Let a be a satisfied number (a > 29), then 2216 is divided by a with remainder 29 or \(2216-29=2187=3^7⋮a\)

\(\Rightarrow a=3^4;3^5;3^6;3^7\)

So, the answer is 4

The bus traveled : 7 + 8 + 8 = 23 (miles)

The length of the straight path from the Kevin's house to the school is 

\(\sqrt{\left(7+8\right)^2+8^2}=17\) (miles)

The answer is : 23 - 17 = 6 (miles)

Let a be the least number of pennies in the jar, then a - 1 is divisible by 2, 3, 4, 5, 6 and 8

\(\Rightarrow a-1\in CM\left(2,3,4,5,6,8\right)=M\left(120\right)=\left\{0;120;240;...\right\}\)

\(\Rightarrow a\in\left\{1;121;241;...\right\}\)

Since 721 is the least multiple of 7 in the set, the answer is 721

We have :

\(\left[{}\begin{matrix}x-2017=0\\x^2-2018^2=0\\x^3-2019^3=0\\x^4-2020^4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2017\\x=\pm2018\\x=2019\\x=\pm2020\end{matrix}\right.\)

So, the equation has 6 roots above

The answer is : \(\dfrac{1}{2}\pi r.2+2\pi r.\dfrac{1}{2}=2\pi r\) (units)

Let $a be the money Arlene had when she started shopping.

After shopping in the first store, she had : \(\left(1-40\%\right)a-4=60\%a-4\)

After shopping in the second store, she had :

\(\left(1-50\%\right)\left(60\%a-4\right)-5=30\%a-7\)

After shopping in the third store, she had :

\(\left(1-60\%\right)\left(30\%a-7\right)-6=12\%a-8.8\)

We have : \(12\%a-8.8=2\Leftrightarrow12\%a=10.8\Leftrightarrow a=90\)

So, the answer is $90

Draw \(BE\perp CD\), then ABED is a rectangle since it has 3 right angles. So, DE = AB = 4 units and :

\(AD=BE=\sqrt{BC^2-EC^2}=\sqrt{15^2-\left(16-4\right)^2}=9\)(units)

The volume of the frustum is :

\(\dfrac{1}{3}\pi\left(AB^2+CD^2+AB.CD\right).AD=\dfrac{1}{3}\pi\left(4^2+16^2+4.16\right).9=1008\pi\) (units³)

The area of the smaller rectangle is : \(\dfrac{400}{3+1}=100\) (ft²)

We have :

 \(Length+Width\ge2\sqrt{Length.Width}\)

\(\Leftrightarrow Perimeter\ge4\sqrt{Area}\)

The answer is : \(4\sqrt{100}=40\) (ft)

The answer is : 83 x 751 = 62333

123_b = b² + 2b + 3. So :

\(b^2+2b+3=363\Leftrightarrow b^2+2b+1=361\Leftrightarrow\left(b+1\right)^2=361\)

\(\Leftrightarrow b+1=19\Leftrightarrow b=18\)

Since \(6^2+8^2=10^2\), the given triangle is a right triangle (the converse of Pythagorean theorem)

We have : \(\dfrac{1}{r^2}=\dfrac{1}{6^2}+\dfrac{1}{8^2}\Leftrightarrow\dfrac{1}{r^2}=\dfrac{25}{576}\Leftrightarrow r^2=\dfrac{576}{25}\Leftrightarrow r^2=\dfrac{24}{5}=4.8\)

The answer is :

\(\dfrac{1}{3}\pi.\left(4.8\right)^2.a+\dfrac{1}{3}\pi.\left(4.8\right)^2.b=\dfrac{1}{3}\pi\left(4.8\right)^2.10=76.8\pi\approx241.3\)

(units³)

The price of a game et each year is : 100% + 4% = 104% of the price of a game et previous year.

The answer is : \(\dfrac{1.100}{\left(104\%\right)^{13}}\%\approx60\%\)

Let a, ab, ab² be the lengths of 2 legs and the hypotenuse of the right triangle respectively (a > 0 ; b > 1). Applying the Pythagorean theorem, we have :

\(a^2+a^2b^2=a^2b^4\Leftrightarrow b^4-b^2-1=0\)

\(\Leftrightarrow\left(b^2\right)^2-2b^2.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{5}{4}=0\Leftrightarrow\left(b^2-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)

Since \(b^2-\dfrac{1}{2}>1^2-\dfrac{1}{2}=\dfrac{1}{2}\), \(b^2-\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\) or \(b^2=\dfrac{\sqrt{5}+1}{2}\)

The answer is : \(\dfrac{a}{ab^2}=\dfrac{1}{b^2}=\dfrac{2}{\sqrt{5}+1}\approx0.62\)

Let a be the satisfied number. We notice that 2a + 1 is divisible by 3, 5, 7 since 1 x 2 + 1 = 3 ; 2 x 2 + 1 = 5 ; 3 x 2 + 1 = 7. 

\(a\ge101\Leftrightarrow2a+1\ge203\). Moreover, 2a + 1 is odd

\(2a+1\in CM\left(3,5,7\right)=\left\{0;105;210;315;...\right\}\)

So, 2a + 1 = 315 or a = 157

So, the answer is 157

Since the first candle burns faster than the second one, the first one can be half as high as the second one. Let a (hours) be the answer.

In 1 hour, \(\dfrac{1}{6}\) of the first one is burned and \(\dfrac{1}{9}\) of th second one is burned. We have :

\(1-\dfrac{1}{9}a=2\left(1-\dfrac{1}{6}a\right)\Leftrightarrow1-\dfrac{1}{9}a=2-\dfrac{1}{3}a\Leftrightarrow\dfrac{2}{9}a=1\)

\(\Leftrightarrow a=\dfrac{9}{2}=4.5\)

So, the answer is 4.5 hours

Each term is as : 729 : 243 = 3 (times) as its next term

The 10^th term is : \(729:3^9=3^6:3^9=\dfrac{1}{3^3}=\dfrac{1}{27}\)

Let a be the positive difference between 2 consecutive terms, then the 18^th term is :

\(4+17a=8.25\Leftrightarrow17a=4.25\Leftrightarrow a=0.25\)

The 35^th term is : 4 + 34 x 0.25 = 12.5

The difference between 2 consecutive terms is :

\(\left(2p+6\right)-p=\left(5p-12\right)-\left(2p+6\right)\) or \(p+6=3p-18\)

\(\Leftrightarrow2p=24\Leftrightarrow p=12\Leftrightarrow p+6=18\)

The 4^th term is : 12 + 18 x 3 = 66

The quotient (greater than 1) of 2 consecutive terms is :

(2n + 6) : (n + 3) = 2(n + 3) : (n + 3) = 2

So, 2n = n + 3 or n = 3

The 4^th term is : 3 x 2³ = 24

Let a be the positive difference between 2 consecutive terms, then the 9^th term is :

\(17+6a=83\Leftrightarrow6a=66\Leftrightarrow a=11\)

So, the 1^st term is : 17 - 11 x 2 = -5