbers including these 3 numbers

With 1, 2, 2, we can have 6 three-digit numbers including these 3 numbers

With 2, 0, 3, we can have 4 three-digit numbers including these 3 numbers

With 5, 0, 0, we can have 1 three-digit number including these 3 numbers

In total, we have 4 + 6 + 6 + 4 + 1 = 21 (numbers)

You can search the Erdos-Anning theorem on Wikipedia.

There's a way to prove but I think you mightn't understand.

\(\dfrac{2011}{1}=2011\) is an integer

2012 is even,so \(\dfrac{2012}{2}\)is an integer

2 + 0 + 1 + 3 \(⋮3\) ,so \(\dfrac{2013}{3}\)is an integer

\(14⋮̸\)\(4\),so \(\dfrac{2014}{4}\)isn't an integer

2015 has 5 as the last digit,so \(\dfrac{2015}{5}\) is an integer

Hence,the answer is D

Just "Calculate" not "Caculader" :) Wrong vocabulary sir :D

x(x + 3) - x - 3 = 0

\(\Leftrightarrow x^2+3x-x-3=0\)

\(\Leftrightarrow x^2-2x-3=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\)

So, \(S=\left\{-1;3\right\}\)

2x - x + 3x - 2x = 0 

\(\Leftrightarrow x+x=0\)

\(\Rightarrow x=-x\)

We just have one answer for this math problem.

That's \(x=0\)

So, \(S=\left\{0\right\}\)

We have : x(x + 3) - x - 3 = 0

=> x(x + 3) - (x + 3) = 0

=> (x - 1)(x + 3) = 0

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

2x - x + 3x - 2x = 0

=> -x + 3x = 0

=> 2x = 0

=> x = 0

\(\dfrac{1}{2}\) + \(\dfrac{1}{6}\) + ... + \(\dfrac{1}{30}\) + \(\dfrac{1}{42}\)

=\(\dfrac{1}{1.2}\) + \(\dfrac{1}{2.3}\) + ... + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)

=\(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\)

=\(\dfrac{1}{1}\) - \(\dfrac{1}{7}\)

=\(\dfrac{6}{7}\)

Thank you. You are very good