A composite number is any number that isn't prime.

\(\displaystyle 105 =21* 5\)

 The remaining numbers are prime.

Finding just one factor eliminates a number from the set of prime numbers. For a large number \(\displaystyle n\), begin with the smallest prime and exhaust all possible prime factors \(\displaystyle 2\), \(\displaystyle 3\), \(\displaystyle 5\), \(\displaystyle 7\), \(\displaystyle 11\), etc. 

\(\displaystyle $111\hspace{1mm}\div\hspace{1mm}3=37\)

\(\displaystyle $121\hspace{1mm}\div\hspace{1mm}11=11\)

\(\displaystyle $141\hspace{1mm}\div\hspace{1mm}3=47\)

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}2=65\) with \(\displaystyle 1\) remaining

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}3=43\) with \(\displaystyle 2\) remaining

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}5=26\) with \(\displaystyle 1\) remaining

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}7=18\) with \(\displaystyle 5\) remaining

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}11=11\) with \(\displaystyle 10\) remaining

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}13=10\) with \(\displaystyle 1\) remaining

For bonus points, explain why it is only necessary to test prime factors up to and including \(\displaystyle \sqrt{n}\).

\(\displaystyle \sqrt{121} = 11\) and \(\displaystyle \sqrt{169} = 13\) so \(\displaystyle 11 < \sqrt{131} < 13\).

\(\displaystyle $131\hspace{1mm}\div\hspace{1mm}13=10\) with \(\displaystyle 1\) remaining.

For any number  \(\displaystyle n\geq13\), \(\displaystyle $131\hspace{1mm}\div\hspace{1mm}n\leq 10\) with \(\displaystyle 1\) remaining \(\displaystyle < 13\).

Thus, for a number \(\displaystyle n\geq13\) to be a factor of \(\displaystyle 131\), it must pair with another factor \(\displaystyle < 13\).

However, we have already exhausted all of the prime numbers \(\displaystyle < 13\) so no prime factor exists.

\(\displaystyle 7 + 0i = 7\) which is the only rational number listed. 

\(\displaystyle \sqrt{7}\) is real but not rational, that is, it cannot be expressed in the form \(\displaystyle \frac{p}{q}\) where \(\displaystyle \mathbb{Z}\) is the set of all integers and \(\displaystyle p, q \hspace{1mm} \epsilon \hspace{1mm} \mathbb{\mathbb{Z}}\).

\(\displaystyle 7i\) and \(\displaystyle ^{\sqrt{-7}}\) are complex numbers and therefore not real. A complex number is any number that contains the expression \(\displaystyle \sqrt{-1}\) or \(\displaystyle i\).

A prime number is a number with with no integer divisors, or factors, other than \(\displaystyle 1\) and the number itself. The prime factors of \(\displaystyle 14\) are \(\displaystyle 2\) and \(\displaystyle 7\). (For technical reasons, \(\displaystyle 1\) is not considered prime and the number itself is not considered a factor.) The prime factors of \(\displaystyle 15\) are \(\displaystyle 3\) and \(\displaystyle 5\). The prime factor of \(\displaystyle 16\) is \(\displaystyle 2\). \(\displaystyle 17\), however, cannot be divided evenly by any integer other than \(\displaystyle 1\) and \(\displaystyle 17\) thus is prime.

A positive number is any number that is greater than zero. \(\displaystyle -3.14\) is a negative decimal, \(\displaystyle -\frac{22}{7}\) is a negative quotient, \(\displaystyle 0\) is non-negative but it is not strictly to the right of zero on the number line. 

\(\displaystyle \pi\), or pi, is a mathematical symbol for the ratio of a circle's circumference \(\displaystyle C\) divided by twice its radius \(\displaystyle r\) or

 \(\displaystyle \pi=\frac{C}{2r}\) 

and, though irrational, is the only number listed that is strictly greater than zero. Irrational numbers are defined more rigorously later in the problem set.

This is a bit of a trick question. \(\displaystyle \pi\) is an irrational number and the incorrect answers listed are all approximations of \(\displaystyle \pi\). However, every number listed can be expressed as a quotient of integers \(\displaystyle \frac{p}{q}\) except \(\displaystyle \sqrt{2}\). The square root of two is irrational as is the square root of any prime number. 

A rational number is a number that can be expressed in the form \(\displaystyle \frac{p}{q}\) where both \(\displaystyle p\) and \(\displaystyle q\) are integers. 

\(\displaystyle \sqrt{9}=\sqrt{3*3}=\sqrt{3^2}=3\) which is not only rational but an integer.

\(\displaystyle \sqrt{8}=\sqrt{2*2*2}=\sqrt{2^3}=2\sqrt{2}\). Any square root of a prime number, the number 2 here, is irrational.

Finally, any square root of a negative number is complex and not real.

Remember that \(\displaystyle i = \sqrt{-1}\) and that any number that includes either \(\displaystyle \sqrt{-1}\) or \(\displaystyle i\) is complex and therefore not real. 

\(\displaystyle 3 - 5 = -2\) which is real but negative.

The other numbers evaluate as

\(\displaystyle 5 - (3 + 3i) = 2 - 3i\) ,

\(\displaystyle \sqrt{-3} = \sqrt{3 * (-1)} = \sqrt{3} * i\) , and 

\(\displaystyle 3i = 0 + 3i\).

Only \(\displaystyle 5 + 3i - (3 + 3i) = (5 - 3) + (3 - 3)i = 2\) is a positive, real number. 

All of the numbers given are complicated, but only one is complex in the mathematical sense. A complex number is a number expressed in the form \(\displaystyle a + bi\) where \(\displaystyle a\) and \(\displaystyle b\) are real numbers (the formal, mathematical expression for this is  \(\displaystyle a, b \hspace{1mm} \epsilon \hspace{1mm} \mathbb{R}\) ) and \(\displaystyle i = \sqrt{-1}\). 

\(\displaystyle 3^7 < 3^{11}\) so \(\displaystyle 3^7 - 3^{11} < 0\) and \(\displaystyle \sqrt{3^7 - 3^{11}}\) is complex.

While this problem can be answered by straight division, an easier method of doing so would be to apply the division tests for 3 and 4.

An integer is divisible by 3 if and only if the sum of its digits is divisible by 3. The digit sum of the number is 

\(\displaystyle 1 + 7 + 6 + 1 + 7 + 6 + 1 + 7 + 6 + 1 + 7 + 6 = 56\)

\(\displaystyle 56 \div 3 = 18\textup{ R }2\)

The digit sum is not divisible by 3. It follows that 176,176,176,176 is not divisible by 3.

An integer is divisible by 4 if and only if the integer formed by its final two digits is divisible by 4. This integer is 76, and

\(\displaystyle 76 \div 4 = 19\).

The two-digit integer is divisible by 4. It follows that 176,176,176,176 is not divisible by 4.