(a) find d largest four digit number which has a total of exactly 3 factors. Assuming that one and the 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
(a) To find the largest four-digit number with exactly 3 factors, we need to consider that 1 and the number itself are factors. A number with exactly 3 factors is a prime number squared, since a prime number only has two factors (1 and itself). So, we can look for the largest prime number less than 100 that can be squared to give us a four-digit number.
Starting with the largest prime less than 100, which is 97, we can square it: 97^2 = 9409. Therefore, the largest four-digit number with exactly 3 factors is 9409.
(b) To find a and b that satisfy the equation a^2 = b^3, we need to find perfect squares that are also perfect cubes. One way to approach this problem is to look for numbers that have the same prime factors in their prime factorization, but with different exponents.
One such pair is 8 and 4. 8 can be written as 2^3, and 4 can be written as 2^2. So, 8^2 = 64 and 4^3 = 64. Hence, a = 8 and b = 4 is a solution to the equation a^2 = b^3.
(c) To find x and y such that x^3 + y^3 = 1729, we can use a technique called "brute force" or "guess and check". We can start by checking for small values of x and y.
Let's start with x = 1 and y = 1:
1^3 + 1^3 = 1 + 1 = 2, which is not equal to 1729.
Next, let's try x = 1 and y = 2:
1^3 + 2^3 = 1 + 8 = 9, which is also not equal to 1729.
Now, let's continue increasing y and checking:
x = 1 and y = 3:
1^3 + 3^3 = 1 + 27 = 28, still not equal to 1729.
x = 1 and y = 4:
1^3 + 4^3 = 1 + 64 = 65, still not equal to 1729.
Finally, let's increase x and reset y to 1:
x = 2 and y = 1:
2^3 + 1^3 = 8 + 1 = 9, still not equal to 1729.
x = 2 and y = 2:
2^3 + 2^3 = 8 + 8 = 16, still not equal to 1729.
After numerous checks, it becomes evident that there are no integer solutions for x and y that satisfy x^3 + y^3 = 1729.