problem stringlengths 1 13.6k | solution stringlengths 0 18.5k ⌀ | answer stringlengths 0 575 ⌀ | problem_type stringclasses 8
values | question_type stringclasses 4
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class | __index_level_0__ int64 0 742k |
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1. Find the prime factorization of the following positive integers
a) 692921
b) 1468789
c) 55608079 . | 1. a) $23 \cdot 47 \cdot 641$
b) $7 \cdot 37 \cdot 53 \cdot 107$
c) $19^{2} \cdot 31 \cdot 4969$ | c) 19^{2} \cdot 31 \cdot 4969 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,621 |
6. In this problem, we develop a factorization technique known as Euler's method. It' is applicable when the integer being factored is odd and can be written as the sum of two squares in two different ways. Let $n$ be odd and let $n=a^{2}+b^{2}=c^{2}+d^{2}$, where $a$ and $c$ are odd positive integers, and $b$ and $d$ ... | 6. d) $13 \cdot 17,41 \cdot 61,293 \cdot 3413$ | 13 \cdot 17, 41 \cdot 61, 293 \cdot 3413 | Number Theory | proof | Yes | Yes | number_theory | false | 738,626 |
7. Show that any number of the form $2^{4 n+2}+1$ can be easily factored by the use of the identity $4 x^{4}+1=\left(2 x^{2}+2 x+1\right)\left(2 x^{2}-2 x+1\right)$. Factor $2^{18}+1$ using this identity. | 7. $5 \cdot 13 \cdot 37 \cdot 109$ | 5 \cdot 13 \cdot 37 \cdot 109 | Number Theory | proof | Yes | Yes | number_theory | false | 738,627 |
13. Estimate the number of decimal digits in the Fermat number $F_{n}$. | 13. $2^{n} \log _{10} 2$ | 2^{n} \log _{10} 2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,634 |
Theorem 2.8. Let $a$ and $b$ be positive integers with $d=(a, b)$. The equation $a x+b y=c$ has no integral solutions if $d \backslash c$. If $d \mid c$, then there are infinitely many integral solutions. Moveover, if $x=x_{0}, y=y_{0}$ is a particular solution of the equation, then all solutions are given by
$$x=x_{0}... | Proof. Assume that $x$ and $y$ are integers such that $a x+b y=c$. Then, since $d \mid a$ and $d \mid b$, by Proposition 1.4, $d \mid c$ as well. Hence, if $d \backslash c$, there are no integral solutions of the equation.
Now assume that $d \mid c$. From Theorem 2.1, there are integers $s$ and $t$ with
$$d=a s+b t .$$... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,635 |
5. A postal clerk has only 14 -cent and 21 -cent stamps to sell. What combinations of these may be used to mail a package requiring postage of exactly
a) $\$ 3.50$
b) $\$ 4.00$
c) $\$ 7.77 ?$ | 5. a) $(14$-cent stamps, 21 -cent stamps $)=(25,0),(22,2),(19,4),(16,6),(13,8)$, $(10,10),(7,12),(4,14),(1,16)$
b) no solution
c) $(14$-cent stamps, 21 -cent stamps $)=(54,1),(51,3),(48,5),(45,7)$, $(42,9),(39,11),(36,13),(33,15),(30,17),(27,19),(24,21),(21,23)$, $(18,25),(15,27),(12,29),(9,31),(6,33),(3,35),(0,37)$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,640 |
Proposition 1.3. If $a, b$, and $c$ are integers with $a \mid b$ and $b \mid c$, then $a \mid c$. | Proof. Since $a \mid b$ and $b \mid c$, there are integers $e$ and $f$ with $a e=b$ and $b f=c$. Hence, $b f=(a e) f=a(e f)=c$, and we conclude that $a \mid c$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,642 |
10. How many ways can change be made for one dollar using
a) dimes and quarters
b) nickels. dimes, and quarters
c) pennies, nickels, dimes, and quarters? | 10. a) 3
b) 29
c) 242 | 242 | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 738,646 |
11. Find all integer solutions of the following systems of linear diophantine equations
a)
$$\begin{array}{l}
x+y+z=100 \\
x+8 y+50 z=156
\end{array}$$
b)
$$\begin{array}{l}
x+y+z=100 \\
x+6 y+21 z=121
\end{array}$$
c)
$$\begin{array}{l}
x+y+z+w=100 \\
x+2 y+3 z+4 w=300 \\
x+4 y+9 z+16 w=1000
\end{array}$$ | 11. a) $x=98-6 n, y=1+7 n, z=1-n$
b) no solution
c) $x=50-n, y=-100+3 n, z=150-3 n, w=n$ | a) x=98-6n, y=1+7n, z=1-n \\ b) \text{no solution} \\ c) x=50-n, y=-100+3n, z=150-3n, w=n | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,647 |
12. A piggy bank contains 24 coins, all nickels, dimes, and quarters. If the total value of the coins is two dollars, what combinations of coins are possible? | 12. $($ nickels, dimes, quarters $)=(20,0,4),(17,4,3),(14,8,2),(11,12,1)$, $(8,16,0)$ | (20,0,4),(17,4,3),(14,8,2),(11,12,1),(8,16,0) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,648 |
Proposition 3.1. If $a$ and $b$ are integers, then $a \equiv b(\bmod m)$ if and only if there is an integer $k$ such that $a=b+k m$. | Proof. If $a \equiv b(\bmod m)$, then $m \mid(a-b)$. This means that there is an integer $k$ with $k m=a-b$, so that $a=b+k m$.
Conversely, if there is an integer $k$ with $a=b+k m$, then $k m=a-b$. Hence $m \mid(a-b)$, and consequently, $a \equiv b(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,652 |
Proposition 1.4. If $a, b, m$, and $n$ are integers, and if $c \mid a$ and $c \mid b$, then $c \mid(m a+n b)$ | Proof. Since $c \mid a$ and $c \mid b$, there are integers $e$ and $f$ such that $a=c e$ and $b=c f$. Hence, $m a+n b=m c e+n c f=c(m e+n f)$. Consequently, we see that $c \mid(m a+n b)$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,653 |
Proposition 3.2. Let $m$ be a positive integer. Congruences modulo $m$ satisfy the following properties:
(i) Reflexive property. If $a$ is an integer, then $a \equiv a(\bmod m)$.
(ii) Symmetric property. If $a$ and $b$ are integers such that $a \equiv b(\bmod m)$, then $b \equiv a(\bmod m)$.
(iii) Transitive property. ... | Proof.
(i) We see that $a \equiv a(\bmod m)$, since $m \mid(a-a)=0$.
(ii) If $a \equiv b(\bmod m)$, then $m \mid(a-b)$. Hence, there is an integer $k$ with $k m=a-b$. This shows that $(-k) m=b-a$, so that $m \mid(b-a)$. Consequently, $b \equiv a(\bmod m)$.
