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It is "true" DS. We have not to solve the problem, we have to say that the problem can be solved.

1. There are total of 100 prime numbers between 1 and p+1 The condition unambiguously says that p is 100th prime number. So, we can write out all 100 consecutive primes numbers and find p. sufficient.

2. There are total of p prime numbers between 1 and 3912. the same logic as in the fist condition. sufficient.

I even dare to say that p=3911. Why? Because I did not remember any DS, in which 2 conditions will be sufficient separately and lead to different values. _________________

I even dare to say that p=3911. Why? Because I did not remember any DS, in which 2 conditions will be sufficient separately and lead to different values.

No. 3911 is a 541th prime number..... _________________

I even dare to say that p=3911. Why? Because I did not remember any DS, in which 2 conditions will be sufficient separately and lead to different values.

No. 3911 is a 541th prime number..... :?

Hey,

The color one is exactly what I want to learn from you, guys. I have the same logic approach as your. I think if it is a PS, maybe we must seat and list all the prime number from 2 to p+1. That is imppossible for me. That is why I need your hand! Thanks

No. 3911 is a 541th prime number. I think you do not need to seat a time to count each number. Right? _________________

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