Imagine you have a number which is not an integer power of 10. If you take
the logarithm to base 10 of your number, how is this linked to the number of
digits in your number?
For example, log10 54  1.73..., log10 412  2.61..., log10 86591  4.93...,
You can see that the integer part of the answer is one less than the number of
digits in the original number.
Applying this to the new large prime:
1012,978,188  2 p  1  1012,978,189
Taking logs:
log10 1012,978,188   log10  2 p  1  log10 1012,978,189 
Since the difference between 2 p  1 and 2 p is negligible:
p log10 2  12,978,188.5
p
This gives
12,978,188.5
 43112 610
log10 2
But p is prime. So the candidates we might check are
p  43112 607, 43112 609, 43112 611, 43112 613
But the sums of the digits of both 43112 607 and 43112 613 are multiples of 3
and so the numbers themselves must be multiples of 3 and therefore not
prime.
This leaves two candidates:
log10  243112 609  1  43112 609 log10 2  12,978,188.5
log10  243112 611  1  43112 611log10 2  12,978,189.1
It follows that p  43112 609 .
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© MEI 2008