Way 1: \(\left\{{}\begin{matrix}GCD\left(48;72\right)=24\\LCM\left(48;72\right)=144\end{matrix}\right.\)

So their product is: 24 . 144 = 3456

Way 2: 

Their product is: 48 . 72 = 3456

\(f\left(x\right)=3\Rightarrow\sqrt{x+4}=3\Rightarrow x+4=9\Rightarrow x=5\)

If A true 

=> B false and C true 

=> D false and A false (Contrary to suppose) 

So A false 

=> C is true statement.

a # b = a + 2b 

=> 3 # 4 = 3 + 2.4 = 3 + 8 = 11

It takes \(\dfrac{35}{30}=\dfrac{7}{6}\left(h\right)=70'\) to drive all this distance at 30mi/h

It takes \(\dfrac{25}{30}=\dfrac{5}{6}\left(h\right)=50'\)to drive all this distance at 25mi/h

So drive at 30mi/h is takes fewer than 25mi/h is: 70' - 50' = 20'

Answer: 20 minutes

\(2015=5\cdot13\cdot31\)

So their sum is: 5 + 13 + 31 = 49

Tell the number to find is x

We have:

x : 8 remainder 7

x : 9 remainder 8

x : 12 remainder 11

==> x + 1 ⋮ 7,8,11

\(\Rightarrow x+1\in CM\left(7,8,11\right)\)

But x is the smallest number 

=> \(x+1\) is \(LCM\left(7,8,11\right)=616\) (Get out 0 because x is positive number) 

So x = 615  

From the 25th day to 284th day has: \(284-25+1=260\left(days\right)\)

   260 days = 37 weeks and 1 day.

So the 284th day is Sunday

The value of (1 + 2 + 3) + (2 + 3 + 4) + (3 + 4 + 5) + … + (38 + 39 + 40) + (39 + 40 + 41) is:

1 + 2.(2 + 3 + ... + 40) + 41 = 42 + 1638 = 1680

\(9x^2+4y^2=20xy\)

\(\Leftrightarrow\left(3x\right)^2-12xy+\left(2y\right)^2=8xy\)

\(\Leftrightarrow\left(3x-2y\right)^2=8xy\)

\(9x^2+4y^2=20xy\)

\(\Leftrightarrow\left(3x\right)^2+12xy+\left(2y\right)^2=32xy\)

\(\Leftrightarrow\left(3x+2y\right)^2=32xy\)

\(\Rightarrow p=\sqrt{\dfrac{\left(3x-2y\right)^2}{\left(3x+2y\right)^2}}=\sqrt{\dfrac{8xy}{32xy}}=\sqrt{\dfrac{1}{4}}=\dfrac{1}{2}\)

\(x^3+y^3+z^3=3xyz\)

\(\Leftrightarrow x^3+y^3+z^3-3xyz=0\)

\(\Leftrightarrow\left(x^3+y^3\right)+z^3-3xyz=0\)

\(\Leftrightarrow\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left[\left(x+y\right)^2-2\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)=0\)

\(\Leftrightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)=0\)

\(x+y+z\ne0\Leftrightarrow x^2+y^2+z^2-xy-yz-zx=0\)

\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yz-2zx=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

\(\Leftrightarrow x=y=z\)

\(\Rightarrow p=\dfrac{x^{2018}+1}{x^{2018}+x^{2018}+x^{2018}+3}=\dfrac{x^{2018}+1}{3\left(x^{2018}+1\right)}=\dfrac{1}{3}\)

\(n^5-n+2=n\left(n^4-1\right)+2=n\left(n^2+1\right)\left(n^2-1\right)+2\)

\(=n\left(n-1\right)\left(n+1\right)\left(n^2+1\right)+2\)

We see: \(\left(n-1\right)\cdot n\cdot\left(n+1\right)⋮3\) (is product of 3 consecutive natural numbers) 

\(\Rightarrow n^5-n+2\equiv2\left(mod3\right)\)

