, 2014

Solution:

Answer: 439 Combining Legendre's Formula and the standard prime approximations, the answer is
$$\prod_{p}\left(1+\frac{2014-s_{p}(2014)}{p-1}\right)$$
where $s_{p}(n)$ denotes the sum of the base $p$-digits of $n$.

Estimate $\ln 1000 \approx 8$, and $\ln 2014 \approx 9$. Using the Prime Number Theorem or otherwise, one might estimate about $150$ primes less than $1007$ and $100$ primes between $1008$ and $2014$. Each prime between $1008$ and $2014$ contributes exactly $\ln 2$. For the other $150$ primes we estimate $\ln 2014 / p$ as their contribution, which gives $\sum_{p<1000}(\ln 2014-\ln p)$. Estimating the average $\ln p$ for $p<1000$ to be $\ln 1000-1 \approx 7$ (i.e. an average prime less than $1000$ might be around $1000 / e$ ), this becomes $150 \cdot 2=300$. So these wildly vague estimates give $300+150 \ln 2 \approx 400$, which is not far from the actual answer.

The following program in Common Lisp then gives the precise answer of $438.50943$.

```
;;;; First, generate a list of all the primes
(defconstant +MAXP+ 2500)
(defun is-prime (p)
(loop for k from 2 to (isqrt p) never (zerop (mod p k))))
(defparameter primes (loop for p from 2 to +MAXP+
if (is-prime p) collect p))
;;;; Define NT functions
```
```
(defconstant +MAXDIGITS+ 15)
(defun base-p-digit (p i n)
(mod (truncate n (expt p i)) p))
(defun sum-base-p-digit (p n)
(loop for i from 0 to +MAXDIGITS+ sum (base-p-digit p i n)))
(defun vp-n-factorial (p n)
(/ (- n (sum-base-p-digit p n)) (1- p)))
;;;; Compute product
(princ (loop for p in primes
sum (log (1+ (vp-n-factorial p 2014)))))
```