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The notation means : $q$ divides $k$ ; There is an integer $m$ with $qm=k$.
$j_1\mid j_2\implies$ either $j_2=0$ or $|j_1|\le |j_2|$. $j_2\mid j_1\implies$ either $j_1=0$ or $|j_2|\le |j_1|$. If $j_2=0$, then $0\mid j_1$, so $j_1=0$, so $j_1=j_2$. If $j_1=0$, then $0\mid j_2$, so $j_2=0$, so $j_1=j_2$. If $j_1j_2\neq 0$, then $|j_1|\le |j_2|$ and $|j_2|\le |j_1|$, so $|j_1|=|j_2|$, so $j_1=\pm j_2$. $j_1\mid j_2\iff ... 4 Use the inclusion-exclusion principle. Let$I=[10000,99999]$and: $$A = \{n\in I:n\equiv 0\pmod{3}\},$$ $$B = \{n\in I:n\equiv 0\pmod{5}\},$$ $$C = \{n\in I:n\equiv 0\pmod{7}\}.$$ Compute$|A|,|B|,|C|,|A\cap B|,|A\cap C|,|B\cap C|,|A\cap B\cap C|$, then recall that: $$|A\cup B \cup C| = |A|+|B|+|C|-|A\cap B|-|A\cap C|-|B\cap C|+|A\cap B\cap C|.$$ You ... 3 I'm willing to do a lot to avoid inclusion-exclusion, so:$3\cdot 5\cdot 7=105$, so the pattern of multiples repeat every 105 numbers. Since you have$90000=857\cdot 105+15$you can get the answer by taking 857 times the number of divisors in a 105-number period, plus those-among$10000, 10001, \cdots, 10014$that are multiples of 3/5/7, which you can count ... 3 $$n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= (n - 4)(n - 3)(n - 2)(n - 1)n(n + 1) \bigl[n (n+1)(3n^2 - n - 6) \bigr]$$ Now, the product of$6$consecutive integers is always divisible by$5$. Also note that $$3n^2 - n - 6=4n(n+1)-(n^2+5n+6)=4n(n+1)-(n+2)(n+3)$$ Thus $$n^2(n - 4)(n - 3)(n - 2)(n - 1)(n + 1)^2(3n^2 - n - 6)\\= 4(n - 4)(n ... 3 Hint: A number is divisible by 3 if the sum of the digits is divisible by 3. 2 Hint: we have 16320=2^6\cdot 3\cdot 5\cdot 17, and already p^{32}\equiv 1 \bmod k for each k\in \{2^6,3,5,17\}. For example, p^2\equiv 1\bmod 3, because \gcd(p,3)=1. 2 First, it is clear that if the answer is positive for some c_0, then it must be for every c < c_0, too: if we keep the same configuration we have A_{n,x} \geq c_0 x > cx. Similarly, if the answer is negative for c_0 it must be negative for every c > c_0, too. We will now prove that the answer is positive for every c < 1. Indeed, let ... 2 It means that p^{\alpha}\mid n but p^{\alpha+1}\nmid n. For example. 2^2\mid 12 but 2^3\nmid 12. Hence 2^2\|12 1 Here is a more inspired solution. I will prove a stronger claim - that$$\frac{m+1}{n}+\frac{n+1}{m}=3$$has infinitely many solutions in positive integers. Setting$$m=2d(d+a), n=2d(d-a)$$We have$$\frac{m+1}{n}+\frac{n+1}{m}=\frac{2d^2+2a^2+1}{d^2-a^2}$$So for \frac{m+1}{n}+\frac{n+1}{m}=3 to hold, we only need d^2-5a^2=1. There are infinitely ... 1 It means that q divides k, or k is divisible by q. Add a little slash and it negates that meaning: q \nmid k means q does not divide k. For example: 3 \mid 1728, 3 \nmid 1729. They are not relatively prime, unless q = 1 or -1. In fact, if q \mid k then \gcd(q, k) = |q|. 1 Write dq = pr+s where s,r \in \Bbb Z[X], \deg s < \deg p and d is a (nonzero) positive integer. (this is always possible by picking for d the dominant coefficient of p to the (\deg q - \deg p+1)th power) Your hypothesis implies that p(n) divides s(n) for infinitely many n. For n large enough, |p(n)| > |s(n)| so you must have ... 1 You have 16320=2^6\cdot3\cdot5\cdot17. Notice that using Fermat's theorem you have: p^{16}\equiv1\pmod{17}, hence also p^{32}=(p^{16})^2\equiv1\pmod{17} p^4\equiv1\pmod5, hence also p^{32} \equiv 1 \pmod5 p^2\equiv1\pmod3. hence also p^{32} \equiv 1\pmod3 Now it only remains to notice that \phi(64)=32 and by Euler's theorem ... 1 Suppose, by contradiction, that 3 divides both n+m and n-m (that is, negate the thesis). Then 3 also divides (n+m)+(n-m)=2n. Since 3\nmid2, we get 3\mid n. Can you show that 3 divides also m? Actually this is not a proof by contradiction, which is not necessary. The statements “A implies B” and “not B implies not A” are ... 1 Let the digits beb,c,d,29-b-c-d As 0\le b,c,d\le9,0<b+c+d\le27<29 Now,$$(29-b-c-d)+10b+100c+1000d=29+9b+99c+999d\equiv-20b+12c-16d$$We need$29\mid(5b-3c+4d)\equiv-24b-3c+33d\iff29\mid(c+8b-11d)$For$29\mid(8b-11d)\implies8b-11d=(29k-c)(11\cdot3-8\cdot4)\iff8(b+116k-4c)=11(87k+d-3c)\iff b+116k=11m+4c\iff b\equiv5k+4c\pmod{11}$and ... 1 If$a \mid c$and if$b \mid c$, then$c = ap$and$c = bq$for some integers$p,q$; if in addition$(a,b) = 1$, then$ap = bq$implies$b \mid p$and$a \mid q\$; but then we are done.