We will prove that \[\sum_{k=2}^n \frac{1}{(k-1)k}=1 - \frac{1}{n}\] for mathematical induction on \(n\). For the basis step, we observe that \[\frac{1}{1 \cdot 2} = \frac{1}{2} = 1 - \frac{1}{1\cdot 2}\] so the statement holds when \(n=2\). For the induction step, we assume that \[\sum_{k=2}^n \frac{1}{(k-1)k}=1 - \frac{1}{n}\] for some integer \(n \ge 2\). Now \[\begin{align*} {}&\sum_{k=2}^{n+1} \frac{1}{(k-1)k}\\ =&\left(\sum_{k=2}^n \frac{1}{(k-1)k}\right) + \frac{1}{n(n+1)}. \end{align*}\] Applying the induction hypothesis, we continue: \[\begin{align*} =&\left(1 - \frac{1}{n}\right) + \frac{1}{n(n+1)}\\ =&1 - \frac{n+1}{n(n+1)} + \frac{1}{n(n+1)}\\ =&1 - \frac{n+1-1}{n(n+1)}\\ =&1 - \frac{1}{(n+1)}. \end{align*}\] Thus, the desired statement follows by induction.

As a consequence of its definition, \(((a,b),b)\) divides \((a,b)\). As a consequence of its definition, \((a,b)\) divides \(b\). Since \((a,b)\) also divides itself, \((a,b)\) is a common divisor of \((a,b)\) and \(b\). Thus, \((a,b)\) divides \(((a,b),b)\). Two positive numbers dividing eachother must be equal, so \(((a,b),b)=(a,b)\).

Since \(a|c\) and \(c|b\), we conclude that \(a|b\). Since \(a|a\) also, we have \(a|(a,b)\) by the definition of gcd. Since \(a|1\) and \(a\) is a positive integer, we conclude that \(a=1\).

We find \(8840 = 2^3 × 5 × 13 × 17\).

Suppose \(n/p\) is composite. Let \(q\) be the least prime factor of \(n\). Then \(n/p=qr\) for some integer \(r\) where \(1 < q \le r < n/p\). Since \(q\) also divides \(n\), we must have \(p\le q\) by the definition of \(p\). From \(p > n^{1/3}\) we find that \[n/p < n/n^{1/3}=n^{2/3} < p^2.\] On the other hand, from \(p \le q \le r\) we conclude that \(p^2 \le qr = n/p\). Thus, we have a contradiction and so \(n/p\) must be prime.

Recall the identity \[x^k-1=(x-1)(x^{k-1}+x^{k-2}+\cdots + x+1)\] which holds for every real number \(x\) and every integer \(k \ge 2\). Since \(n\) is composite, we have \(n=ab\) for integers \(a,b \ge 2\). Thus \(2^n=2^{ab}=(2^a)^b\). Using the identity we obtain \[(2^a)^b-1=\left((2^a)-1\right)\left((2^a)^{b-1}+\cdots + 1\right).\] Since both factors are greater than \(1\), we see that \(2^n-1\) is composite.

In this case, if we divide by \(10\), we find \(94x+68y=15\). The left hand side must be even, while the right hand side is odd. Thus there can be no solutions.

We apply the Euclidean algorithm for \(940\) and \(680\): \[\begin{align*} 940 &= 1(680)+260\\ 680 &= 2(260)+160\\ 260 &= 1(160)+100\\ 160 &= 1(100)+60\\ 100 &= 1(60)+40\\ 60 &= 1(40)+20\\ 40 &= 2(20)+0\\ \end{align*}\] Thus, \(\gcd(940,680)=20\).

Now we apply the extended Euclidean algorithm to find Bezout coefficients: \[\begin{align*} 20 &= (1)(60) + (-1)(40)\\ &= (1)(60) + (-1)[100+(-1)(60)]\\ &= (-1)(100) + (2)(60)\\ &= (-1)(100) + (2)[160+(-1)(100)]\\ &= (2)(160) + (-3)(100)\\ &= (2)(160) + (-3)[260+(-1)(160)]\\ &= (-3)(260) + (5)(160)\\ &= (-3)(260) + (5)[680+(-2)(260)]\\ &= (5)(680) + (-13)(260)\\ &= (5)(680) + (-13)[940+(-1)(680)]\\ &= (-13)(940) + (18)(680)\\ \end{align*}\]

Multiplying everything by \(5\), we obtain one solution \[940(-65)+680(90)=100.\]

To find all solutions, we can add any solution of \(940x+680y=0\) or, after dividing by the gcd, \(47x+34y=0\). Thus the set of all solutions is \[\left\{ (-65+34t,90-47t) \mid t \in \mathbb{Z} \right\}.\]