The last 3 digits of 49^a and 39^c and the last 2 digits od 3993^b have a repeating cycle every 100th term.
303 modulo 100 is 3; 49^3 = 117649
202 modulo 100 is 2; 3993^2 = 15944049
606 modulo 100 is 6; 39^6 = 3518743761.
Multiplying the the last three digits of each of the three exponential products gives: 24200561 (= 649*049*761) Thus, revealing the last three digits of the product of the original three exponential expressions.
Edited on January 16, 2013, 12:47 pm
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Posted by Dej Mar
on 2013-01-16 12:22:30 |
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