Plz need ASAP

Consider the following algorithm:

Step 1: Start with a list of the positive integers from 2 to 100, inclusive (meaning you include 2 and 100 in your list).

Step 2: For each n from 2 to 50, inclusive, do the following: Remove all multiples of n greater than n from the list.

Step 3: Output all of the numbers remaining in the list.

What is the tenth number in the list that gets output?

29

Step-by-step explanation:

If all the multiples are removed, you'll be left with prime numbers only

2, 3, 5, 7, 11, 13, 17, 19, 23, 29..

The tenth prime no. is 29

29

Step-by-step explanation:

2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25,26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100

Delete all even numbers greater than 2 since they are multiples of 2

2, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59, 61, 63, 65, 67, 69, 71, 73, 75, 77, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99.

Now delete multiples of 3 greater than 3

2, 3, 5, 7,  11, 13,  17, 19,  23, 25,  29, 31, 35, 37,  41, 43,  47, 49,  53, 55,  59, 61, 65, 67, 71, 73,  77, 79,  83, 85, , 89, 91, , 95, 97,

We can skip all even numbers since they we all ready taken care of by deleting the evens

Now delete multiples of 5 greater than 5

2, 3, 5, 7,  11, 13,  17, 19,  23,   29, 31,  37,  41, 43,  47, 49,  53,  59, 61,  67, 71, 73,  77, 79,  83,  89, 91,  97,

Now delete multiples of 7 greater than 7

2, 3, 5, 7,  11, 13,  17, 19,  23,   29, 31,  37,  41, 43,  47,   53,  59, 61,  67, 71, 73,  79,  83,  89,   97,

There are no more numbers to delete since these are all prime numbers

29 is the 10th number in the list

a

step-by-step explanation: