Question

Suppose that in standard factored form a = pe1 1 pe2 2 · · · pek k , where k is a positive integer; p1, p2, . . . , pk are prime numbers; and e1, e2, . . . , ek are positive integers. a. What is the standard factored form for a2? b. Find the least positive integer n such that 25 ·3·52 ·73 ·n is a perfect square. Write the resulting product as a perfect square.

Answer 1

To find the standard factored form for a^2, square each prime factor and multiply the resulting powers. For the given expression, the least positive integer n that makes it a perfect square is 7, resulting in 275625 as a perfect square.

Step-by-step explanation:

To find the standard factored form for a^2, we need to square each prime factor and multiply the resulting powers. So, if a = p1^e1 * p2^e2 * ... * pk^ek, then a^2 = (p1^e1 * p2^e2 * ... * pk^ek)^2 = p1^(2e1) * p2^(2e2) * ... * pk^(2ek).

To find the least positive integer n such that 25 * 3 * 5^2 * 7^3 * n is a perfect square, we need to factorize each number and find the missing prime factors. Here's how:

25 = 5^2.

3 is already a prime number.

5^2 = 5^2.

7^3 = 7^2 * 7 = 49 * 7.

So, n = 7 to make the expression a perfect square.

The resulting product as a perfect square is 5^2 * 3 * 7^2 = 525^2 = 275625.

Answer 2

Answer: a . standard factored form for a2= (pe11pe22...pekk)2

b. n = 810540900

Step-by-step explanation:

25 .3 .52 .73=5square .3. (2square.13).73.n,

Making the power of every prime numbers even, we get

n = 3. 13. 73

The resulting number will be

2square. 3square. 5square. 13square.73square=2×2×3×3×5×5×13×13×73×73 = 28470square = 810540900