gowtham28/math_formated · Datasets at Hugging Face

from stringclasses 2 values	value stringlengths 16 6.77k
human	Given that $a$ is an odd multiple of $7767$, find the greatest common divisor of $6a^2+49a+108$ and $2a+9$.
gpt	We can use the Euclidean Algorithm. The closest multiple of $2a+9$ that we can spot to $6a^2 + 49a + 108$ is $6a^2 + 49a + 99 = (2a+9)(3a+11),$ so we have \begin{align} \text{gcd}\,(6a^2+49a+108,2a+9) &=\text{gcd}\,(6a^2+49a+108-(2a+9)(3a+11),2a+9)\\ &=\text{gcd}\,(6a^2+49a+108-(6a^2+49a+99),2a+9)\\ &=\text{gcd}\,(9,2a+9). \end{align}Since $7767$ is a multiple of 9, both $2a$ and $9$ are multiples of $9$, $2a+9$ is also a multiple of $9$ so the greatest common divisor is $\boxed{9}$.
human	What multiple of 15 is closest to 2009?
gpt	An integer is a multiple of 15 if and only if it is a multiple of both 3 and 5. The multiple of 5 closest to 2009 is 2010, and since $2+0+1+0$ is divisible by 3, $\boxed{2010}$ is divisible by 15.
human	What is the greatest 3-digit base 8 positive integer that is divisible by 5? (Express your answer in base 8.)
gpt	The greatest 3-digit base 8 positive integer is $777_8$, which is equal to $7 \cdot 8^2 + 7 \cdot 8 + 7 = 511$. This number leaves a remainder of 1 when divided by 5, so we subtract 1, to get $\boxed{776_8}$.
human	If $a$, $b$, and $c$ are positive integers such that $\gcd(a,b) = 168$ and $\gcd(a,c) = 693$, then what is the smallest possible value of $\gcd(b,c)$?
gpt	Note that $\gcd(168,693) = 21$. Since $\gcd(a,b) = 168 = 8 \cdot 21$, both $a$ and $b$ are divisible by 21. Since $\gcd(a,c) = 693 = 21 \cdot 33$, both $a$ and $c$ are divisible by 21. Therefore, $\gcd(b,c)$ must be at least 21. If we take $a = 5544$ (which is $21 \cdot 8 \cdot 33$), $b = 168$, and $c = 693$, then $\gcd(a,b) = \gcd(5544,168) = 168$, $\gcd(a,c) = \gcd(5544,693) = 693$, and $\gcd(b,c) = \gcd(168,693) = 21$, which shows that the value of 21 is attainable. Therefore, the smallest possible value of $\gcd(b,c)$ is $\boxed{21}$.
human	When four positive integers are divided by $11$, the remainders are $2,$ $4,$ $6,$ and $8,$ respectively. When the sum of the four integers is divided by $11$, what is the remainder?
gpt	We can call the four integers in this problem $a,$ $b,$ $c$, and $d$. Then we have \begin{align} a &\equiv 2\pmod{11}, \\ b &\equiv 4\pmod{11}, \\ c &\equiv 6\pmod{11}, \\ d &\equiv 8\pmod{11}. \end{align}Adding these congruences, we have \begin{align} a+b+c+d &\equiv 2+4+6+8 \\ &\equiv 20\pmod{11}. \end{align}Therefore, $a+b+c+d$ has the same remainder as $20$ upon division by $11$. This remainder is $\boxed{9}$.
human	There are 54 chips in a box. Each chip is either small or large. If the number of small chips is greater than the number of large chips by a prime number of chips, what is the greatest possible number of large chips?
gpt	Let $s$ be the number of small chips and $l$ be the number of large chips. From the given information, we have $s+l=54$ and $s=l+p$ for some prime $p$. Thus, $2l+p=54$. We wish to maximize $l$, so we must minimize $p$. Therefore, we set $p=2$ to get $l=\boxed{26}$.
human	The number $0.428125$ can be written as a fraction $\frac{a}{b}$ for positive integers $a$ and $b$. When this fraction is in simplest terms, what is $a+b$?
gpt	We can write $0.428125$ in the form $\frac{428,\!125}{1,\!000,\!000}$. Note that $428,\!000$ and $125$ are divisible by $5^3=125$. Therefore, we can divide numerator and denominator by 125 to obtain \begin{align} \frac{428,\!125}{1,\!000,\!000} &= \frac{125 \cdot 3425}{125 \cdot 8000}\\ &=\frac{3425}{8000}. \end{align}Since 3425 and 8000 are divisible by 25, we can simplify the fraction further: \begin{align} \frac{428,\!125}{1,\!000,\!000} &= \frac{3425}{8000} \\ &= \frac{5^2\cdot 137}{5^2\cdot 320} \\ &= \frac{137}{320}. \end{align}The sum of the numerator and denominator is $137 + 320 = \boxed{457}$.
human	What is the least positive integer that has a remainder of 0 when divided by 2, a remainder of 1 when divided by 3, and a remainder of 2 when divided by 4?
gpt	Let $a$ be the desired number. We know that \begin{align} a & \equiv 0\pmod 2\\ a & \equiv 1\pmod 3\\ a & \equiv 2\pmod 4 \end{align} Note that $a \equiv 2\pmod 4$ automatically implies $a \equiv 0\pmod 2$, so only $a \equiv 1\pmod 3$ and $a \equiv 2\pmod 4$ need to be considered. The first few positive solutions of $a \equiv 2\pmod 4$ are $2,6,10$. While the first two do not satisfy $a \equiv 1\pmod 3$, luckily $\boxed{10}$ does!
human	Jason borrowed money from his parents to buy a new surfboard. His parents have agreed to let him work off his debt by babysitting under the following conditions: his first hour of babysitting is worth $\$1$, the second hour worth $\$2$, the third hour $\$3$, the fourth hour $\$4$, the fifth hour $\$5$, the sixth hour $\$6$, the seventh hour $\$1$, the eighth hour $\$2$, etc. If he repays his debt by babysitting for 39 hours, how many dollars did he borrow?
gpt	Every $6$ hours, Jason earns $1+2+3+4+5+6=\$21$. Since $39=6(6)+3$, he earns $6\cdot 21=\$126$ from $36$ hours, and in the next $3$ hours, he earns $1+2+3=\$6$. So he borrowed $126+6=\boxed{\$132}$.
human	If $n$ is an integer, what is the remainder when the sum of $7 - n$ and $n + 3$ is divided by $7$?
gpt	We see that $(7 - n) + (n + 3) = 10 \equiv 3 \pmod 7,$ hence the remainder of the sum when divided by $7$ is $\boxed{3}.$
human	Find $73^{-1} \pmod{74}$, as a residue modulo 74. (Give an answer between 0 and 73, inclusive.)
gpt	Since $73^2 \equiv (-1)^2 \equiv 1 \pmod{74}$, $73^{-1} \equiv \boxed{73} \pmod{74}$.
human	What is the $111$th digit after the decimal point when $\frac{33}{555}$ is expressed as a decimal?
gpt	Using long division, we find that $\frac{33}{555}$ can be expressed as a repeating decimal $0.0\overline{594}$. After the first digit, there is a three-digit repeating block. We want to find the $110$th digit after the first digit. The remainder when $110$ is divided by $3$ is $2$. Therefore, the $111$th digit is the second digit in the repeating block, which is $\boxed{9}$.
human	What is the sum of the whole-number factors of 24?
gpt	The sum of the whole-number factors of 24 is $1+24+2+12+3+8+4+6=\boxed{60}$.
human	What is the greatest common divisor of $2^{1001}-1$ and $2^{1012}-1$?
gpt	By the Euclidean algorithm, \begin{align} &\text{gcd}\,(2^{1012}-1, 2^{1001}-1) \\ &\qquad= \text{gcd}\, (2^{1012}-1 - 2^{11}(2^{1001}-1), 2^{1001}-1) \\ &\qquad= \text{gcd}\,(2^{11}-1, 2^{1001}-1) \end{align} Using the divisibility rule for $11$, we know that $11$ divides into $1001$. Writing $2^{1001}$ as $(2^{11})^{91}$ and $1$ as $1^{91}$, we use the difference of odd powers factorization to find that \[ 2^{1001} - 1 = (2^{11})^{91}-1^{91} = (2^{11}-1)((2^{11})^{90} + (2^{11})^{89}+\cdots (2^{11})^1 + 1). \] Thus $2^{1001}-1$ is divisible by $2^{11}-1$, so the greatest common divisor is $2^{11}-1 = \boxed{2047}$.
human	Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.
gpt	The desired integer has at least two digits. Let $d$ be its leftmost digit, and let $n$ be the integer that results when $d$ is deleted. Then for some positive integer $p$, $10^p\cdot d+n=29n$, and so $10^p\cdot d=28n$. Therefore 7 is a divisor of $d$, and because $1\le d\le9$, it follows that $d=7$. Hence $10^p=4n$, so $\displaystyle n={{10^p}\over4}= {{100\cdot10^{p-2}}\over4}=25\cdot10^{p-2}$. Thus every positive integer with the desired property must be of the form $7\cdot10^p+25\cdot10^{p-2}=10^{p-2}(7\cdot10^2+25)=725\cdot10^{p-2}$ for some $p\ge2$. The smallest such integer is $\boxed{725}$.