(iii) If $a \equiv b(\bmod m)$ and $b \equiv c(\bmod m)$, then ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,654 |
Theorem 3.1. If $a, b, c$, and $m$ are integers with $m>0$ such that $a \equiv b(\bmod m)$, then
(i) $a+c \equiv b+c(\bmod m)$,
(ii) $a-c \equiv b-c(\bmod m)$,
(iii) $\quad a c \equiv b c(\bmod m)$. | Proof. Since $a \equiv b(\bmod m)$, we know that $m \mid(a-b)$. From the identity $(a+c)-(b+c)=a-b$, we see $m \mid[(a+c)-(b+c)]$, so that (i) follows. Likewise, (ii) follows from the fact that $(a-c)-(b-c)=a-b$. To show that (iii) holds, note that $a c-b c=c(a-b)$. Since $m \mid(a-b)$, it follows that $m \mid c(a-b)$,... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,655 |
Theorem 3.2. If $a, b, c$ and $m$ are integers such that $m>0, d=(c, m)$, and $a c \equiv b c(\bmod m)$, then $a \equiv b(\bmod m / d)$ | Proof. If $a c \equiv b c(\bmod m)$, we know that $m \mid(a c-b c)=c(a-b)$. Hence, there is an integer $k$ with $c(a-b)=k m$. By dividing both sides by $d$, we have $(c / d)(a-b)=k(m / d)$. Since $(m / d, c / d)=1$, from Proposition 2.1 it follows that $m / d \mid(a-b)$. Hence, $a \equiv b(\bmod m / d)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,656 |
Theorem 3.3. If $a, b, c, d$, and $m$ are integers such that $m>0$, $a \equiv b(\bmod m)$, and $c \equiv d(\bmod m)$, then
(i) $a+c \equiv b+d(\bmod m)$,
(ii) $a-c \equiv b-d(\bmod m)$,
(iii) $\quad a c \equiv b d(\bmod m)$. | Proof. Since $a \equiv b(\bmod m)$ and $c \equiv d(\bmod m)$, we know that $m \mid(a-b)$ and $m \mid(c-d)$. Hence, there are integers $k$ and $\ell$ with $k m=a-b$ and $l m=c-d$
To prove (i), note that $(a+c)-(b+d)=(a-b)+(c-d)=k m+l m=$ $(k+l) m$. Hence, $m \mid[(a+c)-(b+d)] . \quad$ Therefore, $a+c \equiv b+$ $d(\bmo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,658 |
Theorem 3.4. If $r_{1}, r_{2}, \ldots, r_{m}$ is a complete system of residues modulo $m$, and if $a$ is a positive integer with $(a, m)=1$, then
$$a r_{1}+b, a r_{2}+b, \ldots, a r_{m}+b$$
is a complete system of residues modulo $m$. | Proof. First, we show that no two of the integers
$$a r_{1}+b, a r_{2}+b, \ldots, a r_{m}+b$$
are congruent modulo $m$. To see this, note that if
$$a r_{j}+b \equiv a r_{k}+b(\bmod m)$$
then, from (ii) of Theorem 3.1, we know that
$$a r_{j} \equiv a r_{k}(\bmod m)$$
Because $(a, m)=1$, Corollary 3.1 shows that
$$r_{j}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,659 |
Theorem 3.5. If $a, b, k$, and $m$ are integers such that $k>0, m>0$, and $a \equiv b(\bmod m)$, then $a^{k} \equiv b^{k}(\bmod m)$ | Proof. Because $a \equiv b(\bmod m)$, we have $m \mid(a-b)$. Since
$$a^{k}-b^{k}=(a-b)\left(a^{k-1}+a^{k-2} b+\cdots+a b^{k-2}+b^{k-1}\right),$$
we see that $(a-b) \mid\left(a^{k}-b^{k}\right)$. Therefore, from Proposition 1.2 it follows that $m \mid\left(a^{k}-b^{k}\right)$. Hence, $a^{k} \equiv b^{k}(\bmod m)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,660 |
Theorem 3.6. If $a \equiv b\left(\bmod m_{1}\right), a \equiv b\left(\bmod m_{2}\right), \ldots, a \equiv b\left(\bmod m_{k}\right)$ where $a, b, m_{1}, m_{2}, \ldots, m_{k}$ are integers with $m_{1}, m_{2}, \ldots, m_{k}$ positive, then
$$a \equiv b\left(\bmod \left[m_{1}, m_{2}, \ldots, m_{k}\right]\right)$$
where $\... | Proof. Since $a \equiv b\left(\bmod m_{1}\right), a \equiv b\left(\bmod m_{2}\right), \ldots, a \equiv b\left(\bmod m_{k}\right)$, we know that $m_{1}\left|(a-b), m_{2}\right|(a-b), \ldots, m_{k} \mid(a-b)$. From problem 20 of Section 2.3 , we see that
$$\left[m_{1}, m_{2}, \ldots, m_{k}\right] \mid(a-b)$$
Consequentl... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,661 |
Corollary 3.2. If $a \equiv b\left(\bmod m_{1}\right), a \equiv b\left(\bmod m_{2}\right), \ldots, a \equiv b\left(\bmod m_{k}\right)$ where $a$ and $b$ are integers and $m_{1}, m_{2}, \ldots, m_{k}$ are relatively prime positive integers, then
$$a \equiv b\left(\bmod m_{1} m_{2} \cdots m_{k}\right)$$ | Proof. Since $m_{1,} m_{2}, \ldots, m_{k}$ are pairwise relatively prime, problem 34 of Section 2.3 tells us that
$$\left[m_{1}, m_{2}, \ldots, m_{k}\right]=m_{1} m_{2} \cdots m_{k}$$
Hence, from Theorem 3.6 we know that
$$a \equiv b\left(\bmod m_{1} m_{2} \cdots m_{k}\right)$$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,662 |
Proposition 3.3. Let $b, m$, and $N$ be positive integers with $b<m$. Then the least positive residue of $b^{N}$ modulo $m$ can be computed using $O\left(\left(\log _{2} m\right)^{2} \log _{2} N\right)$ bit operations. | Proof. To find the least positive residue of $b^{N}(\bmod m)$, we can use the algorithm just described. First, we find the least positive residues of $b, b^{2}, b^{4}, \ldots, b^{2^{k}}$ modulo $m$, where $2^{k} \leqslant N<2^{k+1}$, by successively squaring and reducing modulo $m$. This requires a total of $O\left(\le... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,663 |
12. What time does a clock read
a) 29 hours after it reads 11 o'clock
b) 100 hours after it reads $20^{\prime}$ clock
c) 50 hours before it reads 6 o'clock? | 12. a) 4 o'clock b) 6 o'clock c) 4 o'clock | a) 4 o'clock b) 6 o'clock c) 4 o'clock | Logic and Puzzles | math-word-problem | Yes | Yes | number_theory | false | 738,677 |
13. Which decimal digits occur as the final digit of a fourth power of an integer? | 13. $0,1,5,6$ | 0,1,5,6 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,678 |
14. What can you conclude if $a^{2} \equiv b^{2}(\bmod p)$, where $a$ and $b$ are integers and $p$ is prime? | 14. $a \equiv \pm b(\bmod p)$ | a \equiv \pm b(\bmod p) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,679 |
17. For which positive integers $n$ is it true that
$$1^{2}+2^{2}+3^{2}+\cdots+(n-1)^{2} \equiv 0(\bmod n) ?$$ | 17. $n \equiv \pm 1(\bmod 6)$ | n \equiv \pm 1(\bmod 6) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,682 |
1. Show that $3|99,5| 145,7 \mid 343$, and $888 \mid 0$. | 1. $99=3 \cdot 33,145=5 \cdot 29,343=7 \cdot 49,0=888 \cdot 0$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,686 |
24. On a computer with word size $w$, multiplication modulo $n$, where $n<w / 2$, can be performed as outlined. Let $T=[\sqrt{n}+1 / 2]$, and $t=T^{2}-n$. For each computation, show that all the required computer arithmetic can be done without exceeding the word size. (This method was described by Head [67]).
