The Unknown Digits (Posted on 2006-04-18)

	Submitted by K Sengupta
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Solution:	(Hide)
The required last six digits are 353607. EXPLANATION: We know that 7^4 = 2401. Accordingly, 7^5000 = 2401^1250 =(1+2400)^1250 = 1 + 3*(10^6) + 44964*(10^8) +.........+(2400^1250) ( Expanding by Binomial Theorm) = 1 + 1,000,000*U, where U is a positive integer. = 1 (Mod 1,000,000); so that: 7^(5,000*M + N) = 7^N ( Mod 1,000,000), where M and N are two positive integers, whether distinct or otherwise...............(#) Now, 3^3000 = 9^1500 = (10-1)^1500 = 10^1500 - 1500*(10^1499)+ ...........-1500*10 + 1 = 5,000*R + 1, where R is a positive integer Hence, 3^3002 = 5,000*M +9, where M = 9*R Consequently, 7^(3^3002) = 7^(5,000*M + 9) = 7^9 ( Mod 1,000,000)( by (# )) = 353,607( Mod 1,000,000) Hence, the required last six digits are 353607.

	Subject	Author	Date
	Binomial Theorem	K Sengupta	2006-05-03 12:27:14
	solution	Charlie	2006-04-18 14:01:35