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Problem 773


Problem 773


Ruff Numbers

Let Sk be the set containing 2 and 5 and the first k primes that end in 7. For example, S3={2,5,7,17,37}.

Define a k-Ruff number to be one that is not divisible by any element in Sk.

If Nk is the product of the numbers in Sk then define F(k) to be the sum of all k-Ruff numbers less than Nk that have last digit 7. You are given F(3)=76101452.

Find F(97), give your answer modulo 1 000 000 007.


仿粗糙数

Sk25和前k个以7结尾的素数所构成的集合。例如,S3={2,5,7,17,37}

如果一个数不能被Sk中的任意元素整除,则称之为k-仿粗糙数

NkSk中的元素之积,则定义F(k)为所有小于Nk且以7结尾的k-仿粗糙数之和。已知F(3)=76101452

F(97),并将你的答案对1 000 000 007取余。


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