Problem 347
Largest integer divisible by two primes
The largest integer ≤ 100 that is only divisible by both the primes 2 and 3 is 96, as 96=32*3=25*3. For two distinct primes p and q let M(p,q,N) be the largest positive integer ≤N only divisible by both p and q and M(p,q,N)=0 if such a positive integer does not exist.
E.g. M(2,3,100)=96.
M(3,5,100)=75 and not 90 because 90 is divisible by 2 ,3 and 5.
Also M(2,73,100)=0 because there does not exist a positive integer ≤ 100 that is divisible by both 2 and 73.
Let S(N) be the sum of all distinct M(p,q,N). S(100)=2262.
Find S(10 000 000).
能被两个素数整除的最大整数
在≤ 100的整数中，能整除的素数只有2和3的最大整数是96，因为96=32*3=25*3。对于两个不同的素数p和q，在≤N的正整数中，能整除的素数只有p和q的最大整数记为M(p,q,N)；如果这样的正整数不存在，则记M(p,q,N)=0。
例如，M(2,3,100)=96。
M(3,5,100)=75而非90，因为90能被2、3、5整除。
此外M(2,73,100)=0，因为不存在≤ 100的正整数使得能整除的素数只有2和73。
记S(N)为所有不同的M(p,q,N)之和。S(100)=2262。
求S(10 000 000)。