Can you please help me with this problem I couldn't find a strategy to answer
the question. "Find two prime numbers that, if multiplied, would generate a 
400-digit number." Please help as soon as you can. Thank you


You couldn't be expected to answer this from scratch without a super-computer.
However, you can go online and see what large primes have been discovered.

You can go to this website at the University of Tennessee at Martin

http://primes.utm.edu/mersenne/index.html

and find a list of what are known as "Mersenne" primes. These are primes of the
form 2p-1.  We find that the largest one of these that has less than 400 digits
is when p = 1279, which has 386-digit,

so 21279-1 is a 386-digit prime. 

Now we know that if we multiply two positive integers together, the number
of digits in the product will either be the sum of the numbers of digits in
the two integers multiplied or one less than that sum. I'll just demonstrate
this fact with a 4 and a 5 digit number: 

The largest 4 digit number times the largest 5 digit number is

9999×99999 = 999890001 which has 9 digits, the sum of 4 and 5.

The smallest 4 digit number times the smallest 5 digit number is

1000×10000 = 10000000 which is 8 digits, one less than the sum
of 4 and 5.

And we know the product of any 4-digit number and any 5-digit number
will be between these, and thus will either be an 8 or 9-digit number.

Similarly the product of any 386-digit number and any 14-digit number
will either be a 399 or 400-digit number.

Also similarly, the product of any 386-digit number and any 15-digit number
will either be a 400 or 401-digit number.

Now since we have found a prime with 386 digits, we need either a
14 or 15 digit prime, but we don't know which. We can write this
inequality:

(21279)N < (21279-1)N < (21280)N

take the common logs (logs base 10) of 
all three sides. The three will be in the same order 
since the log is a strictly increasing function.

Furthermore the next higher integer to log(N) is
the number of digits in N

I'll demonstrate this with the smallest and largest
4-digit numbers:

log(1000) = 3 and log(9999) = 3.000056568

Both 1000 and 9999 are 4-digit numbers and the next
higher integer to both their logs is 4.

(21279)N < (21279-1)N < (21280)N

So take the log of all three sides: 

1279 log(2) + log(N) < log(PRODUCT) < 1280 log 2 + log(N)

385.0173645 + log(N) < log (PRODUCT) < 385.3183944 + log(N)

If we can guarantee the left side to be > 399 and the right side to be
less than 400 then we can be assured that the PRODUCT will have 400
digits

For the left side to be > 399

385.0173645 + log(N) > 399
              log(N) > 13.9826355
                  N  > 9.6 × 1013

For the left side to be < 400

385.0173645 + log(N) < 400
              log(N) < 14.9826355
                  N  < 9.6 × 1014

So any prime between 9.6 × 10 13 and 9.6 × 10^14 will do.

On this AT&T research website: 

http://akpublic.research.att.com/~njas/sequences/table?a=3617&fmt=4

there is a list of the smallest n-digit primes and we find the
smallest 15 digit prime is 100000000000031

That number is between 9.6 × 10 13 and 9.6 × 10^14.

So two primes, the product of which has 400 digits are

21279 - 1 and 100000000000031    

Edwin