human	What is the remainder when $2^{2005}$ is divided by 7?
gpt	We start from $2^1$ and consider the remainder when successive powers of 2 are divided by 7. \begin{align} 2^1 &\text{ leaves a remainder of 2}\\ 2^2 &\text{ leaves a remainder of 4}\\ 2^3 &\text{ leaves a remainder of 1}\\ 2^4 &\text{ leaves a remainder of 2}\\ 2^5 &\text{ leaves a remainder of 4}\\ 2^6 &\text{ leaves a remainder of 1}\\ &\hphantom{\text{ leaves a re}}\vdots \end{align} Since 2004 is divisible by 3 (the digits sum to 6, which is a multiple of 3), we find that $2^{2005}$ leaves a remainder of $\boxed{2}$ when divided by 7.
human	If $a,b,c$ are integers from the set of positive integers less than $7$ such that \begin{align} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align}then what is the remainder when $a+b+c$ is divided by $7$?
gpt	From the second given congruence, we have $$c\equiv 3\cdot 5c\equiv 3\cdot 2\equiv 6\pmod 7.$$From the third given congruence, we have $$5b\equiv 3\pmod 7$$$$\implies b\equiv 3\cdot 5b\equiv 3\cdot 3\equiv 2\pmod 7.$$Then, from the first given congruence, we have $$1\equiv abc a\cdot 6\cdot 2\equiv 12a\equiv 5a\pmod 7$$$$\implies a\equiv 3\cdot 5a\equiv 3\cdot 1\equiv 3\pmod 7.$$Thus, $$a+b+c\equiv 3+2+6\equiv \boxed{4}\pmod 7.$$
human	Express the sum as a common fraction: $.1 + .02 + .003 + .0004 + .00005.$
gpt	The five decimals sum to $0.12345$, which as a fraction is $\frac{12,\!345}{100,\!000}$. Since $100,\!000=2^5\cdot 5^5$, we only have to cancel factors of 2 or 5 from $12,\!345$. Since $12,\!345$ is odd, it has no factors of 2. Dividing by 5, we find that $\dfrac{12,\!345}{100,\!000}=\boxed{\dfrac{2469}{20,\!000}}$.
human	What percent of the prime numbers less than 12 are divisible by 2?
gpt	The prime numbers less than $12$ are $2,3,5,7,11$. Since $2$ is the only even prime out of the five primes less than $12$, the answer is $\frac{1}{5}=\frac{20}{100}=\boxed{20\%}$.
human	Let $N$ be the largest integer whose square has exactly $3$ digits when written in base 9. What is $N$, expressed in base 9?
gpt	Let $n$ be a positive integer. Then $n^2$ has exactly $3$ digits in base 9 if and only if $$9^2\le n^2<9^3.$$Taking square roots, we have $$3^2\le n<3^3.$$We are looking for $N$, the ${\bf largest}$ integer $n$ satisfying the above constraints. So, $$N=3^3-1=3\cdot 9-1 =2\cdot 9+8.$$Written in base $9$, this is $\boxed{28}$ or $\boxed{28_9}$.
human	Convert $135_7$ to a base 10 integer.
gpt	$135_7 = 1\cdot7^2 + 3\cdot7^1 + 5\cdot7^0 = 49 + 21 + 5 = \boxed{75}.$
human	What is the remainder when $333^{333}$ is divided by $11$?
gpt	We use the property that $a \equiv b \pmod{m}$ implies $a^c \equiv b^c \pmod{m}$. $333 \equiv 3 \pmod{11}$, therefore $333^{333} \equiv 3^{333} \pmod{11}$. Since $3^5 \equiv 1 \pmod{11}$, we get that $333^{333} \equiv 3^{333}=3^{5 \cdot 66 +3}=(3^5)^{66} \cdot 3^3 \equiv 1^{66} \cdot 27 \equiv \boxed{5} \pmod{11}$.
human	Four consecutive positive integers have a product of 840. What is the largest of the four integers?
gpt	We have $840=2^3\cdot3\cdot5\cdot7$. From this prime factorization, it is clear that the product of four consecutive positive integers is $840=2^2\cdot5\cdot(2\cdot3)\cdot7=4\cdot5\cdot6\cdot7$. The largest of the four integers is $\boxed{7}$.
human	Find the modular inverse of $27$, modulo $28$. Express your answer as an integer from $0$ to $27$, inclusive.
gpt	We are looking for an integer $a$ such that $27a$ is congruent to 1 modulo 28. In other words, we want to solve \[ 27 a \equiv 1 \pmod{28}. \]We subtract $28a$ from the left-hand side to obtain $-a\equiv 1 \pmod{28}$. This congruence is equivalent to the previous one since $28a$ is a multiple of 28. Next we multiply both sides by $-1$ to obtain $a\equiv -1\pmod{28}$. Thus $28-1=\boxed{27}$ is the modular inverse of 27 (mod 28). (Note that since $(m-1)^2=m^2-2m+1\equiv 1\pmod{m}$, we always have that $m-1$ is its own inverse modulo $m$.)
human	A $\textit{composite number}$ is a number that has two or more prime factors. The number 87 can be expressed as the sum of two composite numbers in many ways. What is the minimum positive difference between two such numbers?
gpt	The minimum difference between two numbers whose sum is 87 is achieved when the numbers are as close as possible to $87\div2=43.5$. These numbers are 43 and 44, but 43 is prime, so we consider the next pair, 42 and 45, both of which are composite. Thus, the minimum positive difference is $45-42=\boxed{3}$.
human	Suppose that $3\triangle_4=\triangle2_{11}$, where $\triangle$ represents a digit 0-9. What is the value of the missing digit?
gpt	After converting everything to base 10, we are able to solve for $\triangle$. We get \begin{align} 3\triangle_4&=\triangle2_{11}\quad\Rightarrow\\ 3\cdot4^1+\triangle\cdot4^0&=\triangle\cdot11^1+2\cdot11^0\quad\Rightarrow\\ 12+\triangle&=11\cdot\triangle+2\quad\Rightarrow\\ 10&=10\cdot\triangle\quad\Rightarrow\\ \boxed{1}&=\triangle. \end{align}
human	An eight-digit integer is formed by repeating a positive four-digit integer. For example, 25,632,563 or 60,786,078 are integers of this form. What is the greatest common divisor of all eight-digit integers of this form?
gpt	If the repeated four-digit integer is $n$, then the eight-digit integer is $10^4n+n=10001n$. So all numbers in this form share the factor 10001. Consider $10001\cdot1000$ and $10001\cdot1001$. After dividing out the factor 10001, 1000 and 1001 share no nontrivial factors, so the greatest common divisor must be exactly $\boxed{10001}$.
human	Three positive integers are each greater than $1$, have a product of $ 27000 $, and are pairwise relatively prime. What is their sum?
gpt	The prime factorization of $27000$ is $2^3\cdot 3^3\cdot 5^3.$ These three factors, $2^3,$ $3^3,$ and $5^3$ are pairwise relatively prime, and this is the only possible triple of positive integers satisfying the given conditions. Therefore, the answer is \[2^3+3^3+5^3=8+27+125=\boxed{160}.\]
human	Nine stones are arranged in a straight line. They are counted from left to right as $1,2,3, \ldots, 9$, and then from right to left, so that the stone previously counted as 8 is counted as 10. The pattern is continued to the left until the stone previously counted as 1 is counted as 17. The pattern then reverses so that the stone originally counted as 2 is counted as 18, 3 as 19, and so on. The counting continues in this manner. Which of the original stones is counted as 99? Express your answer as a single digit which corresponds to the first digit assigned to that stone.
gpt	First we note that 16 stones are enumerated before the pattern repeats. Therefore, if the count enumerates a stone as $n$, then that stone is enumerated $k$ for every \[k\equiv n\pmod{16}\] (though all but the end stones are represented by two residue classes in this way). Since $99\equiv3\pmod{16}$, stone number $\boxed{3}$ is counted as 99.
human	What is $11111111_2+111111_2$? Write your answer in base $10$.
