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): two-thirds 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 second-last 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 third-last 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 2011-10-19 15:05:42 |
re(3): two-thirds of a solution.