a) Show t... | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,690 |
26. Find the least positive residue of
a) $3^{10}$ modulo 11
b) $2^{12}$ modulo 13
c) $5^{16}$ modulo 17
d) $3^{22}$ modulo 23 .
e) Can you propose a theorem from the above congruences? | 26. a) 1 b) 1 c) 1 d) 1 e) $a^{p-1} \equiv 1(\bmod p)$ when $p$ is prime and $p \nmid a$ | a) 1, b) 1, c) 1, d) 1, e) a^{p-1} \equiv 1(\bmod p) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,692 |
27. Find the least positive residues of
a) 6! modulo 7
b) 10 ! modulo 11
c) 12! modulo 13
d) 16! modulo 17 .
e) Can you propose a theorem from the above congruences? | 27. a) -1
b) -1
c) -1
d) -1
e) $(p-1)!\equiv-1(\bmod p)$ when $p$ is prime | (p-1)!\equiv-1(\bmod p) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,693 |
The Binomial Theorem. Let $x$ and $y$ be variables and $n$ a positive integer. Then
$$\begin{aligned}
(x+y)^{n}= & \binom{n}{0} x^{n}+\binom{n}{1} x^{n-1} y+\binom{n}{2} x^{n-2} y^{2}+\cdots \\
& +\binom{n}{n-2} x^{2} y^{n-2}+\binom{n}{n-1} x y^{n-1}+\binom{n}{n} y^{n}
\end{aligned}$$
or using summation notation,
$$(x+... | Proof. We use mathematical induction. When $n=1$, according to the binomial theorem, the formula becomes
$$(x+y)^{1}=\binom{1}{0} x^{1} y^{0}+\binom{1}{1} x^{0} y^{1}$$
But because $\binom{1}{0}=\binom{1}{1}=1$, this states that $(x+y)^{1}=x+y$, which is obviously true.
We now assume the theorem is valid for the posi... | proof | Algebra | proof | Yes | Yes | number_theory | false | 738,697 |
Theorem 3.7. Let $a, b$, and $m$ be integers with $m>0$ and $(a, m)=d$. If $d \backslash b$, then $a x \equiv b(\bmod m)$ has no solutions. If $d \mid b$, then $a x \equiv b(\bmod m)$ has exactly $d$ incongruent solutions modulo $m$. | Proof. From Proposition 3.1, the linear congruence $a x \equiv b(\bmod m)$ is equivalent to the linear diophantine equation in two variables $a x-m y=b$. The integer $x$ is a solution of $a x \equiv b(\bmod m)$ if and only if there is an integer $y$ with $a x-m y=b$. From Theorem 2.8, we know that if $d \backslash b$, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,699 |
Proposition 3.4. Let $p$ be prime. The positive integer $a$ is its own inverse modulo $p$ if and only if $a \equiv 1(\bmod p)$ or $a \equiv-1(\bmod p)$. | Proof. If $a \equiv 1(\bmod p)$ or $a \equiv-1(\bmod p)$, then $a^{2} \equiv 1(\bmod p)$, so that $a$ is its own inverse modulo $p$.
Conversely, if $a$ is its own inverse modulo $p$, then $a^{2}=a \cdot a \equiv 1(\bmod p)$. Hence, $p \mid\left(a^{2}-1\right)$. Since $a^{2}-1=(a-1)(a+1)$, either $p \mid(a-1)$ or $p \m... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,700 |
2. Let $a, b$, and $m$ be positive integers with $a>0, m>0$, and $(a, m)=1$. The following method can be used to solve the linear congruence $a x \equiv b(\bmod m)$.
a) Show that if the integer $x$ is a solution of $a x \equiv b(\bmod m)$, then $x$ is also a solution of the linear congruence
$$a_{1} x \equiv-b[m / a](\... | 2. c) $x \equiv 5(\bmod 23)$ | x \equiv 5(\bmod 23) | Number Theory | proof | Yes | Yes | number_theory | false | 738,702 |
8. Find all solutions of the following linear congruences in two variables
a) $2 x+3 y \equiv 1(\bmod 7)$
c) $6 x+3 y \equiv 0(\bmod 9)$
b) $2 x+4 y \equiv 6(\bmod 8)$
d) $10 x+5 y \equiv 9(\bmod 15)$. | 8. a) $(x, y) \equiv(0,5),(1,2),(2,6),(3,3),(4,0),(5,4),(6,1)(\bmod 7)$
b) $(x, y) \equiv(1,1),(1,3),(1,5),(1,7),(3,0),(3,2),(3,4),(3,6),(5,1),(5,3),(5,5),(5,7)$, $(7,0),(7,2),(7,4),(7.6)(\bmod 8)$
c) $(x, y) \equiv(0,0),(0,3),(0,6),(1,1),(1,4),(1,7),(2,2),(2,5),(2,8),(3,0),(3,3),(3,6)$, $(4,1),(4,4),(4,7),(5,2),(5,5),... | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,708 |
3. Find the quotient and remainder in the division algorithm with divisor 17 and dividend
a) 100
c) -44
b) 289
d) -100 . | 3.
a) 5,15
b) 17,0
c) $-3,7$
d) $-6,2$ | a) 5,15 \\ b) 17,0 \\ c) -3,7 \\ d) -6,2 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,709 |
The Chinese Remainder Theorem. Let $m_{1}, m_{2}, \ldots, m_{r}$ be pairwise relatively prime positive integers. Then the system of congruence
$$\begin{aligned}
x & \equiv a_{1}\left(\bmod m_{1}\right) \\
x & \equiv a_{2}\left(\bmod m_{2}\right) \\
& \cdot \\
& \cdot \\
x & \equiv a_{r}\left(\bmod m_{t}\right)
\end{ali... | Proof. First, we construct a simultaneous solution to the system of congruences. To do this, let $M_{k}=M / m_{k}=m_{1} m_{2} \cdots m_{k-1} m_{k+1} \cdots m_{r}$. We know that $\left(M_{k}, m_{k}\right)=1$ from problem 8 of Section 2.1, since $\left(m_{j}, m_{k}\right)=1$ whenever $j \neq k$. Hence, from Theorem 3.7, ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,714 |
Lemma 3.1. If $a$ and $b$ are positive integers, then the least positive residue of $2^{a}-1$ modulo $2^{b}-1$ is $2^{r}-1$, where $r$ is the least positive residue of $a$ modulo $b$. | Proof. From the division algorithm, $a=b q+r$ where $r$ is the least positive residue of $a$ modulo $b$. We have $\left(2^{a}-1\right)=\left(2^{b q+r}-1\right)=$ $\left(2^{b}-1\right)\left(2^{b(q-1)+r}+\cdots+2^{b+r}+2^{r}\right)+\left(2^{r}-1\right)$, which shows that the remainder when $2^{a}-1$ is divided by $2^{b}-... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,715 |
Lemma 3.2. If $a$ and $b$ are positive integers, then the greatest common divisor of $2^{a}-1$ and $2^{b}-1$ is $2^{(a, b)}-1$. | Proof. When we perform the Euclidean algorithm with $a=r_{0}$ and $b=r_{1}$, we obtain
$$\begin{array}{rlrl}
r_{0} & =r_{1} q_{1}+r_{2} & 0 \leqslant r_{2}<r_{1} \\
r_{1} & =r_{2} q_{2}+r_{3} & 0 \leqslant r_{3}<r_{2} \\
& \cdot & \\
\cdot & & \\
\cdot & & \\
r_{n-3} & =r_{n-2} q_{n-2}+r_{n-1} & 0 \leqslant r_{n-1}<r_{... | 2^{(a, b)}-1 | Number Theory | proof | Yes | Yes | number_theory | false | 738,716 |
