Pizza-No!

Examining the last digits of the numbers in the Fibonacci Sequence, we can see very clearly that they are periodic with period 60 (the last digits repeat themselves every 60 numbers). Hence, we can immediately conclude that, since $2324 = (38 \times 60) + 44$ , there are $(38 \times 8) + 5 = \boxed {309}$ Fibonacci numbers (of the first $2324$ ) which have $3$ as their last digit.

In fact, we can easily establish a general formula: every $60$ consecutive Fibonacci numbers will have $8$ numbers ending in a certain digit $x | x \cong 1 (mod 2)$ and $4$ numbers ending in a digit $y | y \cong 0 (mod 2)$ , where $0 \leq x,y \leq 9$ .

Python one-liner that is pretty much unreadable:

>>> [reduce(lambda x, _: (x[1] % 10, (x[0]+x[1]) % 10), range(i), (1,1))[0] for i in range(2324)].count(3)
309

A more mathematical approach :

Observe that the last digits of the Fibonacci sequence repeats. This is easy to notice: There are only $100$ (read: a finite number) pairs of last digits of $F_{n-1}, F_{n-2}$ , and the sequence of the last digits is uniquely determined from each pair. So once the pair repeats, we have found the cycle.

By bruteforcing from $1,1$ , we see that the sequence of last digits repeat with period $60$ . We can thus mark the terms that have a last digit of $3$ . In this case, there are $8$ such terms in a cycle, and $5$ terms appear up to the $44$ th term.

Since $2324 = 38 \cdot 60 + 44$ , we know that there are $38 \cdot 8$ terms ending with a $3$ in the first $38 \cdot 60$ terms, plus $5$ more for the last $44$ terms, thus we have $38 \cdot 8 + 5 = \boxed{309}$ terms ending with $3$ among the first $2324$ Fibonacci numbers.

We start listing the last digits of the first several Fibonacci numbers (this is not as long as it sounds):

$\begin{aligned}1, 1, 2, (3), 5, 8, (3), 1, 4, 5, 9, 4, (3), 7, 0, 7, 7, 4, 1, 5, 6, 1, 7, 8, 5, (3), 8, 1, 9, 0, 9, 9, 8, \\7, 5, 2, 7, 9, 6, 5, 1, 6, 7, (3), 0, (3), (3), 6, 9, 5, 4, 9, (3), 2, 5, 7, 2, 9, 1, 0, \left(1, 1, 2, 3, ....\right)\end{aligned}$

We notice that the pattern repeats every 60 digits. For every 60 digits, there are eight 3s. For every 60 Fibonacci numbers, there are 8 Fibonacci numbers ending with three. From here, the rest of the problem is simple.

$2324 \div 60 = 38 R 44$

There are 38 complete sets of 8 3s and 44 extra digits. We see from the above list that the first 44 digits of the sequence contain five 3s.

Hence, the total number of Fibonacci numbers ending with 3 is given by:

$38 \times 8 + 5 = \boxed{309}$

This problem can be solved with the following Mathematica command: Total[Array[If[Mod[Fibonacci[#1], 10] == 3, 1, 0] &, {2324}]] .

That said, this problem makes us think about the period of the Fibonacci numbers modulo $m$ -- a fascinating, well-studied number-theoretic object called the Pisano period . In the case $m=10$ , the Pisano period is $60$ ; an alternate solution thus proceeds from the observation that $2324 = 38\cdot 60+44$ .

$\mathbb{QED}$

write down the last digits of the first few fibonacci numbers. we can be sure that a repeating sequence will be found after 100 numbers because max no. of distinct ordered pairs (a,b) where ab are single digit =100.
the sequence repeats after 60th term (61st therm =1,62nd=1).
counting 3s firs 60have 8.
thus first 2280 have 8*38=304.
first 44 consist of 5 3s thus 2281to2324 also have 5 3s.
answer =304+5=309

Why waste time and Ink of your pen when you've got Python. I simply wrote a code and Saved it as Fib2324.py and Ran it. The code is given below:

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a, b = 1, 1
total = 0
for n in range(2324):
    if (a % 10)-3 == 0:
        total += 1
    a, b = b, a+b  # the real formula for Fibonacci sequence
print total

And thus you get your answer as output = 309

In Python, a more readable approach than the one-liner. First, iterate through all the Fibonacci numbers and check if the last digit ends with 3. Here is the code:

a = 0 
b = 1
n = 0
print a
print b
for i in range(1,2325):
    c = a + b
    a = b 
    b = c
if(c % 10 == 3):
        n +=1

print n

The result is:

0
1
309

Therefore, answer is 309!

Assuming g(0) = 0. Fibonacci auxiliar recurrence is cyclic each 60 times, in other hands, periodic each 60 times. I will use a auxiliar recurrence function for look patterns. For example: g(n) = f(n) mod 10,

g(7) = 13 mod 10 = 3. g(1) = 1, g(2) = 1, g(3) = 2 .... g(10) = 5 ... g(59) = . Finish cycle at g(59). g(60) = g(0) = 0 g(60) = g(0)

So, generalizing: g(n) = g(60 * 1+k) = g(60 * 2+k) = . . . = g(60*p+k).

k is cycle position of g(n) and p number of cycles.

The number of 3's in g(n) is 8 per cycle.

2324 mod 60 = 44. Then, g(2324) = g(60 38+44) -> 38 8 + number of 3's between 0 and 44. -> 38*8 + 5 = 309

Put this in python

 list=[]

 k=eval(input("fibonacci"))

 d=1

 a=1

 b=1

 c=a+b

 for i in range(k-2)

     c=a+b

     a=b

     b=c

     if c%10==3:

         list.append(d)

 e=list.count(d)

 print(e)

last digit of fib(n) = last digit of fib(n-1) + last digit (n-2) MOD 10 keep tracing the last digits and you'll find fib(61) =1 and fib(62) = 1 , this means a cycle will be repeated over and over again in the first 60 Fibonacci numbers there are 8 numbers has their last digit = 3, so in the first 2340 there will be 8*39 = 312 but we need to remove all solutions from fib(2324) till fib(2340) they are equivalent to solutions from fib(44) to fib(60) which equal 3 so answer = 312 - 3 = 309

Note that $F_{58} \equiv 9\mod{10}$ , $F_{59} \equiv 1\mod{10}$ . Thus, modulo 10, the Fibonacci numbers are periodic with period 60. In the first 60 Fibonacci numbers, there are 8 which are congruent to 3 modulo 10. There are 38 such sets of 60 in the first 2324 Fibonacci numbers. Because $2324 \equiv 44 \mod{60}$ , we must count the number of fibonacci numbers congruent to 3 modulo 10 in the first 44 (5). Hence, there are $38 \cdot 8 + 5 = \boxed{309}$ Fibonacci numbers congruent to 3 modulo 10 in the first 2324.