A simple calculation will reveal that 72 has 12 factors including 1 and 72. How many natural numbers less than 100 have 12 such factors
To find the number of factors of a number, list its prime factors.
e.g.
12 = 2x2x2x3x3 = (2^3)(3^2)
so now we argue as follows:
the 2's can be used in 4 ways:
that is, we can use none, one, two , or all three of them.
similarly the 2's can be used in 3 ways, none, one of them or both of them
that would include taking neither one ---> represents the factor 1
or taking all, representing the factor 72
then the number of factors including 1 and 72 is 4x3 or 12
So let's work this backwards.
try different combinations of 2's and 3's, staying below a product of 100
e.g. suppose we have 2^5 x 3 , that is 96 which is below 100
we can use the 2's in 6 ways, and the 3 in 2 ways
notice that 6x2 = 12 for 12 factors
So my choice is the number 96, which has 12 factors including 1 and 96 itself
72 = 2^3 * 3^2
So, I'd guess that
96 = 2^5 * 3
will too
the first line of my explanation should have said:
72 = 2x2x2x3x3 = (2^3)(3^2)
Looks like I had 12 on my mind
To find the natural numbers less than 100 that have 12 factors, we need to consider numbers that can be expressed as a product of distinct prime numbers raised to certain powers.
To have 12 factors, the number must be of the form p1^a * p2^b * p3^c, where p1, p2, p3 are distinct prime numbers, and a, b, c are positive integers.
We know that if a number is expressed as a product of prime factors raised to certain powers, the total number of factors is given by (a + 1)(b + 1)(c + 1).
Since the total number of factors is 12, we need to find all the possible combinations of (a + 1)(b + 1)(c + 1) = 12.
The factor pairs of 12 are (1, 12), (2, 6), and (3, 4). Using these factor pairs, we can determine the possible values of (a + 1), (b + 1), and (c + 1).
For (1, 12), (a + 1) = 1, (b + 1) = 3, (c + 1) = 4
For (2, 6), (a + 1) = 2, (b + 1) = 2, (c + 1) = 3
For (3, 4), (a + 1) = 3, (b + 1) = 2, (c + 1) = 2
Now, we need to find the prime numbers that can be raised to these powers.
For (a + 1) = 1, the only prime factor would be 2.
For (b + 1) = 3, the prime factors could be 2, 3, or 5.
For (c + 1) = 4, the prime factors could be 2, 3, 5, or 7.
Combining all the possibilities, we have:
2^1, 2^1 * 3^3, 2^1 * 5^3, 2^1 * 7^3
2^2 * 3^2, 2^2 * 5^2, 2^2 * 7^2
2^3 * 3^1, 2^3 * 5^1, 2^3 * 7^1.
Now, we need to check which of these numbers are less than 100:
2^1 = 2, 2^2 = 4, 2^3 = 8
2^1 * 3^1 = 6, 2^2 * 3^2 = 36, 2^3 * 3^1 = 24
2^1 * 5^1 = 10, 2^2 * 5^2 = 100 (greater than 100)
2^1 * 7^1 = 14, 2^2 * 7^2 = 196 (greater than 100)
2^1 * 3^3 = 54, 2^2 * 3^2 * 5^2 = 180 (greater than 100)
2^1 * 5^3 = 250 (greater than 100)
2^1 * 7^3 = 686 (greater than 100)
2^2 * 3^1 = 12, 2^3 * 5^1 = 40
2^2 * 7^1 = 28, 2^3 * 3^1 * 5^1 = 120 (greater than 100)
2^2 * 3^3 = 216 (greater than 100)
2^3 * 3^2 = 72.
Therefore, there are four natural numbers less than 100 that have 12 factors: 4, 8, 12, and 72.