Question 1127112
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The table below represents the values of  {{{5^n}}} in the second column,  n = 1, 2, 3, 4, . . . , 20.


The third column represents  the last 4 digits of the number  {{{5^n}}}.

1	5	           5
2	25	          25
3	125	         125
4	625	        0625
5	3125	        3125
6	15625	        5625
7	78125	        8125
8	390625	        0625
9	1953125	        3125
10	9765625	        5625
11	48828125	8125
12	244140625	0625
13	1220703125	3125
14	6103515625	5625
15	30517578125	8125
16	152587890625	0625
17	762939453125	3125
18	3814697265625	5625
19	19073486328125	8125
20	95367431640625	0625


An experimental fact is that starting from n = 4, the last 4 digit numbers form a periodical (cyclical) sequence 
with the period (the cycle length) of 4:

    0625,  3125,  5625,  8125.


It can be proved mathematically.  I will not prove it here - see, for example, the lesson  
    <A HREF=https://www.algebra.com/algebra/homework/divisibility/lessons/Find-the-last-three-digits-of-these-numbers.lesson>Find the last three digits of these numbers</A>  in this site.


{{{5^678}}}  is 678-th term of this long sequence, and it is equal to the  (678-3) mod 4  term of the basic cycle.


Since  (678-3) mod 4 = 3,  it means that the 678-th term of the long sequence  {{{5^n}}}  is equal 
to the 3-rd term of the basic cycle, i.e.  5625.


<U>Answer</U>.  {{{5^678}}}  has the last four digits  5625.  
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