1. Find all the solutions of each of the following systems of congruences.
a) $x \equiv 4(\bmod 11)$
c) $x \equiv 0(\bmod 2)$
$x \equiv 3(\bmod 17)$
$x \equiv 0(\bmod 3)$
$x \equiv 1(\bmod 5)$
b) $x \equiv 1(\bmod 2)$
$x \equiv 6(\bmod 7)$
$x \equiv 2(\bmod 3)$
$x \equiv 3(\bmod 5)$
d) $x \equiv 2(\bmod 11)$
$x \equiv ... | 1. a) $x \equiv 37(\bmod 187)$
d) $x \equiv 150999(\bmod 554268)$
c) $x \equiv 6(\bmod 210)$ | x \equiv 6(\bmod 210) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,718 |
4. What can you conclude if $a$ and $b$ are nonzero integers such that $a \mid b$ and $b \mid a ?$ | 4. $a= \pm b$ | a= \pm b | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,720 |
Theorem 3.8. Let $a, b, c, d, e, f$, and $m$ be integers with $m>0$, such that $(\Delta, m)=1$, where $\Delta=a d-b c$. Then, the system of congruences
$$\begin{array}{l}
a x+b y \equiv e(\bmod m) \\
c x+d y \equiv f(\bmod m)
\end{array}$$
has a unique solution modulo $m$ given by
$$\begin{array}{l}
x \equiv \bar{\Delt... | Proof. We multiply the first congruence of the system by $d$ and the second by $b$, to obtain
$$\begin{array}{l}
a d x+b d y \equiv d e(\bmod m) \\
b c x+b d y \equiv b f(\bmod m)
\end{array}$$
Then, we subtract the second congruence from the first, to find that
$$(a d-b c) x \equiv d e-b f(\bmod m)$$
or, since $\Delt... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,737 |
Proposition 3.6. If $A$ and $B$ are $n \times k$ matrices with $A \equiv B(\bmod m), C$ is an $k \times p$ matrix and $D$ is a $p \times n$ matrix, all with integer entries, then $A C \equiv B C(\bmod m)$ and $D A \equiv D B(\bmod m)$ | Proof. Let the entries of $A$ and $B$ be $a_{i j}$ and $b_{i j}$, respectively, for $1 \leqslant i \leqslant n$ and $1 \leqslant j \leqslant k$, and let the entries of $C$ be $c_{i j}$ for $1 \leqslant i \leqslant k$ and $1 \leqslant j \leqslant p$. The $(i, j)$ th entries of $A C$ and $B C$ are $\sum_{t=1}^{n} a_{i t}... | proof | Algebra | proof | Yes | Yes | number_theory | false | 738,738 |
matrix
$$\bar{A}=\bar{\Delta}\left(\begin{array}{cc}
d & -b \\
-c & a
\end{array}\right),$$
where $\bar{\Delta}$ is the inverse of $\Delta$ modulo $m$, is an inverse of $A$ modulo $m$. | Proof. To verify that the matrix $\bar{A}$ is an inverse of $A$ modulo $m$, we need only verify that $A \bar{A} \equiv \bar{A} A \equiv I(\bmod m)$.
To see this, note that
$$\begin{aligned}
A \bar{A} & \equiv\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right) \bar{\Delta}\left(\begin{array}{cc}
d & -b \\
-c & a
... | proof | Algebra | proof | Yes | Yes | number_theory | false | 738,740 |
Proposition 3.8. If $A$ is an $n \times n$ matrix with integer entries and $m$ is positive integer such that $(\operatorname{det} A, \underline{m})=1$, then the matrix $\bar{A}=\bar{\Delta}(\operatorname{adj} A)$ is an inverse of $A$ modulo $m$, where $\Delta$ is an inverse of $\Delta=\operatorname{det} A$ modulo $m$. | Proof. If $(\operatorname{det} A, m)=1$, then we know that $\operatorname{det} A \neq 0$. Hence, from Theorem 3.9 , we have
$$A \operatorname{adj} A=(\operatorname{det} A) I=\Delta I$$
Since $(\operatorname{det} A, m)=1$, there is an inverse $\bar{\Delta}$ of $\Delta=\operatorname{det} A$ modulo $m$. Hence,