gpt	One could add them up by carrying in base $2$. But there is a simpler way. Note that the first number is $2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7$, which, by the formula for geometric series, is $2^8-1=256-1=255$. The second number is $2^0+2^1+2^2+2^3+2^4+2^5=2^6-1=64-1=63$. Thus the sum is $255+63=305+13=\boxed{318}$.
human	What is the modular inverse of $11$, modulo $1000$? Express your answer as an integer from $0$ to $999$, inclusive.
gpt	We know that the modular inverse exists because $11$ and $1000$ are relatively prime. Notice that $1000 = 10^3$ and that $11 = 10 + 1$. Since $11 \cdot 11^{-1} \equiv 1 \pmod{1000}$, it follows that $(10+1) \cdot 11^{-1} = 10^3k + 1$ for some integer $k$. We recognize the potential sum of cubes factorization: if $k=1$, then $$10^3 + 1 = 1001 = 11 \cdot (10^2 - 10 + 1) = 11 \cdot 91.$$Thus, $11^{-1} \equiv \boxed{91} \pmod{1000}$.
human	How many zeroes does $10!$ end with, when $10!$ is written in base 9?
gpt	Let $10!$ be written in base 9 as $a_na_{n-1}\cdots a_1a_0$, where $10! = 9^na_n + 9^{n-1}a_{n-1} + \cdots + 9a_1 + a_0$, and let $k$ be the number of zeroes at the end of the base 9 expansion of $10!$. This means that $9^k$ divides $10!$ without yielding a remainder, because $9^{k-1}a_{k-1} + \cdots + 9a_1 + a_0 = 0$, and every other term on the left-hand side is divisible by $9^k$. However, since $a_k$ is nonzero, $9^{k+1}$ does not divide $10!$. Therefore, we need to find the highest power of $9$ that divides $10!$ without remainder. We can prime factorize $10!$ by prime factorizing each integer between 2 and 10. The exponent of 3 in the prime factorization of $10!$ is 4, since 3 and 6 each contribute one factor of 3 while 9 contributes two. Therefore, $9^2$ divides $10!$ while $9^3$ does not. As a result, when $10!$ is written in base 9, it ends in $\boxed{2}$ zeroes.
human	Add $175_{9} + 714_{9} + 61_9$. Express your answer in base $9$.
gpt	When summing these three numbers, we notice that $5 + 4 + 1$ leaves a residue of $1$ when divided by $9$, so it follows that the sum has a rightmost digit of $1$ and that carry-over must occur. After carrying over to the next digit, we must find the sum of $1 + 7 + 1 + 6 = 16_9$, which leaves a residue of $6$ when divided by $9$. Thus, we write down $6$ as the next digit and carry-over another $1$. Evaluating the next digit, we must find the sum of $1+1+7 = 10_9$, which leaves a residue of $0$ when divided by $9$. Thus, we must carry-over one more time, yielding that: $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & \stackrel{1}{1} & \stackrel{1}{7} & \stackrel{}{5}_{9} \\ & & & 7 & 1 & 4_{9} \\ &+ & & & 6 & 1_{9} \\ \cline{2-6} && 1 & 0 & 6 & 1_{9} \\ \end{array} $$Thus, the answer is $\boxed{1061_{9}}$. Alternatively, we can notice that $175_9 + 714_9 = 1000_9$, so $1000_9 + 61_9 = 1061_9$.
human	What is the sum of all positive integer solutions less than or equal to $20$ to the congruence $13(3x-2)\equiv 26\pmod 8$?
gpt	We can simplify as follows: \begin{align} 13(3x-2)&\equiv 26 &\pmod 8\\ 3x-2&\equiv 2 &\pmod 8\\ 3x&\equiv 4 &\pmod 8\\ 9x&\equiv 4\cdot 3 &\pmod 8\\ x&\equiv 12 &\pmod 8\\ x&\equiv 4 &\pmod 8 \end{align}So $x=4+8n$ is a solution for all $n$ and all solutions are of this form. The solutions in the range $0<x\le 20$ are $4,12,20$, so their sum is $4+12+20=\boxed{36}$.
human	What is the digit in the thousandths place of the decimal equivalent of $\frac{3}{16}$?
gpt	Since the denominator of $\dfrac{3}{16}$ is $2^4$, we multiply numerator and denominator by $5^4$ to obtain \[ \frac{3}{16} = \frac{3\cdot 5^4}{2^4\cdot 5^4} = \frac{3\cdot 625}{10^4} = \frac{1875}{10^4} = 0.1875. \] The digit in the thousandths place is $\boxed{7}$.
human	Express $249_{11}+3AB_{12}$ as a base 10 integer. Here $A$ and $B$ denote the digits whose values are 10 and 11, respectively, in base 12.
gpt	We have that \begin{align} 249_{11} &= 2(11^2)+ 4(11^1) +9(11^0) \\ &= 2(121)+4(11)+9(1)\\ &= 242 + 44 + 9\\ &= 295\\ 3AB_{12} &= 3(12^2)+ 10(12^1) +11(12^0) \\ &= 3(144)+10(12)+11(1)\\ &= 432 + 120 + 11\\ &= 563 \end{align}So, $249_{11}+3AB_{12}=295+563=\boxed{858}$.
human	What is the smallest two-digit integer $n$ such that switching its digits and then adding 3 results in $2n$?
gpt	Let $10a+b$ represent $n$, where $a$ and $b$ are the tens and units digits, respectively. Switching the digits and adding 3 results in $10b+a+3$, which we set equal to $2n$. \begin{align} 2(10a+b)&=10b+a+3\quad\Rightarrow\\ 20a+2b&=10b+a+3\quad\Rightarrow\\ 19a&=8b+3 \end{align}For the smallest $n$, we let the tens digit $a=1$. We have $19=8b+3$, which means $b=2$. So the smallest $n$ is $\boxed{12}$.
human	Kaleb defined a $\emph{clever integer}$ as an even integer that is greater than 20, less than 120, and such that the sum of its digits is 9. What fraction of all clever integers is divisible by 27? Express your answer as a common fraction.
gpt	You can easily list out the numbers between 20 and 120 that have digits that add to 9: 27, 36, 45, 54, 63, 72, 81, 90, 108, 117. To be a $\emph{clever integer}$ the number must be even, which leaves us with 36, 54, 72, 90, and 108, a total of 5 numbers. The problem asks which fraction of these five clever integers are divisible by 27. Only two, 54 and 108 are divisible by 27, thus the total fraction is $\boxed{\frac{2}{5}}$.
human	Find $10110_2\times10100_2\div10_2$. Express your answer in base 2.
gpt	It is much easier to find $10110_2\div10_2$ and then multiply by $10100_2$ than it is to do the calculations in the original order. For $10110_2\div10_2$, since the last digit of $10110_2$ is a 0, we can simply take it off to get $1011_2$. This is similar to base 10, where $10110_{10}\div10_{10}=1011_{10}$. In base 2, each place represents a power of 2, and since we're dividing by 2, each place goes down by a power 2, so each digit shifts to the right. Now we find the product of $1011_2$ and $10100_2$. $$\begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c} & & &1 &0 & 1 & 1_2 & & \\ & & & \times & 1& 0 & 1& 0 & 0_2 \\ \cline{1-9}& & &1 &0 &1 &1 & & \\ & & & & & &0 & & \\ & 1 &\stackrel{1}{0}&1 &1 &0 &0 &\downarrow &\downarrow \\ \cline{1-9} &1 &1 &0 &1 &1 &1 &0 &0_2 \\ \end{array}$$The answer is $\boxed{11011100_2}$.
human	For how many integer values of $n$ between 1 and 120 inclusive does the decimal representation of $\frac{n}{120}$ terminate?
gpt	The decimal representation of a simplified fraction terminates if and only if the denominator is divisible by no primes other than 2 and 5. The prime factorization of $120$ is $2^3 \cdot 5 \cdot 3$. For the fraction to simplify to having only the primes $2$ and $5$ in the denominator, there must be a factor of $3$ in the numerator. There are $\left\lfloor \frac{120-1}{3} \right\rfloor+1=40$ multiples of $3$ between $1$ and $120$, so there are $\boxed{40}$ integers values for $n$.
human	If $n=2^3 \cdot 3^2 \cdot 5$, how many even positive factors does $n$ have?
gpt	A positive integer is a factor of $n$ if and only if its prime factorization is of the form $2^a\cdot 3^b\cdot 5^c$ where $0\leq a\leq 3$, $0\leq b\leq 2$, and $0\leq c\leq 1$. An integer is even if and only if the exponent of 2 in its prime factorization is at least 1. Therefore, we have 3 choices for $a$, 3 choices for $b$ and $2$ choices for $c$, for a total of $(3)(3)(2)=\boxed{18}$ ways to form an even positive factor of $n$.
human	Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?
