Quiz Level-2

Hint

\(
\begin{gathered}
2 \times 2 \times 2 \times 7 \\
=2^3 \times 7^1
\end{gathered}
\)

There is a shortcut to finding how many factors.

Add 1 to each exponent in the prime factorization and multiply = (3+1)(1+1) =8.

Hint

\(
\text { Prime Factorization of } 240: 2 \times 2 \times 2 \times 2 \times 3 \times 5 \text { or } 2^4 \times 3^1 \times 5^1 \text {. }
\)

Number of factors are (4+1)(1+1)(1+1)=20

Hint

An integer with 6 factors has its prime factorization in the form \(a^5\) or \(a^2 \times b^1\). \(2^2 \times 3^1\)=12 is the smallest such number.

Hint

An integer with 10 factors has its prime factorization in the form \(a^9\) or \(a^4 \times b^1\). So, \(2^4 \times 3^1\)=48 is the smallest such number.

Hint

Integers that have an odd number of factors are all perfect squares. There are 9 perfect squares less than 100.

Explanation:

There are 9 integers less than 100 that have an odd number of factors.
Every factor of a number has a pair, eg \(2 \& 3\) are a factor pair of 6 (Factors of 6: 1, 2, 3 and 6 which is 4 that is even); and so it would be expected that every number has an even number of factors.
However, if the factor pair of a number are the same number (eg \(6 \& 6\) are a factor pair of 36), then there will be an odd number of factors.
When there is a repeated factor like this, the number is a perfect square (The factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36 which is 9 and is odd).
Thus only perfect squares have an odd number of factors.
Less than 100 there are 9 perfect squares \((1,4,9,16,25,36,49,64\) & 81) which have an odd number of factors.
100 itself is a perfect square and also has an odd number of factors, but the question asked for those numbers less than 100 with an odd number of factors.

Hint

Both a and b must be the square of a prime number (only the square of a prime number will have exactly 3 factors). We organize our work, looking for products \(x^2 \times y^2 < 1000\), where x <y and both x and y are prime. The following 7 pairs(x,y) are the only ones that satisfy the conditions.

(2,3) (2,5) (2,7) (2.11) (2,13) (3,5) and (3,7)

Hint

To have 31 factors, the prime factorization of a positive integer must be in the form of \(x^{30}\). To minimize this value we use x = 2 to get \(2^{30}\)

Hint

All of the above. In each case number of factors is 12.

Hint

\(
\begin{aligned}
& p: 2,3,5,7,11,13,17,19,23,29 \\
& p=10 \\
& n=30 \\
& f=\frac{p}{n}=\frac{10}{30}=\frac{1}{3}=0.3333
\end{aligned}
\)

Hint

\(2^1 \times 3^1 \times 5^1 \times 7^1=16\)

Number of factors = (1+1) (1+1) (1+1) (1+1) = 16

Factors of 210: 1, 2, 3, 5, 6, 7, 10, 14, 15, 21, 30, 35, 42, 70, 105 and 210.

Prime Factorization of 210: 2 × 3 × 5 × 7.