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Last three digits (Posted on 2011-10-19) Difficulty: 3 of 5
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.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(3): two-thirds of a solution. | Comment 7 of 9 |
(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
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