gpt	Note that $x + 2010 \equiv x \pmod{2010}$. Add 2010 to every negative summand and rearrange the terms to find $S \equiv 0 + 1 + 2 + 3 + \cdots + 2008 + 2009 \pmod{2010}$. The right-hand side is the sum of the integers from 1 to 2010, so $S \equiv \frac{2010 \cdot 2011}{2} \equiv 1005 \cdot 2011 \equiv 1005 \cdot 1 \equiv \boxed{1005} \pmod{2010}$.
human	Sherlock Holmes and Dr. Watson recover a suitcase with a three-digit combination lock from a mathematician turned criminal. Embedded in the suitcase above the lock is the cryptic message "AT SEA BASE. SEAS EBB SEA: BASS. " Dr. Watson comments, "This probably isn't about ocean fish. Perhaps it is an encrypted message. The colon suggests the addition problem $SEAS + EBB + SEA = BASS$, where each letter represents a distinct digit, and the word 'BASE' implies that the problem is in a different base." Holmes calmly turns the combination lock and opens the suitcase. While Dr. Watson gapes in surprise, Holmes replies, "You were indeed right, and the answer was just the value of the word $SEA$, interpreted as decimal digits." What was the lock combination?
gpt	We must solve the addition problem $$ \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c} & & S & E & A & S_d \\ & & & E & B & B_d \\ + & & & S & E & A_d\\ \cline{1-6} & & B & A& S& S_d\end{array},$$ where $d$ is an unknown base. It follows that $S + B + A$ leaves a residue of $S$ upon division by $d$. Thus, $B+A$ must be divisible by $d$. Since $B$ and $A$ cannot both be $0$, and $B+A < (d-1) + (d-1) = 2d-2$, then $B + A = d$. Looking at the $d$s digit, we must carry-over $1$ from the units digit sum, so $1 + A + B + E \equiv S \pmod{d}$. Since $B + A = d$, then $1 + E + d \equiv 1+E \equiv S \pmod{d}$. Thus, $S = E+1$ or $E = d-1$ and $S = 0$. However, the latter is impossible since $S$ is the leftmost digit of 'SEAS' and 'SEA'. Thus, $S = E+1$, and we again carry-over $1$ to the $d^2$ digit. Looking at the $d^2$ digit, after carry-over, it follows that $1 + E + E + S \equiv A \pmod{d}$. Note that $1 + E + E + S < 1 + 3(d-1) = 3d - 2 < 3d$. Then, $2E + S + 1 - A$ is either equal to $0$, $d$, or $2d$. However, we can immediately discard the $0$ case: there would be no carry-over for the leftmost digit, so $S = B$ are not distinct. In the next case, if $2E + S + 1 = A + d$, then there is a carry-over of $1$ to the last digit. It follows that $S + 1 = B$. This gives us the system of equations \begin{align} B + A &= d \\ E + 1 &= S \\ S + 1 &= B \\ 2E + S +1 - A&= d \end{align} Setting the first and fourth equations equal to each other yields that $d = B+A = 2E + S +1 - A$, and since $B = S+1 = E+2$, substituting for $B$ and $S$ yields that $2A = 3E + S + 1 - B = 2E + (E+1) + 1 - (E+2) = 2E$. This contradicts the distinct digits criterion. Thus, $2E + S + 1 - A= 2d = 2(B+A)$, so $2E + S + 1 - 2B = 3A$. Also, we have that $B = S+2$, due to the carry-over in the leftmost digit. Substituting for $B$ and $S$ yields that $3A = 2E + (E+1) + 1 - 2(E + 3) = E - 4$, so $E = 3A+4$. Thus, $S = 3A+5$ and $B=3A+7$. Also, $S,E,$ and $A$ are decimal digits, so it follows that $S = 3A + 5 \le 9 \Longrightarrow A = 0,1$. We can discard the solution $A = 0$, as $d = B+A$ but $B < d$. Thus, $B = 10, S = 8, E = 7$, occurring in base $d = B+A = 11$. The answer is $\boxed{871}$.
human	For a nonnegative integer $n$, let $r_9(n)$ stand for the remainder left when $n$ is divided by $9.$ For example, $r_9(25)=7.$ What is the $22^{\text{nd}}$ entry in an ordered list of all nonnegative integers $n$ that satisfy $$r_9(5n)\le 4~?$$(Note that the first entry in this list is $0$.)
gpt	The condition $r_9(5n)\le 4$ can also be stated as $``5n\equiv 0,1,2,3,\text{ or }4\pmod 9."$' We can then restate that condition again by multiplying both sides by $2:$ $$10n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$This step is reversible (since $2$ has an inverse modulo $9$). Thus, it neither creates nor removes solutions. Moreover, the left side reduces to $n$ modulo $9,$ giving us the precise solution set $$n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$We wish to determine the $22^{\text{nd}}$ nonnegative integer in this solution set. The first few solutions follow this pattern: $$\begin{array}{c c c c c} 0 & 2 & 4 & 6 & 8 \\ 9 & 11 & 13 & 15 & 17 \\ 18 & 20 & 22 & 24 & 26 \\ 27 & 29 & 31 & 33 & 35 \\ 36 & 38 & \cdots \end{array}$$The $22^{\text{nd}}$ solution is $\boxed{38}.$
human	Suppose that $b$ is a positive integer greater than or equal to $2.$ When $197$ is converted to base $b$, the resulting representation has $4$ digits. What is the number of possible values for $b$?
gpt	We would like to find $b$ for which $b^3 \leq 197_{10} < b^4.$ We can see that this inequality is satisfied for $4\leq b \leq 5.$ So, there are $\boxed{2}$ bases for which this holds.
human	A natural number is abundant if it is less than the sum of its proper divisors. What is the smallest abundant number?
gpt	For any prime number, the sum of its proper divisors is equal to $1$, so a prime number cannot be an abundant number. Therefore, it suffices to check only composite numbers: $\bullet$ For $4$, $1 + 2 < 4$, $\bullet$ For $6$, $1 + 2 + 3 = 6$, $\bullet$ For $8$, $1 + 2 + 4 < 8$, $\bullet$ For $9$, $1 + 3 < 9$, $\bullet$ For $10$, $1 + 2 + 5 < 10$, $\bullet$ For $12$, $1 + 2 + 3 + 4 + 6 = 16 > 12$. Thus, the answer is $\boxed{12}$.
human	The time right now is exactly midnight. What time will it be in 1234 minutes?
gpt	There are $60$ minutes in an hour. When $1234$ is divided by $60$, you get $20$ with a remainder of $34$. Therefore, the time in $1234$ minutes will be $\boxed{20\!:\!34}$ or $\boxed{8\!:\!34 \text{ p.m.}}$.
human	Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? $$ \text A. \ \ 21 \qquad \text B. \ \ 60 \qquad \text C. \ \ 119 \qquad \text D. \ \ 180 \qquad \text E. \ \ 231 $$
gpt	There are five prime numbers between $4$ and $18:$ namely $5,$ $7,$ $11,$ $13,$ and $17.$ Hence the product of any two of these is odd and the sum is even. Because $$xy-(x+y)=(x-1)(y-1)-1$$increases as either $x$ or $y$ increases (since both $x$ and $y$ are bigger than $1$), the answer must be an odd number that is no smaller than $$23=5\cdot 7-(5+7)$$and no larger than $$191=13\cdot 17-(13+17).$$The only possibility among the options is $\boxed{119},$ and indeed $119=11\cdot 13-(11+13).$
human	What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{1050}$ is defined?
gpt	For $n$ to have an inverse $\pmod{1050}$, it is necessary for $n$ to be relatively prime to $1050$. Conversely, if $n$ is relatively prime to $1050$, then $n$ has an inverse $\pmod{1050}$. The prime factors of $1050$ include $2$, $3$, $5$, and $7$, so any multiple of any of these primes does not have an inverse $\pmod{1050}$. This rules out all the integers from $2$ to $10$. However, $11$ is relatively prime to $1050$, so $\boxed{11}$ is the smallest integer greater than $1$ that has an inverse $\pmod{1050}$.
human	What is the units digit of $\frac{20 \cdot 21 \cdot 22 \cdot 23 \cdot 24 \cdot 25}{1000}$?
gpt	$\frac{20\cdot 21\cdot 22\cdot 23\cdot 24\cdot 25}{1000} = \frac{2^2\cdot 5\cdot 21\cdot 2\cdot 11\cdot 23\cdot 2^3\cdot 3\cdot 5^2}{2^3\cdot 5^3} = 2^3\cdot 3\cdot 21 \cdot 11\cdot 23 \equiv 2^3\cdot 3^2 \pmod{10} \equiv \boxed{2}\pmod{10}$.