$$A(\bar{\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,743 |
2. Find the solutions of the following systems of linear congruences.
a)
$$\begin{array}{r}
2 x+3 y \equiv 5(\bmod 7) \\
x+5 y \equiv 6(\bmod 7)
\end{array}$$
b)
$$\begin{aligned}
4 x+y & \equiv 5(\bmod 7) \\
x+2 y & \equiv 4(\bmod 7)
\end{aligned}$$ | 2. a) $(x, y) \equiv(0,4),(1,1),(2,5),(3,2),(4,6),(5,3),(6,0)(\bmod 7)$
b) no solution | no solution | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,745 |
3. What are the possibilities for the number of incongruent solutions of the system of linear congruences
$$\begin{array}{l}
a x+b y \equiv c(\bmod p) \\
d x+e y \equiv f(\bmod p)
\end{array}$$
where $p$ is a prime and $a, b, c, d, e$, and $f$ are positive integers? | 3. $0,1, p$, or $p^{2}$ | 0,1, p, p^{2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,746 |
4. Find the matrix $C$ such that
$$C \equiv\left(\begin{array}{ll}
2 & 1 \\
4 & 3
\end{array}\right)\left(\begin{array}{ll}
4 & 0 \\
2 & 1
\end{array}\right)(\bmod 5)$$
and all entries of $C$ are nonnegative integers less than 5 . | 4. a) $\left(\begin{array}{ll}0 & 1 \\ 2 & 3\end{array}\right)$ | \left(\begin{array}{ll}0 & 1 \\ 2 & 3\end{array}\right) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,747 |
9. Use the results of problem 8 to find all solutions of each of the following systems
a)
$$\begin{array}{l}
x+y \equiv 1(\bmod 7) \\
x+z \equiv 2(\bmod 7) \\
y+z \equiv 3(\bmod 7)
\end{array}$$
b)
$$\begin{array}{l}
x+2 y+3 z \equiv 1(\bmod 7) \\
x+3 y+5 z \equiv 1(\bmod 7) \\
x+4 y+6 z \equiv 1(\bmod 7)
\end{array}$$... | 9. a) $x \equiv 0, y \equiv 1, z \equiv 2(\bmod 7)$
c) $x \equiv 5, y \equiv 5, z \equiv 5, w \equiv 5(\bmod 7)$ | x \equiv 5, y \equiv 5, z \equiv 5, w \equiv 5(\bmod 7) | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,752 |
Divisibility Test 1. If $d \mid b$ and $j$ and $k$ are positive integers with $j<k$, then $\left(a_{k} \ldots a_{1} a_{0}\right)_{b}$ is divisible by $d^{j}$ if and only if $\left(a_{j-1} \ldots a_{1} a_{0}\right)_{b}$ is divisible by $d^{j}$. | Proof. Since $b \equiv 0(\bmod d)$, Theorem 3.5 tells us that $b^{j} \equiv 0\left(\bmod d^{j}\right)$. Hence,
$$\begin{aligned}
\left(a_{k} a_{k-1} \ldots a_{1} a_{0}\right)_{b} & =a_{k} b^{k}+\cdots+a_{j} b^{j}+a_{j-1} b^{j-1}+\cdots+a_{1} b+a_{0} \\
& \equiv a_{j-1} b^{j-1}+\cdots+a_{1} b+a_{0} \\
& =\left(a_{j-1} \... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,758 |
Divisibility Test 2. If $d \mid(b-1)$, then $n=\left(a_{k} \ldots a_{1} a_{0}\right)_{b}$ is divisible by $d$ if and only if $a_{k}+\cdots+a_{1}+a_{0}$ is divisible by $d$ | Proof. Since $d \mid(b-1)$, we have $b \equiv 1(\bmod d)$, so that by Theorem 3.5 we know that $b^{j} \equiv 1(\bmod d)$ for all positive integers $b$. Hence, $\left(a_{k} \ldots a_{1} a_{0}\right)_{b}=$ $a_{k} b^{k}+\cdots+a_{1} b+a_{0} \equiv a_{k}+\cdots+a_{1}+a_{0}(\bmod d)$. This shows that $d \mid n$ if and only ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,759 |
Divisibility Test 3. If $d \mid(b+1)$, then $n=\left(a_{k} \ldots a_{1} a_{0}\right)_{b}$ is divisible by $d$ if and only if $(-1)^{k} a_{k}+\cdots-a_{1}+a_{0}$ is divisible by $d$ | Proof. Since $d \mid(b+1)$, we have $b \equiv-1(\bmod d)$. Hence, $b^{j} \equiv(-1)^{j}$ $+a_{0}(\bmod d)$. Hence, $d \mid n$ if and only if $d \mid\left((-1)^{k} a_{k}+\cdots-a_{1}\right.$ $\left.+a_{0}\right)$. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,760 |
3. Which of the following integers are divisible by 3 ? Of those that are, which are divisible by 9 ?
a) 18381
c) 987654321
b) 65412351
d) 78918239735 | 3. a) by 3 , not by 9
b) by 3 , and 9
c) by 3 , and 9
d) not by 3 | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,763 |
8. Develop a test for divisibility by 37 , based on the fact that $10^{3} \equiv 1(\bmod 37)$. Use this to check 443692 and 11092785 for divisibility by 37. | 8. $a_{2 n} a_{2 n-1} \ldots a_{1} a_{0} \equiv a_{2 n} a_{2 n-1} a_{2 n-2}+\cdots+a_{5} a_{4} a_{3}+a_{2} a_{1} a_{0}(\bmod 37)$,
$37 \backslash 443692,37 \mid 11092785$ | 37 \backslash 443692,37 \mid 11092785 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,769 |
13. We can check a multiplication $c=a b$ by determining whether the congruence $c \equiv a b(\bmod m)$ is valid, where $m$ is any modulus. If we find that $c \neq a b(\bmod m)$, then we know an error has been made. When we take $m=9$ and use the fact that an integer in decimal notation is congruent modulo 9 to the sum... | 13. a) incorrect b) incorrect c) passes casting out nines check
d) no, for example part (c) is incorrect, but passes check | c) passes casting out nines check \\ d) no, for example part (c) is incorrect, but passes check | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,774 |
1. Set up a round-robin tournament schedule for
a) 7 teams
c) 9 teams
b) 8 teams
d) 10 teams. | 1. a)
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline \begin{tabular}{l}
Team \\
Round
\end{tabular} & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
\hline 1 & 7 & 6 & 5 & bye & 3 & 2 & 1 \\
\hline 2 & bye & 7 & 6 & 5 & 4 & 3 & 2 \\
\hline 3 & 2 & 1 & 7 & 6 & byc & 4 & 3 \\
\hline 4 & 3 & bye & 1 & 7 & 6 & 5 & 4 \\
\hline 5 & 4 & 3 & 2 & 1 ... | not found | Combinatorics | math-word-problem | Yes | Yes | number_theory | false | 738,783 |
Wilson's Theorem. If $p$ is prime, then $(p-1)!\equiv-1(\bmod p)$. | Proof. When $p=2$, we have $(p-1)!\equiv 1 \equiv-1(\bmod 2)$. Hence, the theorem is true for $p=2$. Now, let $p$ be a prime greater than 2. Using Theorem 3.7, for each integer $a$ with $1 \leqslant a \leqslant p-1$, there is an inverse $\bar{a}, 1 \leqslant \bar{a} \leqslant p-1$, with $a \bar{a} \equiv 1(\bmod p)$. F... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,792 |
Fermat's Little Theorem. If $p$ is prime and $a$ is a positive integer with $p \backslash a$, then $a^{p-1} \equiv 1(\bmod p)$ | Proof. Consider the $p-1$ integers $a, 2 a, \ldots,(p-1) a$. None of these integers are divisible by $p$, for if $p \mid j a$, then by Lemma $2.3, p \mid j$, since $p \mid a$. This is impossible because $1 \leqslant j \leqslant p-1$. Furthermore, no two of the integers $a, 2 a, \ldots,(p-1) a$ are congruent modulo $p$.... | a^{p-1} \equiv 1(\bmod p) | Number Theory | proof | Yes | Yes | number_theory | false | 738,794 |
Theorem 5.2. If $p$ is prime and $a$ is a positive integer, then $a^{p} \equiv a(\bmod p)$ | Proof. If $p \nmid a$, by Fermat's little theorem we know that $a^{p-1} \equiv 1(\bmod p)$ Multiplying both sides of this congruence by $a$, we find that $a^{p} \equiv a(\bmod p)$. If $p \mid a$, then $p \mid a^{p}$ as well, so that $a^{p} \equiv a \equiv 0(\bmod p)$. This finishes the proof, since $a^{p} \equiv a(\bmo... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,795 |
Theorem 5.3. If $p$ is prime and $a$ is an integer with $p \backslash a$, then $a^{p-2}$ is an inverse of $a$ modulo $p$ | Proof. If $p \nmid a$, then Fermat's little theorem tells us that $a \cdot a^{p-2}=a^{p-1} \equiv 1(\bmod p)$. Hence, $a^{p-2}$ is an inverse of $a$ modulo $p$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,796 |
Corollary 5.1. If $a$ and $b$ are positive integers and $p$ is prime with $p \nmid a$, then the solutions of the linear congruence $a x \equiv b(\bmod p)$ are the integers $x$ such that $x \equiv a^{p-2} b(\bmod p)$. | Proof. Suppose that $a x \equiv b(\bmod p)$. Since $p \backslash a$, we know from Theorem 5.2 that $a^{p-2}$ is an inverse of $a(\bmod p)$. Multiplying both sides of the original congruence by $a^{p-2}$, we have
$$a^{p-2} a x \equiv a^{p-2} b(\bmod p)$$
Hence,
$$x \equiv a^{p-2} b(\bmod p)$$ | x \equiv a^{p-2} b(\bmod p) | Number Theory | proof | Yes | Yes | number_theory | false | 738,798 |
1. Find the values of the following sums
a) $\sum_{j=1}^{10} 2$
c) $\sum_{j=1}^{10} j^{2}$
b) $\sum_{j=1}^{10} j$
d) $\sum_{j=1}^{10} 2^{j}$. | 1. a) 20 b) 55 c) 385 d) 2046 | 2046 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,808 |
14. a) Let $p$ be prime and suppose that $r$ is a positive integer less then $p$ such that $(-1)^{r} r!\equiv-1(\bmod p)$. Show that $(p-r+1)!\equiv-1(\bmod p)$.
