Number Theory - SAT Math

Card 0 of 280

Question

Identify the real part of

Answer

A complex number in its standard form is of the form: , where stands for the real part and stands for the imaginary part. The symbol stands for $\sqrt{-1}$ .

The real part in this problem is 1.

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Question

Multiply:

Answer must be in standard form.

Answer

The first step is to distribute which gives us:

$2-2i+3i-3i^{2}$

$2+i-3i^{2}=2+i+3=5+i$

which is in standard form.

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Question

Simplify the expression.

Answer

Combine like terms. Treat $\small i$ as if it were any other variable.

Substitute to eliminate $\small i^2$ .

Simplify.

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Question

$\frac{-5+10i}{3+4i}=?$

Answer

$\frac{-5+10i}{3+4i}$

$=\frac{(-5+10i)(3-4i)}{(3+4i)(3-4i)}$

$=\frac{-15+20i+30i-40i^{2}}{9-16i^{2}}$

$=\frac{-15+50i+40}{9+16}$

$=\frac{25+50i}{25}$

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Question

Which of the following is not an irrational number?

Answer

A root of an integer is one of two things, an integer or an irrational number. By testing all five on a calculator, only $\sqrt[3]{125}$ comes up an exact integer - 5. This is the correct choice.

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Question

$\frac{1-7i}{6-2i}=a-i$

Find .

Answer

Multiply the numerator and denominator by the numerator's complex conjugate.

$\frac{1-7i}{6-2i}\ast \frac{6+2i}{6+2i}=\frac{20-40i}{40}$

Reduce/simplify.

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Question

Multiply:

Answer

Use the FOIL technique:

$= 7 \cdot 1 - 7 \cdot 2i + 3i \cdot 1 - 3i \cdot 2i$

$= 7 - 14i + 3i - 6i ^{2}$

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Question

Evaluate:

$\left (2 + 3i \right )^{3}$

Answer

We can set in the cube of a binomial pattern:

$\left ( A + B\right ) ^{3} = A ^{3} + 3 A^{2 }B + 3AB^{2} + B^{3}$

$= 2 ^{3} + 3 \cdot 2 ^{2 } \cdot 3i + 3 \cdot 2 \cdot (3i)^{2} + (3i)^{3}$

$= 2^{3} + 3 \cdot 2^{2} \cdot 3i + 3 \cdot 2 \cdot 3 ^{2}\cdot i^{2} + 3^{3} \cdot i^{3}$

$= 8 + 36i + 54 i^{2} + 27 i^{3}$

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Question

Simplify by rationalizing the denominator:

$\frac{45}{5 + \sqrt{7}}$

Answer

Multiply the numerator and the denominator by the conjugate of the denominator, which is $5 - \sqrt{7}$ . Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{45}{5 + \sqrt{7}}$

$= \frac{45\left (5 - \sqrt{7} \right )}{\left (5 + \sqrt{7} \right )\left (5 - \sqrt{7} \right )}$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 5 ^{2}- \left ( \sqrt{7} \right ) ^{2} }$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 25 -7}$

$= \frac{45\left (5 - \sqrt{7} \right )}{ 18}$

$= \frac{5\left (5 - \sqrt{7} \right )}{ 2}$

$= \frac{ 25 - 5 \sqrt{7} }{ 2}$

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Question

Solve for and :

$ai^{93}+bi^{35}+ai^{24}-bi^{86}=40+20i$

Answer

Remember that

$i = \sqrt{-1}\ i^2 = \sqrt{-1}^2 = -1\ i^3 = i^2\cdot i = -1 \cdot i = -i\ i^4 = i^2 \cdot i^2 = (-1)\cdot(-1) = 1\ i^5 = i^4\cdot i = 1\cdot i = i = \sqrt{-1}\ i^6 = i^4\cdot i^2 = 1\cdot i^2 = i^2 =-1$

So the powers of are cyclic.This means that when we try to figure out the value of an exponent of , we can ignore all the powers that are multiples of because they end up multiplying the end result by , and therefore do nothing.

This means that

$ai^{93} + bi^{35} + ai^{24} - bi^{86}\ = ai^{92+1} + bi^{32+3} + ai^{24+0} - bi^{84+2}\ = ai^{92}\cdot i^1 + bi^{32}\cdot i^3 + ai^{24}\cdot i^0 - bi^{84}\cdot i^2\ =a(i^4)^{23}\cdot i^1 + b(i^4)^8\cdot i^3 + a(i^4)^6\cdot i^0 - b(i^4)^{21}\cdot i^2\ = a (1^{23})\cdot i + b(1^8)\cdot i^3 + a(1^6) - b(1^{21})\cdot i^2$

Now, remembering the relationships of the exponents of , we can simplify this to:

$ai - bi + a - (-b) = ai-bi+a+b\ = (a+b) + (a-b)i = 40+20i$

Because the elements on the left and right have to correspond (no mixing and matching!), we get the relationships:

No matter how you solve it, you get the values , .

