We construct a sequence of numbers as following :
The first number is equal to 2, and so is the second number. Every following number is the product of the two previous ones. So the first few numbers of the sequence are :
2, 2, 4, 8, 32, .......
What is the final digit of the 2007th number in this sequence?

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Solution

We see that the sequence of final digits is equal to $2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 6, 2, . . . . . . .,$
and that we have a repeating part of length 6, namely $2, 2, 4, 8, 2, 6$ . Since $2007 = 334 \times 6 + 3$ , the final digit of the $2007$ th number must be a $4$ .

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We see that the sequence of final digits is equal to 2,2,4,8,2,6,2,2,4,8,6,2,.......,and that we have a repeating part of length 6, namely 2,2,4,8,2,6. Since 2007=334×6+3, the final digit of the 2007th number must be a 4.

We see that the sequence of final digits is equal to $2, 2, 4, 8, 2, 6, 2, 2, 4, 8, 6, 2, . . . . . . .,$
and that we have a repeating part of length 6, namely $2, 2, 4, 8, 2, 6$ . Since $2007 = 334 \times 6 + 3$ , the final digit of the $2007$ th number must be a $4$ .