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9

Use the fact that $a^2-b^2 = (a+b)(a-b)$ Multiplying $(6-5)$ on your LHS, we obtain: $$\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})=(6-5)\prod\limits_{k=0}^5(5^{2^k}+6^{2^k})$$ $$=(6^2-5^2)\prod\limits_{k=1}^5(5^{2^k}+6^{2^k})=(6^4-5^4)\prod\limits_{k=2}^5(5^{2^k}+6^{2^k})=...$$ Iterating for all the terms in the products, you should get $x=y=2^6=64$, so $x-y=0$


7

Consider the category $\mathsf{C}$ with a single object $x$ and two morphisms: $\operatorname{id}_x$ and $f : x \to x$ such that $f \circ f = f$. Then $x$ is of course the product of all the objects in $\mathsf{C}$. But it's not an initial object, because there are two morphisms $x \to x$.


4

Hint: Assume $a, b \gt 0$. The product $a \times b$ will be minimized when both $a$ and $b$ are made as small as possible. Since the inequality $a^b \ge x$ implies $a \ge x^{1/b}$, the best we can do is to minimize $b \times x^{1/b}$. The result (via derivatives) gives an equation for $b$ and thus the lower bound on $a$. Setting $a$ exactly to the ...


3

$x^T y = y^T x$ always. Note that $x^T y$ is a scalar (i.e. a real number), so its transpose is itself ($y^T x = (x^T y)^T = x^T y$).


3

This is called the geometric mean of the $a_i$. Here is its Wikipedia page.


2

Hint a very interesting fact you can use : $$\dbinom {i+n-1} {i-1}=\frac{\cdots }{n!}$$


2

The idea is to find the greatest power $10^n$ dividing the product of the given integers, first you have to find the greatest powers $2^n$ and $5^n$ dividing each number : $25^5$ gives you $2^{0}$ and $5^{10}$ because $25=5^2$ $150^4$ gives you $2^4$ and $5^{8}$ because $150=3\cdot 5^2\cdot 2$ $2008^3$ gives you $2^{9}$ and $5^{0}$ because $2008=2^3\cdot ...


2

Assuming that the correct expression is $$ P_k = P_0\prod_{i=0}^{m-1}\frac{\alpha}{(i+1)\mu}\prod_{j=m}^{k-1}\frac{\alpha}{m\mu} $$ we just have to count. The total number of factors of $\alpha/\mu$ are $k$, hence we should have $(\alpha/\mu)^k$. The total number of factors of $1/m$ are $k-m$, so we should have $1/m^{k-m}$. The product with $1/(i+1)$, when ...


1

After the latest comment by the OP, I edited again to fix the typo in the indexing of the final product. The two products are indeed equivalent. To see this, I recommend drawing a picture, with $i$ on the horizontal axis and $j$ on the vertical axis. In both cases, the product is taken over all $(i,j)$ coordinates on and above the diagonal. In the first ...


1

Notice that this relates strongly to the so-called choose function, $C(n,k)$, which is defined as $$C(N,k) = \frac{N!}{k!(N-k)!}$$ Since the right-hand side can be written $ \prod_{k=i}^{i+n-1} k = \frac{(i+n-1)!}{(i-1)!} $, the question is really asking you to verify that $$C(n+i-1, n) = \frac{(i+n-1)!}{n!(i-1)!} \in \mathbb Z$$


1

Each factor of the product with exponent $2^k$ is equal to the difference with exponent 2^(k+1) divided by the difference with the same exponent of the considered factor (because an scholar identity with squares). Then the product finally gives the equation 6^(64) - 5^64 = $6^x$ - $5^y$. WARNING: there is an obvious solution x = y but it is not unique! ...



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