## Consider the numbers $15$, $16$ and $18$:

$15=3\times 5$ and $3+5=8$.

$16 = 2\times 2\times 2\times 2$ and $2+2+2+2=8$.

$18 = 2\times 3\times 3$ and $2+3+3=8$.

$15$, $16$ and $18$ are the only numbers that have $8$ as sum of the prime factors (counted with multiplicity).

We define $S(k)$ to be the sum of all numbers $n$ where the sum of the prime factors (with multiplicity) of $n$ is $k$.

Hence $S(8) = 15+16+18 = 49$.

Other examples: $S(1) = 0$, $S(2) = 2$, $S(3) = 3$, $S(5) = 5 + 6 = 11$.

The Fibonacci sequence is $F_1 = 1$, $F_2 = 1$, $F_3 = 2$, $F_4 = 3$, $F_5 = 5$, ….

Find the last nine digits of $\displaystyle\sum_{k=2}^{24}S(F_k)$.

### This is a tricky question because it does not directly lend itself to a mathematical formula.

However, we can solve the question step by step:

First, we find the prime factor for the Fibonacci numbers from 2 to 24.

The Fibonacci sequence from $F_2$ to $F_{24}$ is:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368.

Now, look at the prime factors of each of these numbers:

1 (no prime factors)

2 (one factor: 2)

3 (one factor: 3)

5 (one factor: 1)

8 (four factors: 2, 2, 2, 2)

13 (one factor: 13)

21 (two factors: 3, 7)

34 (two factors: 2, 17)

55 (two factors: 5, 11)

89 (one factor: 89)

144 (seven factors: 2, 2, 2, 2, 3, 3)

233 (one factor: 233)

377 (two factors: 13, 29)

610 (four factors: 2, 5, 61)

987 (two factors: 3, 7, 47)

1597 (one factor: 1597)

2584 (six factors: 2, 2, 2, 2, 7, 23)

4181 (two factors: 37, 113)

6765 (three factors: 3, 5, 5, 11, 41)

10946 (four factors: 2, 13, 13, 41)

17711 (two factors: 89, 199)

28657 (one factor: 28657)

46368 (eight factors: 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 7)

Now, let’s find the sum for every Fibonacci number:

S(1)=0 (because there are no prime factors)

S(2)=2

S(3)=3

S(5)=11

S(8)=49

S(13)=N/A (there are no numbers with a sum of prime factors equalling 13)

S(21)=N/A

S(34)=N/A

S(55)=N/A

S(89)=N/A

S(144)=N/A

S(233)=N/A

S(377)=N/A

S(610)=N/A

S(987)=N/A

S(1597)=N/A

S(2584)=N/A

S(4181)=N/A

S(6765)=N/A

S(10946)=N/A

S(17711)=N/A

S(28657)=N/A

S(46368)=N/A

Finally, find the sum of all these S values:

Sum of S values = 0 + 2 + 3 + 11 + 49 = 65

Therefore, the last nine digits of this sum will be the nine digits of the number itself: 000000065.

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