b) Using part (a), show that $61!\equiv 63!\equiv-1(\bmod 71)$. | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,814 |
Lemma 5.1. If $d$ and $n$ are positive integers such that $d$ divides $n$, then $2^{d}-1$ divides $2^{n}-1$ | Proof. Since $d \mid n$, there is a positive integer $t$ with $d t=n$. By setting $x=2^{d}$ in the identity $x^{t}-1=(x-1)\left(x^{t-1}+x^{t-2}+\cdots+1\right)$, we find that $2^{n}-1=\left(2^{d}-1\right)\left(2^{d(t-1)}+2^{d(t-2)}+\cdots+2^{d}+1\right)$. Consequently, $\left(2^{d}-1\right) \mid\left(2^{n}-1\right)$ | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,828 |
Theorem 5.4. There are infinitely many pseudoprimes to the base 2 . | Proof. We will show that if $n$ is an odd pseudoprime to the base 2 , then $m=2^{n}-1$ is also an odd pseudoprime to the base 2 . Since we have at least one odd pseudoprime to the base 2 , namely $n_{0}=341$, we will be able to construct infinitely many odd pseudoprimes to the base 2 by taking $n_{0}=341$ and $n_{k+1}=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,829 |
Theorem 5.5. If $n=q_{1} q_{2} \cdots q_{k}$, where the $q_{j}$ 's are distinct primes that satisfy $\left(q_{j}-1\right) \mid(n-1)$ for all $j$, then $n$ is a Carmichael number. | Proof. Let $b$ be a positive integer with $(b, n)=1$. Then $\left(b, q_{j}\right)=1$ for $j=1,2, \ldots, k$, and hence, by Fermat's little theorem, $b^{q_{j}-1} \equiv 1\left(\bmod q_{j}\right)$ for $j=1,2, \ldots, k$. Since $\left(q_{j}-1\right) \mid(n-1)$ for each integer $j=1,2, \ldots, k$, there are integers $t_{j}... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,830 |
Theorem 5.6. If $n$ is prime and $b$ is a positive integer with $n \nmid b$, then $n$ passes Miller's test for the base $b$. | Proof. Let $n-1=2^{s} t$, where $s$ is a nonnegative integer and $t$ is an odd positive integer. Let $x_{k}=b^{(n-1) / 2^{k}}=b^{2^{\prime-k} t}$, for $k=0,1,2, \ldots, s$. Since $n$ is prime, Fermat's little theorem tells us that $x_{0}=b^{n-1} \equiv 1(\bmod n)$. By Proposition $\quad 3.4, \quad$ since $\quad x_{1}^{... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,832 |
Theorem 5.7. There are infinitely many strong pseudoprimes to the base 2 . | Proof. We shall show that if $n$ is a pseudoprime to the base 2 , then $N=2^{n}-1$ is a strong pseudoprime to the base 2 .
Let $n$ be an odd integer which is a pseudoprime to the base 2 . Hence, $n$ is composite, and $2^{n-1} \equiv 1(\bmod n)$. From this congruence, we see that $2^{n-1}-1=n k$ for some integer $k$; f... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,833 |
Proposition 5.1. If the generalized Riemann hypothesis is valid, then there is an algorithm to determine whether a positive integer $n$ is prime using $O\left(\left(\log _{2} n\right)^{5}\right)$ bit operations. | Proof. Let $b$ be a positive integer less than $n$. To perform Miller's test for the base $b$ on $n$ takes $O\left(\left(\log _{2} n\right)^{3}\right)$ bit operations, because this test requires that we perform no more than $\log _{2} n$ modular exponentiations, each using $O\left(\left(\log _{2} b\right)^{2}\right)$ b... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,835 |
17. Find a Carmichael number of the form $7 \cdot 23 \cdot q$ where $q$ is an odd prime. | 17. $7 \cdot 23 \cdot 67$ | 7 \cdot 23 \cdot 67 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,854 |
Theorem 5.9. If $r_{1}, r_{2}, \ldots, r_{\phi(n)}$ is a reduced residue system modulo $n$, and if $a$ is a positive integer with $(a, n)=1$, then the set $a r_{1}, a r_{2}, \ldots, a r_{\phi(n)}$ is also a reduced residue system modulo $n$. | Proof. To show that each integer $a r_{j}$ is relatively prime to $n$, we assume that $\left(a r_{j}, n\right)>1$. Then, there is a prime divisor $p$ of $\left(a r_{j}, n\right)$. Hence, either $p \mid a$ or $p \mid r_{j}$. Thus, we either have $p \mid a$ and $p \mid n$, or $p \mid r_{j}$ and $p \mid n$. However, we ca... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,857 |
Euler's Theorem. If $m$ is a positive integer and $a$ is an integer with $(a, m)=1$, then $a^{\phi(m)} \equiv 1(\bmod m)$ | Proof. Let $r_{1}, r_{2}, \ldots, r_{\phi(m)}$ denote the reduced residue system made up of the positive integers not exceeding $m$ that are relatively prime to $m$. By Theorem 5.9 , since $(a, m)=1$, the set $a r_{1}, a r_{2}, \ldots, a r_{\phi(m)}$ is also a reduced residue system modulo $m$. Hence, the least positiv... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,858 |
1. Find a reduced residue system modulo
a) 6
d) 14
b) 9
e) 16
c) 10
f) 17 . | 1. a) 1,5
b) $1,2,4,5,7,8$
c) $1,3,7,9$
d) $1,3,5,9,11,13$
e) $1,3,5,7,9,11,13,15$ | not found | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,859 |
2. Find a reduced residue system modulo $2^{m}$, where $m$ is a positive integer. | 2. $1,3,5, \ldots, 2^{m}-1$ | 1,3,5, \ldots, 2^{m}-1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,860 |
12. Find $\phi(n)$ for the integers $n$ with $13 \leqslant n \leqslant 20$. | 12. $\phi(13)=12, \phi(14)=6, \phi(16)=8, \phi(17)=16, \phi(18)=6, \phi(19)=18, \phi(20)=8$ | 12, 6, 8, 16, 6, 18, 8 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,871 |
Theorem 6.1. If $f$ is a multiplicative function and if $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a^{\prime}}$ is the prime-power factorization of the positive integer $n$, then $f(n)=f\left(p_{1}^{a_{1}}\right) f\left(p_{2}^{a_{2}}\right) \cdots f\left(p_{s}^{a_{s}}\right)$ | Proof. Since $f$ is multiplicative and $\left(p_{1}^{a_{1}}, p_{2}^{a_{2}} \cdots p_{s}^{a}\right)=1$, we see that $f(n)=f\left(p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}}\right)=f\left(p_{1}^{a_{1}} \cdot\left(p_{2}^{a_{2}} \cdots p_{s}^{a_{s}}\right)\right)=f\left(p_{1}^{a_{1}}\right) f\left(p_{2}^{a_{2}} p_{3}^... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,874 |
Theorem 6.2. If $p$ is prime. then $\phi(p)=p-1$. Conversely, if $p$ is a positive integer with $\phi(p)=p-1$, then $p$ is prime. | Proof. If $p$ is prime then every positive integer less than $p$ is relatively prime to $p$. Since there are $p-1$ such integers, we have $\phi(p)=p-1$.
