Let $q$ be a prime and $A \ge B >0$ be two integers with the following properties:
$A$ and $B$ have no prime factor in common, that is $\gcd(A,B)=1$.
The product $AB$ is divisible by every prime less than q.
It can be shown that, given these conditions, any sum $A+B
The problem as it was posed is not a standard form math problem and thus one cannot simply offer a solution as you would a textbook problem.
This is essentially a programming problem or exercise in computational number theory, hence requires the deployment of code to determine the required answer. The standard way of approach to solving this would be through a language like Python, R, or even C.
Below is a simplified Python code to compute S(n) given the details of the problem. The code uses a basic algorithm to compute primes, then it computes the values of V(p), and lastly, it sums them all up. Take note that the primes algorithm is not the most efficient, but it works given the relative small size of 3800.
“`
import math
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def isPrime(n):
if n < 2:
return False
for i in range(2, math.isqrt(n) + 1):
if n % i == 0:
return False
return True
def smallestA(p):
q = 2
while q * q <= p:
if p % q != 0 and isPrime(q):
for A in range(q * q, p):
B1 = A - p
B2 = p - A
if gcd(A, B1) == 1 or gcd(A, B2) == 1 or gcd(B1, B2) == 1:
return min(A, B1, B2)
q += 1
return p - 1
def S(n):
sum = 0
for i in range(2, n):
if isPrime(i):
sum += smallestA(i)
return sum
print(S(3800))
```
Note: This can take a considerable amount of time to run for larger numbers, due to the inefficient prime-checking algorithm. For testing, you may want to use smaller numbers.
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The problem as it was posed is not a standard form math problem and thus one cannot simply offer a solution as you would a textbook problem.
This is essentially a programming problem or exercise in computational number theory, hence requires the deployment of code to determine the required answer. The standard way of approach to solving this would be through a language like Python, R, or even C.
Below is a simplified Python code to compute S(n) given the details of the problem. The code uses a basic algorithm to compute primes, then it computes the values of V(p), and lastly, it sums them all up. Take note that the primes algorithm is not the most efficient, but it works given the relative small size of 3800.
“`
import math
def gcd(a, b):
while b != 0:
a, b = b, a % b
return a
def isPrime(n):
if n < 2:
return False
for i in range(2, math.isqrt(n) + 1):
if n % i == 0:
return False
return True
def smallestA(p):
q = 2
while q * q <= p:
if p % q != 0 and isPrime(q):
for A in range(q * q, p):
B1 = A - p
B2 = p - A
if gcd(A, B1) == 1 or gcd(A, B2) == 1 or gcd(B1, B2) == 1:
return min(A, B1, B2)
q += 1
return p - 1
def S(n):
sum = 0
for i in range(2, n):
if isPrime(i):
sum += smallestA(i)
return sum
print(S(3800))
```
Note: This can take a considerable amount of time to run for larger numbers, due to the inefficient prime-checking algorithm. For testing, you may want to use smaller numbers.
More Answers:
Super Pandigital NumbersIdempotent Matrices
Unfair Race