There are a total of 5 positive three-digit perfect cubes that are even. To find this, we first determine the range of three-digit perfect cubes, which is from 46 to 96. Then, we identify the even perfect cubes within this range, which are 64, 216, 512, 729, and 1000.
The prime factorisation of 248832 is 2¹⁰ × 3⁵ Every perfect square number has a prime factorisation where each prime is to an even power. 2 has an even power 3 has an odd power, so need an extra power → multiple 248832 by 3 which gets (2⁵ × 3³)²
There are 1,963 such integers. Every factor of a number has a pair. The only time there will be an odd number of factors is if one factor is repeated, ie the number is a perfect square. So the question is really asking: how many positive integers less than 2008 (in the range 1 to 2007) are not perfect squares. √2007 = 44 and a bit (it lies between 44 and 45) So there are 44 integers less than (or equal to) 2007 which are perfect squares → 2007 - 44 = 1963 integers are not perfect squares in the range 1-2007 and have an even number of factors (divisors).
The number 16 is an integer (a counting or whole number). It is also a positive number, and it is even. Additionally, it is a perfect square, as it is the product of 4 times 4 (which is 4 squared).
The sum of the first 100 positive even numbers is 10,100.
Positive
8x8x8 6x6x6
When it is of the form x3 + y3 or x3 - y3. x or y can have coefficients that are perfect cubes, or even ratios of perfect cubes eg x3 + (8/27)y3.
No. Not even all hexahedrons are cubes.
All positive integers which are not perfect squares.
0 is even, but not positive.
The prime factorisation of 248832 is 2¹⁰ × 3⁵ Every perfect square number has a prime factorisation where each prime is to an even power. 2 has an even power 3 has an odd power, so need an extra power → multiple 248832 by 3 which gets (2⁵ × 3³)²
To find the smallest positive integer ( n ) such that ( 2n ) is a perfect square, ( 3n ) is a perfect cube, and ( 4n ) is a perfect fourth, we analyze the conditions for each case using prime factorization. Let ( n = 2^a \cdot 3^b \cdot k ), where ( k ) is coprime to 2 and 3. For ( 2n ) to be a perfect square, ( a+1 ) must be even and ( b ) must be even. For ( 3n ) to be a perfect cube, ( a ) must be divisible by 3 and ( b+1 ) must be divisible by 3. For ( 4n ) to be a perfect fourth, ( a+2 ) must be divisible by 4 and ( b ) must be divisible by 4. By solving these conditions simultaneously, the smallest ( n ) that meets all conditions is ( n = 108 ).
Cubing.
Yes they are always even, other wise it would not be a perfect sqare.
No. Consider 15: 1,3,5,15 Every positive whole number has an even number of factors, unless the number is a perfect square.
Any positive odd number.For example for p = 17 positive factors, consider n^(p-1) for any integer n.Since p is odd, p-1 is even and so n^(p-1) is a perfect square number.
Sure cubes can tessellate. It's actually very easy to do so with cubes, as they would all have straight sides of even lengths from any angle.