\(\left(3x+2\right)\left(3x+3\right)\left(3x+4\right)=\left(3x+5-3\right)\left(3x+5-2\right)\left(3x+5-1\right)=\left(13-3\right)\left(13-2\right)\left(13-1\right)=10.11.12=1320\)

 

Draw the equilateral triangle EBC

Consider \(\Delta\)AEB and \(\Delta\)AEC, we have:

- AE is common edge (hypothesis) 

- AB = AC (hypothesis) 

- EB = EC (\(\Delta\)EBC is a equilateral triangle) 

=> \(\Delta\)AEB = \(\Delta\)AEC (S-S-S) 

=> \(\widehat{EAB}=\widehat{EAC}=\dfrac{20^o}{2}=10^o\)

\(\Delta\)ABC balance at A => \(\widehat{ABC}=\widehat{ACB}=\dfrac{180^o-20^o}{2}=80^o\)

Because \(\Delta\)EBC is a equilateral triangle so \(\widehat{EBC}=\widehat{ECB}=60^o\)

\(\Rightarrow\widehat{EBA}=\widehat{ECA}=80^o-60^o=20^o\)

Consider \(\Delta\)AEC and \(\Delta\)CDA, we have:

- AC is common edge (hypothesis) 

- \(\widehat{DAC}=\widehat{ACE}=20^o\)

- AD = EC (= BC) 

=> \(\Delta\)AEC = \(\Delta\)CDA (S-A-S) 

=> \(\widehat{ACD}=\widehat{EAC}=10^o\)

Tell the pencil is a, the paper clips is b and the eraser is c

=> a + 5b = 2c and a = 29b

=> 29b + 5b = 2c

=> 34b = 2c

=> c = 17b

So 17 paper clips weigh the same as an eraser.

Tell these number to find is x

=> 100 < x < 1000

=> x has 3 digits.

The hundred-digit has 3 choice

The ten-digit has 3 choice

The units-digit has 3 choice too

So there are 3.3.3 = 27 intergers satisfy

2²⁰¹⁷ . 7²⁰¹⁷ = 2^504.4+1.7^504.4+1 = ...2*...7 = ...4

So the units digit of 2²⁰¹⁷.7²⁰¹⁷ is 4

Convert: 6 feet = 72 inch ; 8 feet = 96 inch

So the area of this wall is: 72.96 = 6912 (square-inch) 

The area of each tile is: 4.4 = 16 (square-inch) 

So there are 6912: 16 = 432 (tiles) are neede to cover all area of this wall

We have: a^m : a^m = a⁰ = 1

So 0⁵ : 0⁵ = 0⁵ : 0  undetermined

So 0⁰ undetermined

x³−y³−z³=3xyz ⇔ (x−y)³ − z³ + 3xy(x−y) – 3xyz = 0

⇔ (x−y−z)(x²+y²+z²+xy+xz−yz) = 0 ⇔ x=y+z

But 2(y+z) = x² ⇔ x²=2x ⇔ x = 2 (Because x>0)

⇒ y + z = 2 ⇒ y = z = 1

Again have: x² = 2.(y+z)

<=> 2² = 2.(1 + 1) <=> 4 = 4 ---- Good ☺

So (x,y,z) = (2,1,1)

x³−y³−z³=3xyz ⇔ (x−y)³ − z³ + 3xy(x−y) – 3xyz = 0

⇔ (x−y−z)(x²+y²+z²+xy+xz−yz) = 0 ⇔ x=y+z

But 2(y+z) = x² ⇔ x²=2x ⇔ x = 2 (Because x>0)

⇒ y + z = 2 ⇒ y = z = 1

Again have: x² = 2.(y+z)

<=> 2² = 2.(1 + 1) <=> 4 = 4 ---- Good ☺

So (x,y,z) = (2,1,1)

7.

a² = b² + c²

+ If b² and c² \(⋮̸\)3

=> b² + c² = a² divide 3 residuals 2 (unsatisfactory) 

So b², c² has at least a number \(⋮3\) (because 3 is a prime number) 

So abc \(⋮3\)

+ If b² and c² \(⋮̸\)4

=> b² + c² = a² divide 4 residuals 2 (unsatisfactory) 

So b², c² has at least a number \(⋮4\) (because 3 is a prime number) 

So abc \(⋮4\)

+ Because \(b^2,c^2\equiv0,1,4\left(mod5\right)\)

\(\Rightarrow b^2+c^2\equiv0,1,2,3,4\left(mod5\right)\)

But \(a^2\equiv0,1,4\left(mod5\right)\)

So the equal cause happen when and only when b², c² have at least a number \(⋮5\)

\(\Rightarrow abc⋮5\)

\(\left(3,4,5\right)=1\Rightarrow abc⋮60\)

Tell the number to find is a

Because a is a palindrome so p have odd digits

But a > 2018 and a is the first number so a has 5 digits.

So the number to find is 10301

The number of degrees in ^oF is:

13 - (-5) = 18 (^oF)

So the number of degrees in ^oC is: -7,78^oC

\(3\text{®}\left(2\text{®}1\right)=3\text{®}\left(2^2-1^2\right)=3\text{®}3=3^2-3^2=0\)

1.2 + 3:6.5 - 4

= 2 + 5/2 - 4 = 9/2 - 4 = 1/2

\(\sqrt{2.3.4.5.6.7.10}=\sqrt{50400}=60\sqrt{14}\)

5-5.5 + 5 : 5

= 5 - 25 + 1 = -20 + 1 = -19

\(7^{107}=7^{20.5+7}=7^{20.5}\cdot7^7\equiv01\cdot823543\equiv43\left(mod10\right)\)

So the last two digits of 7¹⁰⁷ is 43

Suppose 2^m + 3ⁿ ⋮ 23

=> \(8^n\left(2^m+3^n\right)\text{⋮}23\)

\(\Rightarrow2^{m+3n}+24^n\text{⋮}23\)

Because \(24\equiv1\left(mod23\right)\Rightarrow24^n\equiv1\left(mod23\right)\)

\(\Rightarrow2^{m+3n}+24^n\equiv2^{m+3n}+1\left(mod23\right)\)

We must prove that 2^m+3n + 1 doesn't divisible by 23

Simple we can have a example: 2¹¹ \(⋮̸23\)

So \(2^m+3^n⋮23̸\)

The first number is 1 and the last is: \(51\cdot2+1=101\)

So their sum is \(\dfrac{\left(101+1\right)\cdot51}{2}=2601\)

1 + 4 + 16 + ... + 1024 

= 2⁰ + 2² + 2⁴ + ... + 2¹⁰

= 2⁰ + 2² + 2⁴ + 2⁶ + 2⁸ + 2¹⁰

= 1 + 4 + 16 + 64 + 256 + 1024 

= 1365