But no square numbers are in the form 3k + 2

So \(n\notin\varnothing\)

We have:

f(5) - f(4) = (a.5³ + b.5² + c.5 + d) - (a.4³ + b.4² + c.4 + d) 

= (125a + 25b + 5c + d) - (64a + 16b + 4c + d) 

= 61a + 9b + c 

f(7) - f(2) = (a.7³ + b.7² + c.7 + d) - (a.2³ + b.2² + c.2 + d) 

= (343a + 49b + 7c + d) - (8a + 4b + 2c + d) 

= 335a + 45b + 5c \(⋮5\) (1) 

Because a > 0  

=> 61a + 9b + c < 335 + 45b + 5c 

=> f(7) - f(2) > 2012 (2) 

(1), (2) => f(7) - f(2) isn't a prime number 

We have: \(36,5,m,n\in Z^+\)

\(\Rightarrow36^m,5^n\in Z^+\)

\(\Rightarrow36^m\ge36^1=36;5^n\ge5^1=5\)

\(\Rightarrow36^m-5^n\ge36-5=31\)

\(\Rightarrow\left(36^m-5^n\right)_{Min}=31\Leftrightarrow m=n=1\)

We known: A odd square number divide for 8 has the remaining is 1

So to 3ⁿ - 1⋮ 8 <=> 3ⁿ is square number <=> n is an even number.

Last case: n is odd number

=> \(3^n-1=3^{2k}\cdot3-1=3\cdot\left[\left(3^k\right)^2-1\right]+2\equiv2\left(mod8\right)\)

So with all even n then 3ⁿ-1⋮8

* With n = 4k+1 or n = 4k+3 \(\left(k\in N\right)\)

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1^n+2^n+\left(-2\right)^n+\left(-1\right)^n\equiv0\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮5\)

* With n = 4k+2

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1+2^{4k+2}+3^{4k+2}+4^{4k+2}\equiv1+4\cdot16^k+9\cdot81^k+16\cdot256^k\equiv1+4+9+16\equiv30\equiv0\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮5\)

* With n = 4k

\(\Rightarrow1^n+2^n+3^n+4^n\equiv1+2^{4k}+3^{4k}+4^{4k}\equiv1+16^k+81^k+256^k\equiv1+1+1+1\equiv4\left(mod5\right)\)

So \(1^n+2^n+3^n+4^n⋮̸5\)

==> With \(n\ne4k\Leftrightarrow1^n+2^n+3^n+4^n⋮5\) 

We have:

\(3^{10}=59049\equiv049\left(mod1000\right)\)

\(3^{50}=\left(3^{10}\right)^5\equiv49^5\equiv249\left(mod1000\right)\)

\(3^{100}=\left(3^{50}\right)^2\equiv249^2\equiv001\left(mod1000\right)\)

So last 3 digits of 3¹⁰⁰ is 001

\(3^{1998}+5^{1998}=\left(3^3\right)^{666}+\left(5^2\right)^{999}=27^{666}+25^{999}=\left(26+1\right)^{666}+\left(26-1\right)^{999}=\left(13k+1\right)^{666}+\left(13k-1\right)^{999}\equiv1+\left(-1\right)\equiv0\left(mod13\right)\)

Answer: 0

We have: \(1995⋮7\)

\(\Rightarrow1992^{1993}+1994^{1995}\equiv\left(7k-3\right)^{1993}+\left(7k-1\right)^{1995}\equiv\left(7-3\right)^{1993}+\left(7-1\right)\equiv4+6\equiv10\equiv3\left(mod7\right)\)

Answer: 3

\(2^{1994}=2^2\cdot\left(2^3\right)^{664}=4\cdot\left(7k+1\right)^{664}\equiv4\cdot1\equiv4\left(mod7\right)\)

Answer: 4

\(3^{1993}=3\cdot\left(3^3\right)^{664}=3\cdot\left(7k-1\right)^{664}\equiv3\cdot1\equiv3\left(mod7\right)\)

Answer: 3