Exercises: Simple fractions

Calculate (without a calculator).

$3^{5}$

$\left[400-\left(7+3\cdot2^7\right)\right]:3$

Divide the power expression with base 75 and exponent 3 by the power expression with base 5 and exponent 5.

3
Divide the power expression with base 6 and exponent 4 by the product of 9 and 16.

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
Set up the term
First, you need to "translate" the text in the task into a mathematical term:
"Power with base 6 and exponent $4$ "
results in: $6^{4}$
"divide"
divided by
"Product of 9 and 16"
results in the term: $9\cdot 16$
Now you put this together to a term.
If you write the division with the divide-by sign, you get:
$6^4 : (9\cdot 16)$
(Important: Around " $9\cdot16$ " you have to write a parenthesis, because otherwise it would be divided only by $9$ and not by $9\cdot 16$ ).
If you write the division with the fraction bar, you get:
$\dfrac{6^4}{9\cdot16}$
(You don't need to put parentheses here because the fraction bar acts like a parenthesis).
Calculate term
1st way:
Now calculate the term:
$\dfrac{6^4}{9\cdot16}$ $=\dfrac{1296}{144}$
Since 1296 is a multiple of 144, you can shorten the fraction by 144.
$\dfrac{6^4}{9\cdot 16}= \dfrac{1296}{144}=9$
2nd way:
$\dfrac{6^4}{9\cdot16}\ =\ \dfrac{(2\cdot3)^4}{3^2\cdot2^4}$
$\dfrac{ 2^4\cdot3^4}{3^2\cdot2^4}\ =\ \dfrac{\cancel {2^4}\cdot3^4}{3^2\cdot\cancel{2^4}}\ =\ \dfrac{3^4}{3^2}\ =\ 3^{(4-2)}$ apply the power laws
$3^2\ =\ 9$


"Power with base 6 and exponent $4$ "	results in: $6^{4}$
"divide"	divided by
"Product of 9 and 16"	results in the term: $9\cdot 16$

Divide the power expression with base 18 and exponent 4 by the power expression with base 3 and exponent 7. Calculate as advantageously as possible!

Multiply the power expression with base 6 and exponent 4 by the power expression with base 5 and exponent 3.

Hint: Try to use some clever transformations.

6
In how many two-digit square numbers is the last digit odd?

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
We try the square numbers one by one. Since $10^{2} = 100$ is already a three-digit number, we should not get too many two-digit square numbers:
$\def\arraystretch{1.25} \begin{array}{rll}1^2 & = 1 & | \text{ not two-digit} \\2^2 &= 4 & | \text{ not two-digit} \\3^2 &= 9 & | \text{ not two-digit} \\4^2 &= 16 & | \text{ even square number} \\\color{Green}{5^2} &\color{Green}{= 25} & \color{Green}{|\text{ 1st odd square number}} \\6^2 &= 36 & | \text{ even square number} \\\color{Green}{7^2} &\color{Green}{= 49} & \color{Green}{|\text{ 2nd odd square number}} \\8^2 &= 64 & | \text{ even square number} \\\color{Green}{9^2} &\color{Green}{= 81} & \color{Green}{|\text{ 3rd odd square number}} \\10^2 &= 100 & | \text{ already three-digit}\end{array}$
All square numbers from 100 on are no longer two-digit numbers. Thus there are 3 square numbers, whose final digit is odd, namely 25, 49 and 81.
7
Martin would like to create a flower bed. He is planning a square-shaped bed with a length of $0.5 m$ . Right now, he is heading to the store, buying some flower soil.
His mother is quite experienced in gardening. Before Martin left, she told him that one bag of soil covers $0.15 m^{2}$ . Is one bag enough for Martin, or does he need to buy more? (if yes, how many?)

