To find the number of consecutive zeros at the end of the sum \((P + Q)\), we must analyze the prime factorization of each expression \(P\) and \(Q\).

Step 1: Prime Factorization

Firstly, let's express \(P = 5^5 \times 15^{15} \times 25^{25} \times 35^{35}\) in terms of its prime factors:

• \(5^5\) is already a power of 5.
• \(15 = 3 \times 5\), thus \(15^{15} = (3 \times 5)^{15} = 3^{15} \times 5^{15}\).
• \(25 = 5^2\), so \(25^{25} = (5^2)^{25} = 5^{50}\).
• \(35 = 5 \times 7\), therefore \(35^{35} = (5 \times 7)^{35} = 5^{35} \times 7^{35}\).

Combining these, the prime factorization of \(P\) is:

\(P = 5^{5 + 15 + 50 + 35} \times 3^{15} \times 7^{35}\)

\(P = 5^{105} \times 3^{15} \times 7^{35}\)

Step 2: Prime Factorization

Now, let's express \(Q = 10^{10} \times 20^{20} \times 30^{30} \times 40^{40}\):

• \(10 = 2 \times 5\), thus \(10^{10} = (2 \times 5)^{10} = 2^{10} \times 5^{10}\).
• \(20 = 2^2 \times 5\), hence \(20^{20} = (2^2 \times 5)^{20} = 2^{40} \times 5^{20}\).
• \(30 = 3 \times 2 \times 5\), so \(30^{30} = (3 \times 2 \times 5)^{30} = 3^{30} \times 2^{30} \times 5^{30}\).
• \(40 = 2^3 \times 5\), therefore \(40^{40} = (2^3 \times 5)^{40} = 2^{120} \times 5^{40}\).

Simplifying, we get:

\(Q = 2^{10 + 40 + 30 + 120} \times 3^{30} \times 5^{10 + 20 + 30 + 40}\)

\(Q = 2^{200} \times 3^{30} \times 5^{100}\)

Step 3: Analyzing the Sum for Consecutive Zeros

To determine the trailing zeros in the sum \((P + Q)\), we find the minimum power of 10, which is expressed as \(10^n = 2^n \times 5^n\). This needs both factors 2 and 5 in equal number.

In \(P\), there are no 2's, and in \(Q\), there are:

• \(200\) powers of 2.
• \(100\) powers of 5.

The minimum of powers of 2 or 5 in \(Q\) is 100, suggesting that \(Q\) can have up to 100 trailing zeros. However, note that \(P\) contributes no factors of 2, so it cannot contribute to the trailing zeros when added with \(Q\).

Thus, since \(P + Q\) lacks matching powers of 2 for those of 5, the number of zeros at the end of \((P + Q)\) is actually Zero.