MHB Factorisation Related Question

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Factoring large numbers, such as a 1050-digit number, poses significant computational challenges, particularly with traditional methods. The discussion highlights that multiplying primes sequentially may be computationally easier than direct factorization. A reference to the time taken to factor a 232-digit number illustrates the increasing difficulty with larger digits. The consensus is that as numbers grow, factorization becomes increasingly problematic for current algorithms. The initial question about the comparative ease of the two calculations remains central to the discussion.
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A slightly odd layman's question about factoring large numbers and comparing two calculations.

Call N a number with 1050 digits.

1) Factorise N
2) Multiply the primes sequentially from 2 onwards until the product is as close as possible to N.

Would calculation 2 be significantly easier computationally than calculation 1?

How many digits would a number has to have before factorisation becomes a problem for our current methods?

Thanks
 
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Thanks Greg.

"... to factor a 232-digit number (RSA-768) utilizing hundreds of machines took two years and the researchers estimated that a 1024-bit RSA modulus would take about a thousand times as long.[1]"

This helps with the second question but not the first, which is my main question.
 
Doh! On reflection the answer to the second question is obvious. Consider it answered.
 
Thread 'Erroneously  finding discrepancy in transpose rule'
Obviously, there is something elementary I am missing here. To form the transpose of a matrix, one exchanges rows and columns, so the transpose of a scalar, considered as (or isomorphic to) a one-entry matrix, should stay the same, including if the scalar is a complex number. On the other hand, in the isomorphism between the complex plane and the real plane, a complex number a+bi corresponds to a matrix in the real plane; taking the transpose we get which then corresponds to a-bi...

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