The comments are that they both are multidigits numbers
Answer:
x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)
OR x^9/(∛y)
Step-by-step explanation:
Given the indicinal equation
(x^27/y)^1/3
To find the corresponding expression, we will simplify the equation as shown:
(x^27/y)^⅓
= (x^27)^⅓/y⅓
= {x^(3×9)}^⅓/y⅓
= x^9/y⅓
= x^9/(∛y)
The right answer is x Superscript 9 Baseline (RootIndex 3 StartRoot y EndRoot)
For the first digit you are choosing from 2 digits
for the second digit you are choosing from 2 digits
for the third digit you are choosing from 2 digits
2*2*2=8
111
110
011
101
000
001
100
010
However......
A number doesn't usually start with a 0 or 0s.
Therefore, if you want 3-digit numbers and not just permutations using 0 and 1, then you must eliminate
011, 000, 001, and 010
Seeing that the first digit can't be 0, you choose from 1, then 2, and then 2 digits again; 1*2*2=4 numbers
You choose which answer best suits your problem.
So we can start with the full of possibilities and eliminate them one by one.
The full set is {0,1,2,3,4,5,6,7,8,9}.
Now we know that any prime greater than 2 is odd as otherwise it would have 2 as a factor, so we can eliminate all of these digits that would be an even number, leaving:
{1,3,5,7,9}
We also know that any prime greater than 5 cannot be a multiple of 5 and that all numbers with 5 in the digits are a multiple of 5, so we can eliminate 5.
{1,3,7,9}
We know that 11,13,17 and 19 are all primes, so we cannot eliminate any more of these, leaving the set:
{1,3,7,9} as our answer.