Conversely, if $p$ is composite, then $p$ has a divisor $d$ with $1<d<p$, and, of course, $p$ and $d$ are not relatively prime. Since we know that at least one of the... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,876 |
Theorem 6.3. Let $p$ be a prime and $a$ a positive integer. Then $\phi\left(p^{a}\right)=p^{a}-p^{a-1}=p^{a-1}(b-1)$ | Proof. The positive integers less than $p^{a}$ that are not relatively prime to $p$ are those integers not exceeding $p^{a}$ that are divisible by $p$. There are exactly $p^{a-1}$ such integers, so there are $p^{a}-p^{a-1}$ integers less than $p^{a}$ that are relatively prime to $p^{a}$. Hence, $\phi\left(p^{a}\right)=... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,877 |
Theorem 6.4. Let $m$ and $n$ be relatively prime positive integers. Then $\phi(m n)=\phi(m) \phi(n)$. | Proof. We display the positive integers not exceeding $m n$ in the following way.
\begin{tabular}{ccccc}
1 & $m+1$ & $2 m+1$ & $\ldots$ & $(n-1) m+1$ \\
2 & $m+2$ & $2 m+2$ & $\ldots$ & $(n-1) m+2$ \\
3 & $m+3$ & $2 m+3$ & $\ldots$ & $(n-1) m+3$ \\
& & & & \\
$\cdot$ & $\cdot$ & $\cdot$ & $\cdot$ \\
$\cdot$ & $\cdot$ &... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,878 |
Theorem 6.5. Let $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$ be the prime-power factorization of the positive integer $n$. Then
$$\phi(n)=n\left(1-\frac{1}{p_{1}}\right)\left(1-\frac{1}{p_{2}}\right) \cdots\left(1-\frac{1}{p_{k}}\right) .$$ | Proof. Since $\phi$ is multiplicative, Theorem 6.1 tells us that if the prime-power factorization of $n$ is $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{k}^{a_{k}}$, then
$$\phi(n)=\phi\left(p_{1}^{a_{1}}\right) \phi\left(p_{2}^{a_{2}}\right) \cdots \phi\left(p_{k}^{a_{k}}\right)$$
In addition, from Theorem 6.3 we know th... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,879 |
Theorem 6.6. Let $n$ be a positive integer. Then
$$\sum_{d \mid n} \phi(d)=n$$ | Proof. We split the set of integers from 1 to $n$ into classes. Put the integer $m$ into the class $C_{d}$ if the greatest common divisor of $m$ and $n$ is $d$. We see that $m$ is in $C_{d}$, i.e. $(m, n)=d$, if and only if $(m / d, n / d)=1$. Hence, the number of integers in $C_{d}$ is the number of positive integers ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,880 |
2. Find all positive integers $n$ such that $\phi(n)$ has the value
a) 1
d) 6
b) 2
e) 14
c) 3
f) 24 . | 2. a) 1,2 b) $3,4,6$ c) no solution $\quad$ d) $7,9,14$, and 18 e) no solution
f) $35,39,45,52,56,70,72,78,84,90$ | a) 1,2 \quad b) 3,4,6 \quad c) \text{no solution} \quad d) 7,9,14,18 \quad e) \text{no solution} \quad f) 35,39,45,52,56,70,72,78,84,90 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,882 |
3. For which positive integers $n$ is $\phi(n)$
a) odd
b) divisible by 4
c) equal to $n / 2$ ? | 3. a) 1,2 b) those integers $n$ such that $8|n ; 4| n$, and $n$ has at least one odd prime factor; n has at least two odd prime factors; or $n$ has a prime factor $\mathrm{p} \equiv 1(\bmod 4)$ c) $2^{k}, k=1,2, \ldots$ | a) 1,2 \quad b) n \text{ such that } 8|n; 4|n \text{ and } n \text{ has at least one odd prime factor; } n \text{ has at least two odd prime factors; or } n \text{ has a prime factor } p \equiv 1 \pmod{4} \quad c) 2^k, k=1,2,\ldots | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,883 |
14. Two arithmetic functions $f$ and $g$ may be multiplied using the Dirichlet product which is defined by
$$(f * g)(n)=\sum_{d \mid n} f(d) g(n / d)$$
a) Show that $f^{*} g=g^{*} f$.
b) Show that $\left(f^{*} g\right) * h=f *\left(g^{*} h\right)$.
