SOLUTION: A “simple square” is any integer greater than 1 that has only three positive integer factors—itself,its square root, and 1. Which of the following is a simple square? A. 121 B.

Algebra ->  Problems-with-consecutive-odd-even-integers -> SOLUTION: A “simple square” is any integer greater than 1 that has only three positive integer factors—itself,its square root, and 1. Which of the following is a simple square? A. 121 B.      Log On


   

Question 338592: A “simple square” is any integer greater than 1 that has only three positive integer factors—itself,its square root, and 1.
Which of the following is a simple square?
A. 121 B. 100 C. 81 D. 64 E. 33

Found 2 solutions by Edwin McCravy, CharlesG2:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
A. 121 B. 100 C. 81 D. 64 E. 33

121 has only the factors 1, 11 (itself), and its square root, 11.  
So that's the correct answer.

100 has factors 1,2,4,5,10,20,25,50, and 100. So that's not it.

81 has factors 1,3,9,27, and 81, so that's not it.

64, has factors 1,2,4,8,16,32, and 64, so that's not it.

33 isn't even a square at all, and has factors 1,3,11, and 33, so that's
not it.

Edwin

Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
A “simple square” is any integer greater than 1 that has only three positive integer factors—itself,its square root, and 1.
Which of the following is a simple square?
A. 121 B. 100 C. 81 D. 64 E. 33


factor each of the choices
121 = 11^2 = 1 * 121 = 1 * 11 * 11 (3 positive integer factors)
100 = 10^2 = 2^2 * 5^2 = 1 * 4 * 25 = 1 * 2 * 2 * 5 * 5
81 = 9^2 = 1 * 3 * 27 = 1 * 3 * 3 * 3 * 3
64 = 8^2 = 2^2 * 4^2 = 1 * 8 * 8 = 1 * 2 * 2 * 2 * 2 * 2 * 2
33 = 1 * 3 * 11
33 is not a product of a whole number times itself like the other choices


the answer is A. 121