Taking things series-ly
Consider a recurrence relation with a i + 1 = a i 5 + 1 and initial term a 1 = 0 . What is the last digit of a 1 2 ?
The answer is 1.
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1 solution
Moderator note:
Can you prove the very first line?
Sure. 0 5 = 0, 1 5 = 1, 2 5 = 32, 3 5 = 243, 4 5 = 1024, 5 5 = 3125, any power of 6 ends in 6 since 6 * 6 is 36, 7 5 = 16807, 8 5 = 32768, 9 5 ends in 9 because 9 4 = 6561, and for any number that's 2 or more digits, we can write it as (10a + b) where b is the units digit, and ( 1 0 a + b ) 5 has the same units digit as b 5 because everything else when you expand it out has at least 1 factor of 10a, rendering its units digit 0.
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why last digit of n 5 always same with n? Is that just a coincidence or it have a theory behind it?
For positive integers, the last digit of n 5 is the same as the last digit of n .
Since the problem only concerns last digits, we can substitute a i for ( a i ) 5 in the iteration.
With the substitution, a 1 2 = 11, which has a last digit of 1.