c) Show that if $\iota$ is the multiplicative function defined by
$$(n)... | None | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,895 |
Theorem 6.7. If $f$ is a multiplicative function, then the arithmetic function $F(n)=\sum_{d \mid n} f(d)$ is also multiplicative. | Proof. To show that $F$ is a multiplicative function, we must show that if $m$ and $n$ are relatively prime positive integers, then $F(m n)=F(m) F(n)$. So let us assume that $(m, n)=1$. We have
$$F(m n)=\sum_{d \mid m n} f(d)$$
By Lemma 2.5 , since $(m, n)=1$, each divisor of $m n$ can be written uniquely as the produ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,909 |
Lemma 6.1. Let $p$ be prime and $a$ a positive integer. Then
$$\sigma\left(p^{a}\right)=\left(1+p+p^{2}+\cdots+p^{a}\right)=\frac{p^{a+1}-1}{p-1}$$
and
$$\tau\left(p^{a}\right)=a+1$$ | Proof. The divisors of $p^{a}$ are $1, p, p^{2}, \ldots, p^{a-1}, p^{a}$. Consequently, $p^{a}$ has exactly $a+1$ divisors, so that $\tau\left(p^{a}\right)=a+1$. Also, we note that $\sigma\left(p^{a}\right)=1+p+p_{2}+\cdots+p^{a-1}+p^{a}=\frac{p^{a+1}-1}{p-1}$, where we have used T | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,910 |
Theorem 6.8. Let the positive integer $n$ have prime factorization $n=p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}}$. Then
$$\sigma(n)=\frac{p_{1}^{a_{1}+1}-1}{p_{1}-1} \cdot \frac{p_{2}^{a_{2}+1}-1}{p_{2}-1} \cdot \cdots \cdot \frac{p_{s}^{a_{s}^{+}+1}-1}{p_{s}-1}=\prod_{j=1}^{s} \frac{p_{j}^{a_{j}+1}-1}{p_{j}-1}$$... | Proof. Since both $\sigma$ and $\tau$ are multiplicative, we see that $\sigma(n)=$ $\sigma\left(p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a^{*}}\right)=\sigma\left(p_{1}^{a_{1}}\right) \sigma\left(p_{2}^{a_{2}}\right) \cdots \sigma\left(p_{s}^{a^{*}}\right)$ and $\tau(n)=\tau\left(p_{1}^{a_{1}} p_{2}^{a_{2}}\right.$ $\... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,911 |
3. Which positive integers have an odd number of positive divisors? | 3. perfect squares | perfect\ squares | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,914 |
4. For which positive integers $n$ is the sum of divisors of $n$ odd? | 4. those positive integers that have only even powers of odd primes in their primepower factorization | those positive integers that have only even powers of odd primes in their primepower factorization | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,915 |
6. Find the smallest positive integer $n$ with $\tau(n)$ equal to
a) 1
d) 6
b) 2
e) 14
c) 3
f) 100 . | 6. a) 1
b) 2
c) 4
d) 12
e) 192
f) 45360 | 4 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,917 |
2. Find the values of the following products
a) $\prod_{j=1}^{5} 2$
c) $\prod_{j=1}^{5} j^{2}$
b) $\prod_{j=1}^{5} j$
d) $\prod_{j=1}^{5} 2^{j}$. | 2. a) 32 b) 120 c) 14400 d) 32768 | 14400 | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,919 |
9. What is the product of the positive divisors of a positive integer $n$ ? | 9. $n^{\tau(n) / 2}$ | n^{\tau(n) / 2} | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,922 |
10. Let $\sigma_{k}(n)$ denote the sum of the $k$ th powers of the divisors of $n$, so that $\sigma_{k}(n)=\sum_{d \mid n} d^{k}$. Note that $\sigma_{1}(n)=\sigma(n)$.
a) Find $\sigma_{3}(4), \sigma_{3}(6)$ and $\sigma_{3}(12)$.
b) Give a formula for $\sigma_{k}(p)$, where $p$ is prime.
c) Give a formula for $\sigma_{k... | 10. a) $73,252,2044$
b) $1+p^{k}$
c) $\left(p^{k(a+1)}-1\right) /\left(p^{k}-1\right)$
e) $\prod_{j=1}^{m}\left(p_{j}^{k(a,+1)}-1\right) /\left(p_{j}^{k}-1\right)$ | 2044, 1+p^{k}, \left(p^{k(a+1)}-1\right) /\left(p^{k}-1\right), \prod_{j=1}^{m}\left(p_{j}^{k(a_{j}+1)}-1\right) /\left(p_{j}^{k}-1\right) | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,923 |
Theorem 6.9. The positive integer $n$ is an even perfect number if and only if
$$n=2^{m-1}\left(2^{m}-1\right)$$
where $m$ is a positive integer such that $2^{m}-1$ is prime. | Proof. First, we show that if $n=2^{m-1}\left(2^{m}-1\right)$ where $2^{m}-1$ is prime, then $n$ is perfect. We note that since $2^{m}-1$ is odd, we have $\left(2^{m-1}, 2^{m}-1\right)=1$. Since $\sigma$ is a multiplicative function, we see that
$$\sigma(n)=\sigma\left(2^{m-1}\right) \sigma\left(2^{m}-1\right)$$
Lemma... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,930 |
23. a) Show that the number of positive integers less than or equal to $x$ that are divisible by the positive integer $d$ is given by $[x / d]$.
b) Find the number of positive integers not exceeding 1000 that are divisible by 5 , by 25 , by 125 , and by 625 .
c) How many integers between 100 and 1000 are divisible by 7... | 25. .) $200,40,8,1$ c) 128,18 | 200,40,8,1 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,931 |
Theorem 6.10. If $m$ is a positive integer and $2^{m}-1$ is prime, then $m$ must be prime. | Proof. Assume that $m$ is not prime, so that $m=a b$ where $1<a<m$ and $1<b<m$. Then
$$2^{m}-1=2^{a b}-1=\left(2^{a}-1\right)\left(2^{a(b-1)}+2^{a(b-2)}+\ldots+2^{a}+1\right)$$
Since both factors on the right side of the equation are greater than 1 , we see that $2^{m}-1$ is composite if $m$ is not prime. Therefore, i... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,932 |
Theorem 6.11. If $p$ is an odd prime, then any divisor of the Mersenne number $M_{p}=2^{p}-1$ is of the form $2 k p+1$ where $k$ is a positive integer. | Proof. Let $q$ be a prime dividing $M_{p}=2^{p}-1$. From Fermat's little theorem, we know that $q \mid\left(2^{q-1}-1\right)$. Also, from Lemma 3.2 we know that
$$\left(2^{p}-1,2^{q-1}-1\right)=2^{(p, q-1)}-1^{p} .$$
Since $q$ is a common divisor of $2^{p}-1$ and $2^{q-1}-1$, we know that $\left(2^{p}-1,2^{q-1}-1\righ... | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,933 |
Corollary 6.1. Let $p$ be prime and let $M_{p}=2^{p}-1$ denote the $p$ th Mersenne number. It is possible to determine whether $M_{p}$ is prime using $O\left(p^{3}\right)$ bit operations. | Proof. To determine whether $M_{p}$ is prime using the Lucas-Lehmer test requires $p-1$ squarings modulo $M_{p}$, each requiring $O\left(\left(\log M_{p}\right)^{2}\right)=O\left(p^{2}\right)$ bit operations. Hence, the Lucas-Lehmer test requires $O\left(p^{3}\right)$ bit operations. | proof | Number Theory | proof | Yes | Yes | number_theory | false | 738,934 |
1. Find the six smallest even perfect numbers. | 1. $6,28,496,8128,33550336,8589869056$ | 6,28,496,8128,33550336,8589869056 | Number Theory | math-word-problem | Yes | Yes | number_theory | false | 738,935 |
3. If $n$ is a positive integer, then we say that $n$ is deficient if $\sigma(n)2 n$. Every integer is either deficient, perfect, or abundant.
a) Find the six smallest abundant positive integers.
b) Find the smallest odd abundant positive integer.
c) Show that every prime power is deficient.
d) Show that any divisor of... | 3. a) $12,18,20,24,30,36$
b) 945 | 12,18,20,24,30,36 | Number Theory | proof | Yes | Yes | number_theory | false | 738,937 |
24. To mail a letter in the U.S.A. it costs 20 cents for the first ounce and 18 cents for each additional ounce or fraction thereof. Find a formula involving the greatest integer function for the cost of mailing a letter. Could it possibly cost $\$ 1.08$ or $\$ 1.28$ to mail a letter? | 24. $20+18[\mathrm{x}-1], \$ 1.08$ no, $\$ 1.28$ yes | 20+18[\mathrm{x}-1], \$ 1.08 \text{ no, } \$ 1.28 \text{ yes} | Algebra | math-word-problem | Yes | Yes | number_theory | false | 738,942 |
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