Which numerals can appear as the last digit in two consecutive Fibonacci numbers?

Question

MathMan TG, I like your observation.

Answer 1

It occurs every 15 numbers, another Fibonacci wonder!

1: 1 and 1

16: 987 and 1597

31: 1346269 and 2178309

46: 1836311903 and 2971215073

61: 2504730781961 and 4052739537881

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Answer 2

The first number of the sequence is 0, the second number is 1, and each subsequent number is equal to the sum of the previous two numbers of the sequence itself, yielding the sequence 0, 1, 1, 2, 3, 5, 8, etc.

Every 3rd number of the sequence is even and more generally, every kth number of the sequence is a multiple of Fk.

Since the sequence extends out to infinity, it would be logical to conclude that any numeral (0 through 9) could appear as the last digit in two consecutive Fibonacci numbers.

Answer 3

at the beginning of the series you have 1,1 so that's a start.

except for that, you can see that the last digits are of the recurring parities: odd,odd,even,...

the explanation is such:

odd+odd=even

odd+even=odd

even+odd=odd

odd+odd+...

Therefore, only odd digits can appear as consecutive last.

Additionally for 5 to appear twice as the last digit it would mean that the digit before was 0, and the digit before was 5, and the digit before was also 5, so you get the following recurring sequence as the only way to get consecutive 5:

5,5,0,5,5,...

since we do not begin the fibonacci series with this sequence,  we will never get to it, and therefore the possible digits are:

1,3,7, and 9

Answer 4

56

If not 85

If not...I overcurriculated things wrong. O.0

Answer 5

1, 7, 9, 3 in that order for the first 60 numbers.

Answer 6

1,3,7,9