human

Given that $a$ is an odd multiple of $7767$, find the greatest common divisor of $6a^2+49a+108$ and $2a+9$.

gpt

We can use the Euclidean Algorithm. The closest multiple of $2a+9$ that we can spot to $6a^2 + 49a + 108$ is $6a^2 + 49a + 99 = (2a+9)(3a+11),$ so we have \begin{align*} \text{gcd}\,(6a^2+49a+108,2a+9) &=\text{gcd}\,(6a^2+49a+108-(2a+9)(3a+11),2a+9)\\ &=\text{gcd}\,(6a^2+49a+108-(6a^2+49a+99),2a+9)\\ &=\text{gcd}\,(9,2a+9). \end{align*}Since $7767$ is a multiple of 9, both $2a$ and $9$ are multiples of $9$, $2a+9$ is also a multiple of $9$ so the greatest common divisor is $\boxed{9}$.

human

What multiple of 15 is closest to 2009?

gpt

An integer is a multiple of 15 if and only if it is a multiple of both 3 and 5. The multiple of 5 closest to 2009 is 2010, and since $2+0+1+0$ is divisible by 3, $\boxed{2010}$ is divisible by 15.

human

What is the greatest 3-digit base 8 positive integer that is divisible by 5? (Express your answer in base 8.)

gpt

The greatest 3-digit base 8 positive integer is $777_8$, which is equal to $7 \cdot 8^2 + 7 \cdot 8 + 7 = 511$. This number leaves a remainder of 1 when divided by 5, so we subtract 1, to get $\boxed{776_8}$.

human

If $a$, $b$, and $c$ are positive integers such that $\gcd(a,b) = 168$ and $\gcd(a,c) = 693$, then what is the smallest possible value of $\gcd(b,c)$?

gpt

Note that $\gcd(168,693) = 21$. Since $\gcd(a,b) = 168 = 8 \cdot 21$, both $a$ and $b$ are divisible by 21. Since $\gcd(a,c) = 693 = 21 \cdot 33$, both $a$ and $c$ are divisible by 21. Therefore, $\gcd(b,c)$ must be at least 21. If we take $a = 5544$ (which is $21 \cdot 8 \cdot 33$), $b = 168$, and $c = 693$, then $\gcd(a,b) = \gcd(5544,168) = 168$, $\gcd(a,c) = \gcd(5544,693) = 693$, and $\gcd(b,c) = \gcd(168,693) = 21$, which shows that the value of 21 is attainable. Therefore, the smallest possible value of $\gcd(b,c)$ is $\boxed{21}$.

human

When four positive integers are divided by $11$, the remainders are $2,$ $4,$ $6,$ and $8,$ respectively. When the sum of the four integers is divided by $11$, what is the remainder?

gpt

We can call the four integers in this problem $a,$ $b,$ $c$, and $d$. Then we have \begin{align*} a &\equiv 2\pmod{11}, \\ b &\equiv 4\pmod{11}, \\ c &\equiv 6\pmod{11}, \\ d &\equiv 8\pmod{11}. \end{align*}Adding these congruences, we have \begin{align*} a+b+c+d &\equiv 2+4+6+8 \\ &\equiv 20\pmod{11}. \end{align*}Therefore, $a+b+c+d$ has the same remainder as $20$ upon division by $11$. This remainder is $\boxed{9}$.

human

There are 54 chips in a box. Each chip is either small or large. If the number of small chips is greater than the number of large chips by a prime number of chips, what is the greatest possible number of large chips?

gpt

Let $s$ be the number of small chips and $l$ be the number of large chips. From the given information, we have $s+l=54$ and $s=l+p$ for some prime $p$. Thus, $2l+p=54$. We wish to maximize $l$, so we must minimize $p$. Therefore, we set $p=2$ to get $l=\boxed{26}$.

human

The number $0.428125$ can be written as a fraction $\frac{a}{b}$ for positive integers $a$ and $b$. When this fraction is in simplest terms, what is $a+b$?

gpt

We can write $0.428125$ in the form $\frac{428,\!125}{1,\!000,\!000}$. Note that $428,\!000$ and $125$ are divisible by $5^3=125$. Therefore, we can divide numerator and denominator by 125 to obtain \begin{align*} \frac{428,\!125}{1,\!000,\!000} &= \frac{125 \cdot 3425}{125 \cdot 8000}\\ &=\frac{3425}{8000}. \end{align*}Since 3425 and 8000 are divisible by 25, we can simplify the fraction further: \begin{align*} \frac{428,\!125}{1,\!000,\!000} &= \frac{3425}{8000} \\ &= \frac{5^2\cdot 137}{5^2\cdot 320} \\ &= \frac{137}{320}. \end{align*}The sum of the numerator and denominator is $137 + 320 = \boxed{457}$.

human

What is the least positive integer that has a remainder of 0 when divided by 2, a remainder of 1 when divided by 3, and a remainder of 2 when divided by 4?

gpt

Let $a$ be the desired number. We know that \begin{align*} a & \equiv 0\pmod 2\\ a & \equiv 1\pmod 3\\ a & \equiv 2\pmod 4 \end{align*} Note that $a \equiv 2\pmod 4$ automatically implies $a \equiv 0\pmod 2$, so only $a \equiv 1\pmod 3$ and $a \equiv 2\pmod 4$ need to be considered. The first few positive solutions of $a \equiv 2\pmod 4$ are $2,6,10$. While the first two do not satisfy $a \equiv 1\pmod 3$, luckily $\boxed{10}$ does!

human

Jason borrowed money from his parents to buy a new surfboard. His parents have agreed to let him work off his debt by babysitting under the following conditions: his first hour of babysitting is worth $\$1$, the second hour worth $\$2$, the third hour $\$3$, the fourth hour $\$4$, the fifth hour $\$5$, the sixth hour $\$6$, the seventh hour $\$1$, the eighth hour $\$2$, etc. If he repays his debt by babysitting for 39 hours, how many dollars did he borrow?

gpt

Every $6$ hours, Jason earns $1+2+3+4+5+6=\$21$. Since $39=6(6)+3$, he earns $6\cdot 21=\$126$ from $36$ hours, and in the next $3$ hours, he earns $1+2+3=\$6$. So he borrowed $126+6=\boxed{\$132}$.

human

If $n$ is an integer, what is the remainder when the sum of $7 - n$ and $n + 3$ is divided by $7$?

gpt

We see that $(7 - n) + (n + 3) = 10 \equiv 3 \pmod 7,$ hence the remainder of the sum when divided by $7$ is $\boxed{3}.$

human

Find $73^{-1} \pmod{74}$, as a residue modulo 74. (Give an answer between 0 and 73, inclusive.)

gpt

Since $73^2 \equiv (-1)^2 \equiv 1 \pmod{74}$, $73^{-1} \equiv \boxed{73} \pmod{74}$.

human

What is the $111$th digit after the decimal point when $\frac{33}{555}$ is expressed as a decimal?

gpt

Using long division, we find that $\frac{33}{555}$ can be expressed as a repeating decimal $0.0\overline{594}$. After the first digit, there is a three-digit repeating block. We want to find the $110$th digit after the first digit. The remainder when $110$ is divided by $3$ is $2$. Therefore, the $111$th digit is the second digit in the repeating block, which is $\boxed{9}$.

human

What is the sum of the whole-number factors of 24?

gpt

The sum of the whole-number factors of 24 is $1+24+2+12+3+8+4+6=\boxed{60}$.

human

What is the greatest common divisor of $2^{1001}-1$ and $2^{1012}-1$?

gpt

By the Euclidean algorithm, \begin{align*} &\text{gcd}\,(2^{1012}-1, 2^{1001}-1) \\ &\qquad= \text{gcd}\, (2^{1012}-1 - 2^{11}(2^{1001}-1), 2^{1001}-1) \\ &\qquad= \text{gcd}\,(2^{11}-1, 2^{1001}-1) \end{align*} Using the divisibility rule for $11$, we know that $11$ divides into $1001$. Writing $2^{1001}$ as $(2^{11})^{91}$ and $1$ as $1^{91}$, we use the difference of odd powers factorization to find that \[ 2^{1001} - 1 = (2^{11})^{91}-1^{91} = (2^{11}-1)((2^{11})^{90} + (2^{11})^{89}+\cdots (2^{11})^1 + 1). \] Thus $2^{1001}-1$ is divisible by $2^{11}-1$, so the greatest common divisor is $2^{11}-1 = \boxed{2047}$.

human

Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is 1/29 of the original integer.

gpt

The desired integer has at least two digits. Let $d$ be its leftmost digit, and let $n$ be the integer that results when $d$ is deleted. Then for some positive integer $p$, $10^p\cdot d+n=29n$, and so $10^p\cdot d=28n$. Therefore 7 is a divisor of $d$, and because $1\le d\le9$, it follows that $d=7$. Hence $10^p=4n$, so $\displaystyle n={{10^p}\over4}= {{100\cdot10^{p-2}}\over4}=25\cdot10^{p-2}$. Thus every positive integer with the desired property must be of the form $7\cdot10^p+25\cdot10^{p-2}=10^{p-2}(7\cdot10^2+25)=725\cdot10^{p-2}$ for some $p\ge2$. The smallest such integer is $\boxed{725}$.

