210
21 10
7 3 5 2
Now add 1 to each of the exponents and multiply together
7^1 3^1 5^ 1 2^1
so total of four exponents
(1+1) * (1+1) * (1+1) * ( 1+ 1)
= 16
Trying to write out and determine each integer factor of 210 would take a lot of time and it would be very easy to make a mistake. Instead of doing that, you can use a short cut called prime factorization.
To calculate the prime factorization of a number, first we choose any two factors of our number, for this problem use 2 and 105:
210 = 2 * 105
Now we can choose factors of those numbers until we break it down fully to prime numbers. 2 because has already been reduced down to a prime number, so we will take factors of 105, 5 and 21.
210 = 2 * 5 * 21
We can repeat the process with 21, but again we can leave 5 because it has been reduced down to a prime number.
210 = 2 * 5 * 3 * 7
Now that we are left with only prime numbers, combine and find the exponents of those numbers (in some problems you may have multiple 2s and combine them as 2x). For this problem, we have the exponents {1, 1, 1, 1}:
210 = 21 * 51 * 31 * 71
Next, add 1 to each exponent, giving {2, 2, 2, 2}, and then multiply the exponents together to give you the answer.
(1+1)*(1+1)*(1+1)*(1+1)=2 * 2 * 2 * 2 = 16, [D]
Extended Explanation:
Why do we add 1 to each exponent? The reason why we add 1 to each exponent is because there are two possibilities for that exponent. The exponent of each prime factor can be either 0 or 1 in order to get all possible integer factors.
For example, the prime factors we have for 210 are:
2, 5, 3, 7
We can set the exponent of 2 to “0” in order to get one possible integer factor. 2^0 = 1, so therefore:
1 * 5 * 3 *7 = 105, which is one of the possible integer factors.
We can also leave the exponent of 2 as “1” in order to get another possible integer factor:
2 * 5 * 3 * 7 = 210, which is another possible integer factor.
Therefore, we have two exponent possibilities for each prime factor (2, 5, 3, 7). Since there are 4 total prime factors, there must be 2 * 2 * 2 * 2 = 16 total combinations of possibilities.
Try setting the exponent to either 0 or 1 for one or more of the other prime factors (5, 3, or 7) if you’re still having trouble seeing how this method works.
For your reference, here is the complete list of factors:
1, 2, 3, 5, 6, 7,10,14,15, 21, 30, 35, 42, 70,105, 210 (for a total of 16)
Topic: Numerical Calculations