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3 years ago

Prove that 17^16+17^14 is divisible by 5

Mathematics

1 answer:

fredd [130]3 years ago

6 0

We'll use the exponent rule that a^(b+c) = a^b*a^c

to break up 17^16 into 17^(14+2) = 17^14*17^2

That allows us to factor like so

17^16+17^14

17^14*17^2+17^14*1

17^14(17^2 + 1)

Computing 17^14 is a lot more difficult compared to computing 17^2 + 1, assuming you only could use paper and pencil.

Note that 17^2 = 17*17 = 289, so 17^2 + 1 = 289+1 = 290

Because 290 ends in a 0, this means that 5 is a factor

290 = 5*58

This makes 5 to be a factor of the original expression.

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So overall,

17^16+17^14 = 17^14(17^2 + 1) = 17^14*290 = 17^14*5*58 = 5*17^14*58

The stuff "17^14*58" doesn't matter. All we care about is that factor of 5.

Since 5 is a factor, this means that 17^16+17^14 is a multiple of 5, or 17^16+17^14 is divisible by 5.

Phrased another way, 17^16+17^14 all divided over 5 leads to some integer. In this case, the integer would be the result of 17^14*58.

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In isosceles right triangle ABC, point is on hypotenuse \overline{BC} such that \overline{AD} is an altitude of \triangle ABC an

Dmitriy789 [7]

Answer:

Area of triangle is 25.

Step-by-step explanation:

We have been given an isosceles right triangle

Isosceles triangle is the triangle having two sides equal.

Figure is shown in attachment

By Pythagoras theorem

$BC^2=AC^2+AB^2$

AD is altitude which divides the triangle into two parts

DC=5 implies BC =10 since D equally divides BC

Let AC=a implies AB=a being Isosceles

On substituting the values in the Pythagoras theorem:

$10^2=a^2+a^2$

$100=2a^2$

$\Rightarrow a^2=50$

$\Rightarrow a=\pm5\sqrt{2}$

WE can find area of right triangle by considering height AB and AD

Area of triangle ABC is:

$\frac{1}{2}\cdot BC\cdot AD$ (1)

$\Rightarrow \frac{1}{2}\cdot 10\cdot AD$

And other method of area of triangle is:

$\frac{1}{2}\cdot AB\cdot BC$ (2)

Equating (1) and (2) we get:

$\frac{1}{2}\cdot 10\cdot AD=\frac{1}{2}\cdot a\cdot a$

$\Rightarrow AD=\frac{a^2}{10}$

$\Rightarrow AD=\frac{50}{10}=5$

Using area of triangle is: $\frac{1}{2}\cdot BC\cdot AD$

Now, the area of triangle ABC= $\frac{1}{2}\cdot 5\cdot 10$

$\Rightarrow 25$

4 0

3 years ago

What is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)?

Rasek [7]

The side lengths of triangle are 6 units, 8 units and 10 units.

SOLUTION:

Given that, we have to find what is the length side of a triangle that has vertices at (-5, -1), (-5, 5), and (3, -1)

We know that, distance between two points $P\left(x_{1}, y_{1}\right) \text { and } Q\left(x_{2}, y_{2}\right)$ is given by

$P Q=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$

Now,

$\begin{array}{l}{\text { Distance between }(-5,-1) \text { and }(-5,5)=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {\qquad \begin{array}{l}{=\sqrt{(-5-(-5))^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(0)^{2}+(5+1)^{2}}=\sqrt{(6)^{2}}=6} \\\\ {=\sqrt{(-5)^{2}+(5+1)^{2}+(5-(-1))^{2}}} \\\\ {=\sqrt{(-8)^{2}+(5+1)^{2}}=\sqrt{64+36}=\sqrt{100}=10} \\\\ {=\sqrt{(3-1)^{2}+(-1-(-1))^{2}}} \\\\ {=\sqrt{(5+3)^{2}+(0)^{2}}=\sqrt{(8)^{2}}=8}\end{array}}\end{array}$