101
16,36,64,100
No, all perfect square numbers are not even numbers. Eg. the square of 3 is 9. (32=9) To generalize the proof: If p is odd then p=2n+1 and p2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1 So odd numbers have odd square
Since all the numbers in that range start with "11", there is really only one option!
1,4,9,16,25,36,49,64,81,100,121,144,169,196,225,256,289,324,361,400 All the square numbers up to 20!
No - but it is a square number. 1,3,6,10,15,21 & 28 are all triangular numbers. 1,4,9,16,25 & 36 are all square numbers.
numbers! An endless string or 'series' of even numbers!
No. The square roots 8 are irrational, as are the square roots of most even numbers.
1100
No; most are not, not even close. 2 is not a square, 6 is not a square, 8 is not a square, 10 is not a square, 12 is not a square, and 14 is not a square, just for starters. Only a very small proportion of even numbers are squares: 4, 16, 36, 64, 100, 144, 196, 256, 324, 400....
The factors of all numbers can be written in pairs. With square numbers, one of those pairs is the same number twice. When listed singly, square numbers have an odd number of factors. All others are even.
No.
No. Perfect square numbers have an odd number of factors.
All non-even numbers are odd. All squares of even numbers are even. All squares of odd numbers are odd. Therefore all numbers which are odd and not the square of an odd number are the solution set. It contains: {3, 5, 7, 11, 13, 15, 17, 19, 21, 23, 27, 29, 31, ...}
No, all perfect square numbers are not even numbers. Eg. the square of 3 is 9. (32=9) To generalize the proof: If p is odd then p=2n+1 and p2=(2n+1)2=4n2+4n+1=2(2n2+2n)+1 So odd numbers have odd square
All numbers, even imaginary or complex ones, have squares.
All nonzero numbers have factors. Some factors are even numbers, some factors are odd numbers.
False. eg 3 * 3 = 9 --> an odd number.
Square numbers have an odd number of factors. All others are even.