This is one of those topics where questions can use different phrases that all mean the same thing …
Find the prime factors of 360.
Give your answer in the form , where , and are integers to be found.
For each number find any two numbers, (not 1), which are factors and write those as the next pair of numbers in the tree.
If a number is prime, put a circle around it.
When all the end numbers are circled, you are done!
For any numbers that are repeated, write them as powers of the number.
You don’t usually have to write a “1” as a power if there is a number that isn’t repeated, but in this question, it has asked for it.
When a number has been written as in its prime factor decomposition (PFD), it can be used to find out if that number is a square or cube number, or to find the square root of that number without using a calculator.
Substitute N = 2^{3 }× 3^{2 }× 5^{7} into the formula AN = B.
A(2^{3 }× 3^{2 }× 5^{7} ) = B
2, 3 and 5 are all prime numbers, so for A(2^{3 }× 3^{2 }× 5^{7} ) to be a square number, its prime factors must all have even powers.
Consider the prime factors A needs to have to make all the values on the left hand side have even powers.
(2 × 5) (2^{3 }× 3^{2 }× 5^{7}) = B
2^{4 }× 3^{2 }× 5^{8 }= B
So A when written as a product of its prime factors, is 2 × 5.
Make sure you A as an integer value in the answer.
A = 10