N is the product of all the prime numbers less than 100.
Determine the last three digits of N.
*** For an extra challenge, solve this puzzle without the aid of a computer program.
(In reply to
re(2): twothirds of a solution. by Jer)
Of course you are correct. I shall give this another go.
Taking the last two digits from each of your power equations:
1^5 = 01
3^7 = ...87
7^6 = ...49
9^5 = ...49
49*49 = 2401 (easy as = 50*48 + 1)
Multiplying 87 by the last two digits 01 = 87. (The 7 here is the secondlast digit you determined earlier.)
In addition to this '8' contribution is the sum of the 'tens' in the primes multiplied by the 8.
This turns out to be 94*8 = ...2.
As I said before, multiplication of the 'tens' ends in a 0 so there is 0 contribution from this source.
8+2+0 = 10 so the thirdlast digit is 0.
Making the solution 070 when combined with earlier.
Alternatively, I'm talking nonsense. Apologies if this is the case.

Posted by Matt
on 20111019 15:05:42 