We know that the highest common factor (HCF) of N and 500 is 2^2 x 5^2. So we can write:
N = (2^2 x 5^2) x m (where m is a product of any remaining prime factors in N)
500 = (2^2 x 5^2) x n (where n is a product of any remaining prime factors in 500)
We can simplify these expressions by dividing both sides by 2^2 x 5^2:
N/(2^2 x 5^2) = m
500/(2^2 x 5^2) = n
So, we have:
N/(2^2 x 5^2) = 2^p x 5^q x 7^r/(2^2 x 5^2) = m
500/(2^2 x 5^2) = 2^2 x 5^3/(2^2 x 5^2) = n
Simplifying further:
N/(2^2 x 5^2) = 2^(p-2) x 5^(q-2) x 7^r = m
500/(2^2 x 5^2) = 5 x n
Now, we can compare the prime factors on both sides to find the values of p, q, and r.
Comparing the powers of 2:
For N: p - 2 = 2
=> p = 4
Comparing the powers of 5:
For N: q - 2 = 3
=> q = 5
Comparing the powers of 7:
For N: r = 0 (since there is no 7 in 500)
So we have:
N = 2^4 x 5^5
To find r, we can use the fact that the lowest common multiple (LCM) of N and 500 is 2^3 x 5^3 x 7. Since the LCM is the smallest number that is a multiple of both N and 500, it must contain all the prime factors of N and 500, raised to their highest powers:
LCM = 2^4 x 5^3 x 7^1
Comparing the powers of 7:
For N: r = 1
So we have:
N = 2^4 x 5^5 x 7^0 = 5000
Therefore, p = 4, q = 5, and r = 0, and the prime factorization of N is 2^4 x 5^5 x 7^0 = 5000.