Für diese Aufgabe benötigst Du folgendes Grundwissen: Powers
The area for the bed can be calculated using the formula for the area of a square if you substitute $0.5 m$ for $a$ :
$\def\arraystretch{1.25} \begin{array}{rcl} A_{bed}&=&a^2 \\\\&=&(0.5m)^2 \\\\&=& 0.25m^2 \end{array}$
Since one bag is only enough for $0.15 m^{2}$ , he must take more.
Two bags are sufficient for $2\cdot0.15m^2=0.3m^2$ .
So two bags are enough, because Martin only needs $0.25 m^{2}$ .

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$\displaystyle 3^5$	$=$	$\displaystyle 3\cdot3\cdot3\cdot3\cdot3$
	↓	Calculate step by step.
	$=$	$\displaystyle \underbrace{3\cdot3}_9\cdot3\cdot3\cdot3$
	$=$	$\displaystyle \underbrace{9\cdot3}_{27}\cdot3\cdot3$
	$=$	$\displaystyle \underbrace{27\cdot3}_{81}\cdot3$
	$=$	$\displaystyle 81\cdot 3$
	$=$	$\displaystyle 243$

$\displaystyle \left[400-\left(7+3\cdot2^7\right)\right]:3$	$=$	$\displaystyle \left[400-\left(7+3\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\right)\right]:3$
	↓	Conclude the product
	$=$	$\displaystyle \left[400-\left(7+3\cdot4\cdot4\cdot4\cdot2\right)\right]:3$
	↓	Conclude again
	$=$	$\displaystyle \left[400-\left(7+3\cdot16\cdot4\cdot2\right)\right]:3$
	↓	and conclude...
	$=$	$\displaystyle \left[400-\left(7+3\cdot16\cdot8\right)\right]:3$
	↓	...
	$=$	$\displaystyle \left[400-\left(7+3\cdot128\right)\right]:3$
	↓	... and conclude once more
	$=$	$\displaystyle \left[400-\left(7+384\right)\right]:3$
	↓	Now evaluate the sum
	$=$	$\displaystyle \left[400-391\right]:3$
	↓	Evaluate the difference
	$=$	$\displaystyle 9:3$
	↓	This can easily be divided
	$=$	$\displaystyle 3$

$\displaystyle \frac{75^3}{5^5}$	$=$	$\displaystyle \frac{15^3\cdot5^3}{5^5}$
	$=$	$\displaystyle \frac{3^3\cdot5^3\cdot5^3}{5^5}$
	↓	Shorten by $5^{5}$ .
	$=$	$\displaystyle 3^3\cdot5^1$
	↓	Now calculate the individual powers. $3^3 = 3\cdot 3\cdot 3 = 27$ and $5^{1} = 5$
	$=$	$\displaystyle 27\cdot5$
	$=$	$\displaystyle 135$

$\displaystyle \frac{18^4}{3^7}$	$=$	$\displaystyle \frac{\left(2\cdot3^2\right)^4}{3^7}$
	↓	Apply power laws.
	$=$	$\displaystyle \frac{2^4\cdot\left(3^2\right)^4}{3^7}$
	↓	Apply power laws.
	$=$	$\displaystyle \frac{2^4\cdot3^8}{3^7}$
	↓	Apply power laws.
	$=$	$\displaystyle 2^4\cdot3^1$
	$=$	$\displaystyle 16\cdot3$
	$=$	$\displaystyle 48$

$\displaystyle 6^4\cdot5^3$	$=$	$\displaystyle 3^4\cdot2^4\cdot5^3$
	↓	$2^4=2\cdot2^3$
	$=$	$\displaystyle 3^4\cdot2\cdot2^3\cdot5^3$
	↓	$2^3\cdot5^3=(2\cdot5)^3$
	$=$	$\displaystyle 3^4\cdot2\cdot\left(2\cdot5\right)^3$
	↓	Multiply.
	$=$	$\displaystyle 162\cdot1000$
	$=$	$\displaystyle 162000$

Exercises: Simple fractions

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