human

What is the remainder when $2^{2005}$ is divided by 7?

gpt

We start from $2^1$ and consider the remainder when successive powers of 2 are divided by 7. \begin{align*} 2^1 &\text{ leaves a remainder of 2}\\ 2^2 &\text{ leaves a remainder of 4}\\ 2^3 &\text{ leaves a remainder of 1}\\ 2^4 &\text{ leaves a remainder of 2}\\ 2^5 &\text{ leaves a remainder of 4}\\ 2^6 &\text{ leaves a remainder of 1}\\ &\hphantom{\text{ leaves a re}}\vdots \end{align*} Since 2004 is divisible by 3 (the digits sum to 6, which is a multiple of 3), we find that $2^{2005}$ leaves a remainder of $\boxed{2}$ when divided by 7.

human

If $a,b,c$ are integers from the set of positive integers less than $7$ such that \begin{align*} abc&\equiv 1\pmod 7,\\ 5c&\equiv 2\pmod 7,\\ 6b&\equiv 3+b\pmod 7, \end{align*}then what is the remainder when $a+b+c$ is divided by $7$?

gpt

From the second given congruence, we have $$c\equiv 3\cdot 5c\equiv 3\cdot 2\equiv 6\pmod 7.$$From the third given congruence, we have $$5b\equiv 3\pmod 7$$$$\implies b\equiv 3\cdot 5b\equiv 3\cdot 3\equiv 2\pmod 7.$$Then, from the first given congruence, we have $$1\equiv abc a\cdot 6\cdot 2\equiv 12a\equiv 5a\pmod 7$$$$\implies a\equiv 3\cdot 5a\equiv 3\cdot 1\equiv 3\pmod 7.$$Thus, $$a+b+c\equiv 3+2+6\equiv \boxed{4}\pmod 7.$$

human

Express the sum as a common fraction: $.1 + .02 + .003 + .0004 + .00005.$

gpt

The five decimals sum to $0.12345$, which as a fraction is $\frac{12,\!345}{100,\!000}$. Since $100,\!000=2^5\cdot 5^5$, we only have to cancel factors of 2 or 5 from $12,\!345$. Since $12,\!345$ is odd, it has no factors of 2. Dividing by 5, we find that $\dfrac{12,\!345}{100,\!000}=\boxed{\dfrac{2469}{20,\!000}}$.

human

What percent of the prime numbers less than 12 are divisible by 2?

gpt

The prime numbers less than $12$ are $2,3,5,7,11$. Since $2$ is the only even prime out of the five primes less than $12$, the answer is $\frac{1}{5}=\frac{20}{100}=\boxed{20\%}$.

human

Let $N$ be the largest integer whose square has exactly $3$ digits when written in base 9. What is $N$, expressed in base 9?

gpt

Let $n$ be a positive integer. Then $n^2$ has exactly $3$ digits in base 9 if and only if $$9^2\le n^2<9^3.$$Taking square roots, we have $$3^2\le n<3^3.$$We are looking for $N$, the ${\bf largest}$ integer $n$ satisfying the above constraints. So, $$N=3^3-1=3\cdot 9-1 =2\cdot 9+8.$$Written in base $9$, this is $\boxed{28}$ or $\boxed{28_9}$.

human

Convert $135_7$ to a base 10 integer.

gpt

$135_7 = 1\cdot7^2 + 3\cdot7^1 + 5\cdot7^0 = 49 + 21 + 5 = \boxed{75}.$

human

What is the remainder when $333^{333}$ is divided by $11$?

gpt

We use the property that $a \equiv b \pmod{m}$ implies $a^c \equiv b^c \pmod{m}$. $333 \equiv 3 \pmod{11}$, therefore $333^{333} \equiv 3^{333} \pmod{11}$. Since $3^5 \equiv 1 \pmod{11}$, we get that $333^{333} \equiv 3^{333}=3^{5 \cdot 66 +3}=(3^5)^{66} \cdot 3^3 \equiv 1^{66} \cdot 27 \equiv \boxed{5} \pmod{11}$.

human

Four consecutive positive integers have a product of 840. What is the largest of the four integers?

gpt

We have $840=2^3\cdot3\cdot5\cdot7$. From this prime factorization, it is clear that the product of four consecutive positive integers is $840=2^2\cdot5\cdot(2\cdot3)\cdot7=4\cdot5\cdot6\cdot7$. The largest of the four integers is $\boxed{7}$.

human

Find the modular inverse of $27$, modulo $28$. Express your answer as an integer from $0$ to $27$, inclusive.

gpt

We are looking for an integer $a$ such that $27a$ is congruent to 1 modulo 28. In other words, we want to solve \[ 27 a \equiv 1 \pmod{28}. \]We subtract $28a$ from the left-hand side to obtain $-a\equiv 1 \pmod{28}$. This congruence is equivalent to the previous one since $28a$ is a multiple of 28. Next we multiply both sides by $-1$ to obtain $a\equiv -1\pmod{28}$. Thus $28-1=\boxed{27}$ is the modular inverse of 27 (mod 28). (Note that since $(m-1)^2=m^2-2m+1\equiv 1\pmod{m}$, we always have that $m-1$ is its own inverse modulo $m$.)

human

A $\textit{composite number}$ is a number that has two or more prime factors. The number 87 can be expressed as the sum of two composite numbers in many ways. What is the minimum positive difference between two such numbers?

gpt

The minimum difference between two numbers whose sum is 87 is achieved when the numbers are as close as possible to $87\div2=43.5$. These numbers are 43 and 44, but 43 is prime, so we consider the next pair, 42 and 45, both of which are composite. Thus, the minimum positive difference is $45-42=\boxed{3}$.

human

Suppose that $3\triangle_4=\triangle2_{11}$, where $\triangle$ represents a digit 0-9. What is the value of the missing digit?

gpt

After converting everything to base 10, we are able to solve for $\triangle$. We get \begin{align*} 3\triangle_4&=\triangle2_{11}\quad\Rightarrow\\ 3\cdot4^1+\triangle\cdot4^0&=\triangle\cdot11^1+2\cdot11^0\quad\Rightarrow\\ 12+\triangle&=11\cdot\triangle+2\quad\Rightarrow\\ 10&=10\cdot\triangle\quad\Rightarrow\\ \boxed{1}&=\triangle. \end{align*}

human

An eight-digit integer is formed by repeating a positive four-digit integer. For example, 25,632,563 or 60,786,078 are integers of this form. What is the greatest common divisor of all eight-digit integers of this form?

gpt

If the repeated four-digit integer is $n$, then the eight-digit integer is $10^4n+n=10001n$. So all numbers in this form share the factor 10001. Consider $10001\cdot1000$ and $10001\cdot1001$. After dividing out the factor 10001, 1000 and 1001 share no nontrivial factors, so the greatest common divisor must be exactly $\boxed{10001}$.

human

Three positive integers are each greater than $1$, have a product of $ 27000 $, and are pairwise relatively prime. What is their sum?

gpt

The prime factorization of $27000$ is $2^3\cdot 3^3\cdot 5^3.$ These three factors, $2^3,$ $3^3,$ and $5^3$ are pairwise relatively prime, and this is the only possible triple of positive integers satisfying the given conditions. Therefore, the answer is \[2^3+3^3+5^3=8+27+125=\boxed{160}.\]

human

Nine stones are arranged in a straight line. They are counted from left to right as $1,2,3, \ldots, 9$, and then from right to left, so that the stone previously counted as 8 is counted as 10. The pattern is continued to the left until the stone previously counted as 1 is counted as 17. The pattern then reverses so that the stone originally counted as 2 is counted as 18, 3 as 19, and so on. The counting continues in this manner. Which of the original stones is counted as 99? Express your answer as a single digit which corresponds to the first digit assigned to that stone.

gpt

First we note that 16 stones are enumerated before the pattern repeats. Therefore, if the count enumerates a stone as $n$, then that stone is enumerated $k$ for every \[k\equiv n\pmod{16}\] (though all but the end stones are represented by two residue classes in this way). Since $99\equiv3\pmod{16}$, stone number $\boxed{3}$ is counted as 99.

human

What is $11111111_2+111111_2$? Write your answer in base $10$.

gpt

One could add them up by carrying in base $2$. But there is a simpler way. Note that the first number is $2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7$, which, by the formula for geometric series, is $2^8-1=256-1=255$. The second number is $2^0+2^1+2^2+2^3+2^4+2^5=2^6-1=64-1=63$. Thus the sum is $255+63=305+13=\boxed{318}$.

human

What is the modular inverse of $11$, modulo $1000$? Express your answer as an integer from $0$ to $999$, inclusive.

