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5

From the Prime number theorem, the prime numbers are asymptotically distributed according to $$ \pi(n) \approx \frac n{\ln n}\ , $$ where $\pi(n)$ is the number of primes less than or equal to $n$. This shows that for $n=2^{2^{100}}$ we'd have $$ \pi(n) \approx \frac {2^{2^{100}}}{\ln 2^{2^{100}}}=\frac {2^{2^{100}}}{2^{100}\ln 2} \approx 2^{(2^{100}-99)} $...


3

Since $15 = 1\cdot15 = 3 \cdot 5$, the only possibilities are $$(x_1 + x_2 + x_3, y_1 + y_2 + y_3 + y_4) = (1, 15), (15, 1), (3, 5), (5, 3)$$ Since the variables are all at least $1$, we can rule out the case where $x_1 + x_2 + x_3 = 1$ or where $y_1 + y_2 + y_3 + y_4 = 1, 3$ for the minimum values of each sum exceeds the respective assignments. This ...


3

There are $9\cdot10^4$ five-digit numbers, since there are $9$ choices for the first digit, and $10$ choices for the next $4$ digits. Of these numbers, $8\cdot 9^4$ do not contain $4$ as a digit. Therefore the number of $5$-digit numbers with at least one $4$ as a digit is equal to $$ 9\cdot 10^4-8\cdot 9^4=37512$$


1

$\frac{x}{ln(x)}$=number of primes until x (for large enough x, the error becomes almost nill) i think the rest is easy, btw search prime number theorem :)


1

The answer to your question is no. You can try to do the following exercise (taken from Serre, Local Fields, Ch. III, section 6): Define $A,B,K,L$ in the way you did above, and assume $B$ is 'completely decomposed', i.e., there are $n = [L:K]$ primes of $B$ above the prime $\mathfrak m$ of $A$. Then $B$ is of the form $A[x]$ (for some $x\in B$) if and only ...


1

The easiest is to just subtract all of the ones that don't: 1 digit numbers: 8 possibilities (1,2,3,5,6,7,8,9) 2 digit numbers: 8 possibilities for the first * 9 possibilities for the second (0,1,2,3,5,6,7,8,9) = 72 3 digit numbers: 8 possibilities for the first * 9 possibilities for the second (0,1,2,3,5,6,7,8,9) * 9 possibilities for the third (0,1,2,3,...


1

The beauty of this problem is that it can be studied with any level of mathematics. Solving it however, is a different matter. The fact that so many have tried and failed proves the following: If proof with elementary mathematics is possible, then it requires some exceptional feat of logic, reasoning, or intuition, which has eluded so many people before. ...



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