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Question

If $\small x$ and $\small y$ are real numbers, and $\small \small 3x+y+(2x+2y)i=13+10i$ , what is $\small z$ if $\small z=x-4y$ ?

Answer

To solve for $\small z$ , we must first solve the equation with the complex number for $\small x$ and $\small y$ . We therefore need to match up the real portion of the compex number with the real portions of the expression, and the imaginary portion of the complex number with the imaginary portion of the expression. We therefore obtain:

$\small 3x+y=13$ and $\small 2x+2y=10$

We can use substitution by noticing the first equation can be rewritten as $\small y=13-3x$ and substituting it into the second equation. We can therefore solve for $\small x$ :

$\small 2x+2(13-3x)=10$

$\small 2x+26-6x=10$

$\small -4x=-16$

$\small x=4$

With this $\small x$ value, we can solve for $\small y$ :

$\small 3(4)+y=13$

$\small 12+y=13$

$\small y=1$

Since we now have $\small x$ and $\small y$ , we can solve for $\small z$ :

$\small z=x-4y$

$\small \small z=(4)-4(1)=0$

Our final answer is therefore $\small z=0$

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Question

Evaluate $\small \frac{-6+2i}{10-3i}$

Answer

To divide by a complex number, we must transform the expression by multiplying it by the complex conjugate of the denominator over itself. In the problem, $\small 10-3i$ is our denominator, so we will multiply the expression by $\small \frac{10+3i}{10+3i}$ to obtain:

$\small \frac{-6+2i}{10-3i}\frac{10+3i}{10+3i}=\frac{-60+20i-18i+6i^2}{100-30i+30i-9i^2}$ .

We can then combine like terms and rewrite all $\small i^2$ terms as $\small -1$ . Therefore, the expression becomes:

$\small \frac{-60+2i+6i^2}{100-9i^2}=\frac{-66+2i}{109}$

Our final answer is therefore $\small \frac{-66+2i}{109}$

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Question

Simplify the following product:

Answer

Multiply these complex numbers out in the typical way:

${}(5+3i)(-2+i) = -10+5i-6i+3i^2$

and recall that by definition. Then, grouping like terms we get

which is our final answer.

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Question

Which pair of number sets have no intersection?

Answer

An intersection of two sets is defined as the set of elements that are members of both sets. The correct answer is the pair of sets that has no overlap.

Rationals are defined $Rational =\frac{a}{b}$ if a and b are integers. The irrationals are defined as any real number that is not rational. By definition a number cannot be both rational and irrational.

Prime numbers are divisible only by 1 and themselves. Even numbers are defined as integer multiples of . A number common to both sets is .

$Prime \left { 2,3,5,7,11...\right }$

$Even \left { 0,2,4,6...\right }$

Natural numbers are the counting numbers, while the integers are all the naturals and their opposites. There are many elements common to both sets, such as :

$Natural\left {1,2,3...\right }$

$Integer \left { ...-2,-1,0,1,2...\right }$

The negatives are all the integers less than zero. And remember that an even number is defined as any integer multiple of two. Again there are many elements in common including :

$Integer \left { ...-2,-1,0,1,2...\right }$

$Negative\left { 0,-1,-2,-3...\right }$

The naturals are the counting numbers, while the wholes are all the naturals and zero. These two sets share many elements in common such as .

$Natural\left {1,2,3...\right }$

$Whole\left { 0,1,2,3...\right }$

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Question

What is the value of ?

Answer

When dealing with imaginary numbers, we multiply by foiling as we do with binomials. When we do this we get the expression below:

Since we know that we get which gives us .

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Question

Simplify:

Answer

To add complex numbers, find the sum of the real terms, then find the sum of the imaginary terms.

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Question

Simplify:

Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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Question

Simplify:

Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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Question

Evaulate:

$\frac{45-35i}{5i}$

Answer

Multiply both numerator and denominator by , then divide termwise:

$\frac{45-35i}{5i}$

$= \frac{\left (45-35i \right )\cdot i}{5i \cdot i}$

$= \frac{ 45\cdot i-35i\cdot i }{5i ^{2}}$

$= \frac{ 45 i-35i ^{2}}{5i ^{2}}$

$= \frac{ 45 i-35(-1) }{5 (-1)}$

$= \frac{ 45 i- (-35) }{-5 }$

$= \frac{35+ 45 i }{-5 }$

$= \frac{35 }{-5 } + \frac{ 45 i }{-5 }$

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Question

Simplify:

Answer

Start by using FOIL. Which means to multiply the first terms together then the outer terms followed by the inner terms and lastly, the last terms.

Remember that $i=\sqrt-1$ , so .

Substitute in for

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