gpt

We know that the modular inverse exists because $11$ and $1000$ are relatively prime. Notice that $1000 = 10^3$ and that $11 = 10 + 1$. Since $11 \cdot 11^{-1} \equiv 1 \pmod{1000}$, it follows that $(10+1) \cdot 11^{-1} = 10^3k + 1$ for some integer $k$. We recognize the potential sum of cubes factorization: if $k=1$, then $$10^3 + 1 = 1001 = 11 \cdot (10^2 - 10 + 1) = 11 \cdot 91.$$Thus, $11^{-1} \equiv \boxed{91} \pmod{1000}$.

human

How many zeroes does $10!$ end with, when $10!$ is written in base 9?

gpt

Let $10!$ be written in base 9 as $a_na_{n-1}\cdots a_1a_0$, where $10! = 9^na_n + 9^{n-1}a_{n-1} + \cdots + 9a_1 + a_0$, and let $k$ be the number of zeroes at the end of the base 9 expansion of $10!$. This means that $9^k$ divides $10!$ without yielding a remainder, because $9^{k-1}a_{k-1} + \cdots + 9a_1 + a_0 = 0$, and every other term on the left-hand side is divisible by $9^k$. However, since $a_k$ is nonzero, $9^{k+1}$ does not divide $10!$. Therefore, we need to find the highest power of $9$ that divides $10!$ without remainder. We can prime factorize $10!$ by prime factorizing each integer between 2 and 10. The exponent of 3 in the prime factorization of $10!$ is 4, since 3 and 6 each contribute one factor of 3 while 9 contributes two. Therefore, $9^2$ divides $10!$ while $9^3$ does not. As a result, when $10!$ is written in base 9, it ends in $\boxed{2}$ zeroes.

human

Add $175_{9} + 714_{9} + 61_9$. Express your answer in base $9$.

gpt

When summing these three numbers, we notice that $5 + 4 + 1$ leaves a residue of $1$ when divided by $9$, so it follows that the sum has a rightmost digit of $1$ and that carry-over must occur. After carrying over to the next digit, we must find the sum of $1 + 7 + 1 + 6 = 16_9$, which leaves a residue of $6$ when divided by $9$. Thus, we write down $6$ as the next digit and carry-over another $1$. Evaluating the next digit, we must find the sum of $1+1+7 = 10_9$, which leaves a residue of $0$ when divided by $9$. Thus, we must carry-over one more time, yielding that: $$ \begin{array}{c@{}c@{\;}c@{}c@{}c@{}c} & & & \stackrel{1}{1} & \stackrel{1}{7} & \stackrel{}{5}_{9} \\ & & & 7 & 1 & 4_{9} \\ &+ & & & 6 & 1_{9} \\ \cline{2-6} && 1 & 0 & 6 & 1_{9} \\ \end{array} $$Thus, the answer is $\boxed{1061_{9}}$. Alternatively, we can notice that $175_9 + 714_9 = 1000_9$, so $1000_9 + 61_9 = 1061_9$.

human

What is the sum of all positive integer solutions less than or equal to $20$ to the congruence $13(3x-2)\equiv 26\pmod 8$?

gpt

We can simplify as follows: \begin{align*} 13(3x-2)&\equiv 26 &\pmod 8\\ 3x-2&\equiv 2 &\pmod 8\\ 3x&\equiv 4 &\pmod 8\\ 9x&\equiv 4\cdot 3 &\pmod 8\\ x&\equiv 12 &\pmod 8\\ x&\equiv 4 &\pmod 8 \end{align*}So $x=4+8n$ is a solution for all $n$ and all solutions are of this form. The solutions in the range $0<x\le 20$ are $4,12,20$, so their sum is $4+12+20=\boxed{36}$.

human

What is the digit in the thousandths place of the decimal equivalent of $\frac{3}{16}$?

gpt

Since the denominator of $\dfrac{3}{16}$ is $2^4$, we multiply numerator and denominator by $5^4$ to obtain \[ \frac{3}{16} = \frac{3\cdot 5^4}{2^4\cdot 5^4} = \frac{3\cdot 625}{10^4} = \frac{1875}{10^4} = 0.1875. \] The digit in the thousandths place is $\boxed{7}$.

human

Express $249_{11}+3AB_{12}$ as a base 10 integer. Here $A$ and $B$ denote the digits whose values are 10 and 11, respectively, in base 12.

gpt

We have that \begin{align*} 249_{11} &= 2(11^2)+ 4(11^1) +9(11^0) \\ &= 2(121)+4(11)+9(1)\\ &= 242 + 44 + 9\\ &= 295\\ 3AB_{12} &= 3(12^2)+ 10(12^1) +11(12^0) \\ &= 3(144)+10(12)+11(1)\\ &= 432 + 120 + 11\\ &= 563 \end{align*}So, $249_{11}+3AB_{12}=295+563=\boxed{858}$.

human

What is the smallest two-digit integer $n$ such that switching its digits and then adding 3 results in $2n$?

gpt

Let $10a+b$ represent $n$, where $a$ and $b$ are the tens and units digits, respectively. Switching the digits and adding 3 results in $10b+a+3$, which we set equal to $2n$. \begin{align*} 2(10a+b)&=10b+a+3\quad\Rightarrow\\ 20a+2b&=10b+a+3\quad\Rightarrow\\ 19a&=8b+3 \end{align*}For the smallest $n$, we let the tens digit $a=1$. We have $19=8b+3$, which means $b=2$. So the smallest $n$ is $\boxed{12}$.

human

Kaleb defined a $\emph{clever integer}$ as an even integer that is greater than 20, less than 120, and such that the sum of its digits is 9. What fraction of all clever integers is divisible by 27? Express your answer as a common fraction.

gpt

You can easily list out the numbers between 20 and 120 that have digits that add to 9: 27, 36, 45, 54, 63, 72, 81, 90, 108, 117. To be a $\emph{clever integer}$ the number must be even, which leaves us with 36, 54, 72, 90, and 108, a total of 5 numbers. The problem asks which fraction of these five clever integers are divisible by 27. Only two, 54 and 108 are divisible by 27, thus the total fraction is $\boxed{\frac{2}{5}}$.

human

Find $10110_2\times10100_2\div10_2$. Express your answer in base 2.

gpt

It is much easier to find $10110_2\div10_2$ and then multiply by $10100_2$ than it is to do the calculations in the original order. For $10110_2\div10_2$, since the last digit of $10110_2$ is a 0, we can simply take it off to get $1011_2$. This is similar to base 10, where $10110_{10}\div10_{10}=1011_{10}$. In base 2, each place represents a power of 2, and since we're dividing by 2, each place goes down by a power 2, so each digit shifts to the right. Now we find the product of $1011_2$ and $10100_2$. $$\begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c@{}c@{}c@{}c} & & &1 &0 & 1 & 1_2 & & \\ & & & \times & 1& 0 & 1& 0 & 0_2 \\ \cline{1-9}& & &1 &0 &1 &1 & & \\ & & & & & &0 & & \\ & 1 &\stackrel{1}{0}&1 &1 &0 &0 &\downarrow &\downarrow \\ \cline{1-9} &1 &1 &0 &1 &1 &1 &0 &0_2 \\ \end{array}$$The answer is $\boxed{11011100_2}$.

human

For how many integer values of $n$ between 1 and 120 inclusive does the decimal representation of $\frac{n}{120}$ terminate?

gpt

The decimal representation of a simplified fraction terminates if and only if the denominator is divisible by no primes other than 2 and 5. The prime factorization of $120$ is $2^3 \cdot 5 \cdot 3$. For the fraction to simplify to having only the primes $2$ and $5$ in the denominator, there must be a factor of $3$ in the numerator. There are $\left\lfloor \frac{120-1}{3} \right\rfloor+1=40$ multiples of $3$ between $1$ and $120$, so there are $\boxed{40}$ integers values for $n$.

human

If $n=2^3 \cdot 3^2 \cdot 5$, how many even positive factors does $n$ have?

gpt

A positive integer is a factor of $n$ if and only if its prime factorization is of the form $2^a\cdot 3^b\cdot 5^c$ where $0\leq a\leq 3$, $0\leq b\leq 2$, and $0\leq c\leq 1$. An integer is even if and only if the exponent of 2 in its prime factorization is at least 1. Therefore, we have 3 choices for $a$, 3 choices for $b$ and $2$ choices for $c$, for a total of $(3)(3)(2)=\boxed{18}$ ways to form an even positive factor of $n$.

human

Let $S = 1 - 2 + 3 - 4 + \cdots + 2009 - 2010$. What is the residue of $S$, modulo 2010?

