Quiz Level-3

Hint

\(
\text { 480= } 2^5 \times 3^1 \times 5^1
\)

So it has (5+1)(1+1)(1+1)=24 factors. There are (1+1)(1+1)=4 odd factors, so 20 of its factors are even.

Hint

\(
900= 2^2 \times 3^2 \times 5^2
\)

Odd factor can not include any 2’s, so 900 has (2+1)(2+1)=9 odd factors.

Hint

\(1296 = 2^4 . 3^4\). To make a perfect square, we can use any even power of 2 and even power of 3. There are three of each (\(2^0, 2^2, 2^4\)) and (\(3^0, 3^2, 3^4\)), giving us 3 . 3=9 ways to create perfect square factors (1, 4, 16, 3, 9, 81, 36, 324, and 1296).

Hint

A number with 7 factors must have a prime factorization \(n^6\), so \((n^6)^2=n^{12}\) will have 13 factors.

Hint

The prime factorization of an integer with exactly 3 factors is of the form \(a^2\), so we are asked to find primes which have a square that is 3 -digits long. Beginning with \(11^2=121\) and ending with \(31^2=961\), we have \(11^2, 13^2, 17^2, 19^2, 23^2, 29^2\), and \(31^2\) for a total of 7.

Hint

The only multiple of 7 having 5 factors is \(7^4\), so \(n=7^4\) and \(3 n\) is \(3 \cdot 7^4\), which has \(\mathbf{1 0}\) factors.

Hint

For a multiple of 6 to have 9 factors, it must be a perfect square, making our integer \(2^2 \cdot 3^2\). Multiplying by 10 we get \(2^3 \cdot 3^2 \cdot 5\), which has 24 factors.

Hint

Perfect squares have three factors only when they are the square of a prime. There are 19 perfect squares less than 400. Excluding 1 and all perfect squares of primes: \(2^2, 3^2, 5^2, 7^2, 11^2, 13^2, 17^2\), and \(19^2\) leaves us with \(19-9=10\) perfect squares. (We could have just as easily counted composites).

Hint

An integer with 18 factors will have a prime factorization of the form \(a^{17}, a^8 \cdot b, a^5 \cdot b^2\), or \(a^2 \cdot b^2 \cdot c\). It is safe to quickly rule out the first two options, and we \(\operatorname{try} 2^5 \cdot 3^2=288\) against \(2^2 \cdot 3^2 \cdot 5=180.180\) is the smallest integer which has 18 factors.

Hint

The prime factorization can have a 2 or a 3 , but not both. It makes sense to use only the 2 . We quickly arrive at \(2^3 \cdot 5^1 \cdot 7^1=\mathbf{2 8 0}\), which has \(4 \cdot 2 \cdot 2=16\) factors and is the smallest such integer not divisible by 6 . Other candidates: \(2^7 \cdot 5^1=640\), \(2^1 \cdot 5^1 \cdot 7^1 \cdot 11^1=770\), and \(2^3 \cdot 5^3=1,000\)

Hint

We are looking for an integer with a prime factorization in the form \(a^2 \cdot b \cdot \mathrm{c}\), or \(a^3 \cdot b^2\), or perhaps \(a^5 \cdot b\) using prime factors greater than 2. \(3^2 \cdot 5 \cdot 7=315\), \(3^3 \cdot 5^2=675\), and \(3^5 \cdot 5=1,215\) so 315 is the smallest odd integer with 12 factors.

Hint

For \(a\) to have 5 factors, its prime factorization must be in the form \(x^4\) for some prime integer \(x\), and for \(b\) to have 6 factors its prime factorization must be in the form \(y^2 z\) (or \(y^5\) ) for some prime integers \(y\) and \(z\). The product \(a b\) must therefore have prime factorization in the form \(x^4 \cdot y^5\) or \(x^4 \cdot y^2 \cdot z\). To minimize this product we use \(2^4 \cdot 3^2 \cdot 5=720\left(2^5 \cdot 3^4=2,592\right)\)

Hint

\(
(1+2)(1+3+9+27)(1+7)=(3)(40)(8)=960
\)

Hint

\(
\begin{aligned}
& 6=2 \cdot 3:(1+2)(1+3)=12 \\
& \mathbf{8}=2^3:(1+2+4+8)=15 \\
& 9=3^2:(1+3+9)=13
\end{aligned}
\)

Hint

\(2^{30}\) has a factor sum: \(\left(2^0+2^1+2^2+2^3+\ldots+2^{30}\right)=2^{31}-1\)

Hint

Guess-and-check gives us \(\left(1+7+7^2+7^3\right)=(1+7+\) \(49+343)=400\), so our answer is \(7^3\) or 343 .

Hint

To find the average of the factors of 48, we find the sum and divide it by the number of factors. \(48=2^4 \cdot 3\), which has 10 factors and a factor sum of \((1+2+4+8+16)(1+3)=124\). The average factor of 48 is \(124 / 10=\mathbf{1 2} .4\)

Hint

Look for numbers that are less than half of 100 , this is really an educated guess-and-check, choosing numbers which have a lot of factors. Nothing below 36 comes close, so we begin there: \(36=2^2 \cdot 3^2\). Factor sum: \((1+2+4)(1+3+9)=91\). \(40=2^3 \cdot 3\). Factor sum: \((1+2+4+8)(1+5)=90\). \(42=2 \cdot 3 \cdot 7\). Factor sum \((1+2)(1+3)(1+7)=96\) \(44=2^2 \cdot 11\). Factor \(\operatorname{sum}(1+2+4)(1+11)=84\). \(45=3^2 \cdot 5\). Factor sum \((1+3+9)(1+5)=78\) \(48=2^4 \cdot 3\). Factor sum \((1+2 \ldots+16)(1+3)=124\) Somewhat surprisingly, 48 is the smallest.

Hint

Prime numbers greater than 2 will be both odd. So their sum will be even since odd \(+\) odd \(=\) even and their product will be odd since odd x odd \(=\) odd
Product \(–\) Sum \(=\) odd \(–\) even \(=\) odd
The greater the numbers, the greater the product though the sum will increase slowly. Hence Product – Sum will increase as the numbers increase. Let’s find the max and min value of Product – Sum
Min: The prime numbers are 5 and 7
Product \(–\) Sum \(=23\)
Max: The prime numbers are 13 and 17
Product – Sum \(=221-30=191\)
The only value lying in this range is 119.

Hint

If \(p=3\) :
\(1+3+3^2+3^3+3^4=121\)
(it’s a square!)