(a) find the largest four digit number which has a total of exactly 3 factors assuming that one and d number itself are factors. (b) find a and b greater than zero such that a^2 = b^3. (c) find x and y such that x^3 y^3 =1729
sounds like you want a perfect square. Any other kind of number will have at least 4 factors: 1,n,q, and n/q or be a prime, with only 2 factors.
So, what's the largest square n <= 9999?
√9999 = 99.99, so
99^2 = 9801 has just 3 factors: 1,99,9801
64 = 8^2 = 4^3
1729 = 1^3 + 12^3 = 9^3 + 10^3
It is the smallest number which can be expressed as the sum of two cubes in two different ways.
b)8^2=4^3 (c) 1729=12^3+1^3=10^3+9^3
(a) To find the largest four-digit number with exactly 3 factors, we need to find a number that is a perfect square. The reason for this is that a perfect square has an odd number of factors. Let's start by finding the largest perfect square less than 10000 (since we're looking for a four-digit number).
The largest perfect square less than 10000 is 9801, which is equal to 99 * 99. To check if it has exactly 3 factors, we need to see if it is a square of a prime number.
Since 99 is not a prime number (it can be divided by 3), we need to find the next largest perfect square. The perfect square of the next prime number, 101, is 10201, which is larger than 9999 (the largest four-digit number).
Therefore, there is no four-digit number that has exactly 3 factors.
(b) To find values of a and b greater than zero such that a^2 = b^3, we need to find a perfect square on one side that can be expressed as a perfect cube on the other side.
Let's take a=4 as an example:
4^2 = 16
4^3 = 64
In this example, a=4 and b=2, as 4^2=2^3.
There are other possible combinations, such as a=1 and b=1, as 1^2=1^3.
Therefore, possible values for a and b greater than zero such that a^2=b^3 include:
- a=1, b=1
- a=4, b=2
(c) To find values of x and y such that x^3 - y^3 = 1729, we can use the concept of Fermat's Last Theorem. This theorem states that there are no three positive integers a, b, and c such that a^n + b^n = c^n for any integer value of n greater than 2.
However, we can still find values that satisfy the equation x^3 - y^3 = 1729 by manually trying different values:
1^3 - 12^3 = -1727
-1^3 - (-12)^3 = -1727
9^3 - 10^3 = 170
19^3 - 18^3 = 6859
-19^3 - (-18)^3 = -6859
25^3 - 24^3 = 2903
By testing these values, we find that the equation x^3 - y^3 = 1729 is satisfied when x = 12 and y = 1.
Therefore, one possible solution for x and y in the equation x^3 - y^3 = 1729 is x = 12 and y = 1.