gpt

Note that $x + 2010 \equiv x \pmod{2010}$. Add 2010 to every negative summand and rearrange the terms to find $S \equiv 0 + 1 + 2 + 3 + \cdots + 2008 + 2009 \pmod{2010}$. The right-hand side is the sum of the integers from 1 to 2010, so $S \equiv \frac{2010 \cdot 2011}{2} \equiv 1005 \cdot 2011 \equiv 1005 \cdot 1 \equiv \boxed{1005} \pmod{2010}$.

human

Sherlock Holmes and Dr. Watson recover a suitcase with a three-digit combination lock from a mathematician turned criminal. Embedded in the suitcase above the lock is the cryptic message "AT SEA BASE. SEAS EBB SEA: BASS. " Dr. Watson comments, "This probably isn't about ocean fish. Perhaps it is an encrypted message. The colon suggests the addition problem $SEAS + EBB + SEA = BASS$, where each letter represents a distinct digit, and the word 'BASE' implies that the problem is in a different base." Holmes calmly turns the combination lock and opens the suitcase. While Dr. Watson gapes in surprise, Holmes replies, "You were indeed right, and the answer was just the value of the word $SEA$, interpreted as decimal digits." What was the lock combination?

gpt

We must solve the addition problem $$ \begin{array}{@{}c@{\;}c@{}c@{}c@{}c@{}c} & & S & E & A & S_d \\ & & & E & B & B_d \\ + & & & S & E & A_d\\ \cline{1-6} & & B & A& S& S_d\end{array},$$ where $d$ is an unknown base. It follows that $S + B + A$ leaves a residue of $S$ upon division by $d$. Thus, $B+A$ must be divisible by $d$. Since $B$ and $A$ cannot both be $0$, and $B+A < (d-1) + (d-1) = 2d-2$, then $B + A = d$. Looking at the $d$s digit, we must carry-over $1$ from the units digit sum, so $1 + A + B + E \equiv S \pmod{d}$. Since $B + A = d$, then $1 + E + d \equiv 1+E \equiv S \pmod{d}$. Thus, $S = E+1$ or $E = d-1$ and $S = 0$. However, the latter is impossible since $S$ is the leftmost digit of 'SEAS' and 'SEA'. Thus, $S = E+1$, and we again carry-over $1$ to the $d^2$ digit. Looking at the $d^2$ digit, after carry-over, it follows that $1 + E + E + S \equiv A \pmod{d}$. Note that $1 + E + E + S < 1 + 3(d-1) = 3d - 2 < 3d$. Then, $2E + S + 1 - A$ is either equal to $0$, $d$, or $2d$. However, we can immediately discard the $0$ case: there would be no carry-over for the leftmost digit, so $S = B$ are not distinct. In the next case, if $2E + S + 1 = A + d$, then there is a carry-over of $1$ to the last digit. It follows that $S + 1 = B$. This gives us the system of equations \begin{align*} B + A &= d \\ E + 1 &= S \\ S + 1 &= B \\ 2E + S +1 - A&= d \end{align*} Setting the first and fourth equations equal to each other yields that $d = B+A = 2E + S +1 - A$, and since $B = S+1 = E+2$, substituting for $B$ and $S$ yields that $2A = 3E + S + 1 - B = 2E + (E+1) + 1 - (E+2) = 2E$. This contradicts the distinct digits criterion. Thus, $2E + S + 1 - A= 2d = 2(B+A)$, so $2E + S + 1 - 2B = 3A$. Also, we have that $B = S+2$, due to the carry-over in the leftmost digit. Substituting for $B$ and $S$ yields that $3A = 2E + (E+1) + 1 - 2(E + 3) = E - 4$, so $E = 3A+4$. Thus, $S = 3A+5$ and $B=3A+7$. Also, $S,E,$ and $A$ are decimal digits, so it follows that $S = 3A + 5 \le 9 \Longrightarrow A = 0,1$. We can discard the solution $A = 0$, as $d = B+A$ but $B < d$. Thus, $B = 10, S = 8, E = 7$, occurring in base $d = B+A = 11$. The answer is $\boxed{871}$.

human

For a nonnegative integer $n$, let $r_9(n)$ stand for the remainder left when $n$ is divided by $9.$ For example, $r_9(25)=7.$ What is the $22^{\text{nd}}$ entry in an ordered list of all nonnegative integers $n$ that satisfy $$r_9(5n)\le 4~?$$(Note that the first entry in this list is $0$.)

gpt

The condition $r_9(5n)\le 4$ can also be stated as $``5n\equiv 0,1,2,3,\text{ or }4\pmod 9."$' We can then restate that condition again by multiplying both sides by $2:$ $$10n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$This step is reversible (since $2$ has an inverse modulo $9$). Thus, it neither creates nor removes solutions. Moreover, the left side reduces to $n$ modulo $9,$ giving us the precise solution set $$n \equiv 0,2,4,6,\text{ or }8\pmod 9.$$We wish to determine the $22^{\text{nd}}$ nonnegative integer in this solution set. The first few solutions follow this pattern: $$\begin{array}{c c c c c} 0 & 2 & 4 & 6 & 8 \\ 9 & 11 & 13 & 15 & 17 \\ 18 & 20 & 22 & 24 & 26 \\ 27 & 29 & 31 & 33 & 35 \\ 36 & 38 & \cdots \end{array}$$The $22^{\text{nd}}$ solution is $\boxed{38}.$

human

Suppose that $b$ is a positive integer greater than or equal to $2.$ When $197$ is converted to base $b$, the resulting representation has $4$ digits. What is the number of possible values for $b$?

gpt

We would like to find $b$ for which $b^3 \leq 197_{10} < b^4.$ We can see that this inequality is satisfied for $4\leq b \leq 5.$ So, there are $\boxed{2}$ bases for which this holds.

human

A natural number is abundant if it is less than the sum of its proper divisors. What is the smallest abundant number?

gpt

For any prime number, the sum of its proper divisors is equal to $1$, so a prime number cannot be an abundant number. Therefore, it suffices to check only composite numbers: $\bullet$ For $4$, $1 + 2 < 4$, $\bullet$ For $6$, $1 + 2 + 3 = 6$, $\bullet$ For $8$, $1 + 2 + 4 < 8$, $\bullet$ For $9$, $1 + 3 < 9$, $\bullet$ For $10$, $1 + 2 + 5 < 10$, $\bullet$ For $12$, $1 + 2 + 3 + 4 + 6 = 16 > 12$. Thus, the answer is $\boxed{12}$.

human

The time right now is exactly midnight. What time will it be in 1234 minutes?

gpt

There are $60$ minutes in an hour. When $1234$ is divided by $60$, you get $20$ with a remainder of $34$. Therefore, the time in $1234$ minutes will be $\boxed{20\!:\!34}$ or $\boxed{8\!:\!34 \text{ p.m.}}$.

human

Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained? $$ \text A. \ \ 21 \qquad \text B. \ \ 60 \qquad \text C. \ \ 119 \qquad \text D. \ \ 180 \qquad \text E. \ \ 231 $$

gpt

There are five prime numbers between $4$ and $18:$ namely $5,$ $7,$ $11,$ $13,$ and $17.$ Hence the product of any two of these is odd and the sum is even. Because $$xy-(x+y)=(x-1)(y-1)-1$$increases as either $x$ or $y$ increases (since both $x$ and $y$ are bigger than $1$), the answer must be an odd number that is no smaller than $$23=5\cdot 7-(5+7)$$and no larger than $$191=13\cdot 17-(13+17).$$The only possibility among the options is $\boxed{119},$ and indeed $119=11\cdot 13-(11+13).$

human

What is the smallest integer $n$, greater than $1$, such that $n^{-1}\pmod{1050}$ is defined?

gpt

For $n$ to have an inverse $\pmod{1050}$, it is necessary for $n$ to be relatively prime to $1050$. Conversely, if $n$ is relatively prime to $1050$, then $n$ has an inverse $\pmod{1050}$. The prime factors of $1050$ include $2$, $3$, $5$, and $7$, so any multiple of any of these primes does not have an inverse $\pmod{1050}$. This rules out all the integers from $2$ to $10$. However, $11$ is relatively prime to $1050$, so $\boxed{11}$ is the smallest integer greater than $1$ that has an inverse $\pmod{1050}$.

human

What is the units digit of $\frac{20 \cdot 21 \cdot 22 \cdot 23 \cdot 24 \cdot 25}{1000}$?

gpt

$\frac{20\cdot 21\cdot 22\cdot 23\cdot 24\cdot 25}{1000} = \frac{2^2\cdot 5\cdot 21\cdot 2\cdot 11\cdot 23\cdot 2^3\cdot 3\cdot 5^2}{2^3\cdot 5^3} = 2^3\cdot 3\cdot 21 \cdot 11\cdot 23 \equiv 2^3\cdot 3^2 \pmod{10} \equiv \boxed{2}\pmod{10}$.