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Last three digits (Posted on 2013-01-16) Difficulty: 2 of 5
What are the last three digits of 49303 * 3993202 * 39606?

No Solution Yet Submitted by Danish Ahmed Khan    
Rating: 2.0000 (3 votes)

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Solution answer | Comment 2 of 3 |

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
  Posted by Dej Mar on 2013-01-16 12:22:30

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