gowtham28/math_formated · Datasets at Hugging Face

from stringclasses 2 values	value stringlengths 16 6.77k
human	What is the greatest whole number that must be a divisor of the product of any three consecutive positive integers?
gpt	We know that at least one of our three consecutive positive integers must be a multiple of $2$ since every other integer in a list of consecutive integers is divisible by $2$. Similarly, one of our three consecutive integers must also be divisible by $3$. Thus, the product of our three integers must be divisible by $2 \cdot 3 = 6$. By choosing the example where our three consecutive integers are $1$, $2$, and $3$ and their product is $6$, we see that $\boxed{6}$ is indeed the greatest whole number that must be a factor of the product of any three consecutive positive integers.
human	If $A,B$ and $C$ are non-zero distinct digits in base $6$ such that $\overline{ABC}_6 + \overline{BCA}_6+ \overline{CAB}_6 = \overline{AAA0}_6$, find $B+C$ in base $6$.
gpt	By definition of a base, it follows that $\overline{ABC}_6 = 6^2 \cdot A + 6 \cdot B + C$. Noting that every digit appears in every possible slot once, it follows that $\overline{ABC}_6 + \overline{BCA}_6+ \overline{CAB}_6 = (6^2 + 6 + 1)(A + B + C).$ The value is equal to the sum, $\overline{AAA0}_6 = 6^3 \cdot A + 6^2 \cdot A + 6 \cdot A = (6^2 + 6 + 1) \cdot (6 \cdot A)$. Setting them equal, $$(6^2 + 6 + 1)(A + B + C) = (6^2 + 6 + 1) \cdot (6 \cdot A) \Longrightarrow B+C = 5 \cdot A.$$Since $B,C < 6$, then $B+C < 2 \cdot 6$, so $A = 1,2$. However, there do not exist distinct base $6$ digits such that $B + C = 2 \cdot 5$, so it follows that $A = 1_6$, and $B+C = \boxed{5}_6$.
human	How many distinct, natural-number factors does $4^3 \cdot 5^4 \cdot 6^2$ have?
gpt	We prime factorize the given number as $2^8\cdot 3^2\cdot 5^4$. A factor of this number takes the form $2^a3^b5^c$ for integers $a$ between 0 and 8, $b$ between 0 and 2, and $c$ between 0 and 4. There are $9$ ways to choose $a$, 3 ways to choose $b$, and 5 ways to choose $c$. In total, there are $9\cdot3\cdot5 = \boxed{135}$ factors.
human	Find the units digit of the following within the indicated number base: $52_7 + 62_7$
gpt	Since $2_7 + 2_7 = 4_7$, the units digit is $\boxed{4}$.
human	Divide the product of the first five positive composite integers by the product of the next five composite integers. Express your answer as a common fraction.
gpt	The first ten positive composite integers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. The desired quotient is $\frac{4\cdot6\cdot8\cdot9\cdot10}{12\cdot14\cdot15\cdot16\cdot18}$. After cancellation, we get $\frac{1}{14\cdot3}=\boxed{\frac{1}{42}}$.
human	How many of the positive divisors of 3240 are multiples of 3?
gpt	$$ 3240 = 2^3 \cdot 3^4 \cdot 5^1 $$A positive divisor of 3240 is a multiple of 3 when it has a prime factorization in the form $2^a \cdot 3^b \cdot 5^c$ where $0 \le a \le 3$, $1 \le b \le 4$, and $0 \le c \le 1$. There are $4 \cdot 4 \cdot 2 = \boxed{32}$ choices for $a$, $b$, and $c$, giving the number of positive divisors of 3240 that are multiples of 3.
human	Convert $\rm{A}03_{16}$ to a base 10 integer, where the 'digits' A through F represent the values 10, 11, 12, 13, 14, and 15 in order.
gpt	In base 16, $A = 10$, $B = 11$, $C = 12$, $D = 13$, $E = 14$, and $F = 15$. So $A03 = 10\cdot16^2 + 0\cdot16 + 3 = \boxed{2563}.$
human	Find the integer $n$, $0 \le n \le 5$, such that \[n \equiv -3736 \pmod{6}.\]
gpt	Since $-3736 \equiv 2 \pmod{6}$, the integer $n$ we seek is $n = \boxed{2}$.
human	When the set of natural numbers is listed in ascending order, what is the smallest prime number that occurs after a sequence of five consecutive positive integers all of which are nonprime?
gpt	Consider the differences between consecutive prime numbers and look for the first difference of 6 or greater. The first several prime numbers are \[ 2,3,5,7,11,13,17,19,23,29,31, 37,\ldots, \] and the differences between consecutive terms of this sequence are \[ 1,2,2,4,2,4,2,4,6,2,\ldots. \] The first appearance of a difference of 6 or greater occurs between 23 and $\boxed{29}$.
human	Suppose that $n$ is a positive integer such that in base $7$, then $n$ can be expressed as $\overline{ABC}_7$, and in base $11$, then $n$ can be expressed as $\overline{CBA}_{11}$. Find the largest possible value of $n$ in base $10$.
gpt	We convert $n$ to base $10$. The base $7$ expression implies that $n = 49A + 7B + C$, and the base $11$ expression implies that $n = 121C + 11B + A$. Setting the two expressions equal to each other yields that $$n = 49A + 7B + C = 121C + 11B + A \Longrightarrow 48A - 4B - 120C = 0.$$Isolating $B$, we get $$B = \frac{48A - 120C}{4} = 12A - 30C = 6(2A - 5C).$$It follows that $B$ is divisible by $6$, and since $B$ is a base $7$ digit, then $B$ is either $0$ or $6$. If $B$ is equal to $0$, then $2A - 5C = 0 \Longrightarrow 2A = 5C$, so $A$ must be divisible by $5$ and hence must be either $0$ or $5$. Since $n$ is a three-digit number in base $7$, then $A \neq 0$, so $A = 5$ and $C = 2$. Thus, $n = 502_7 = 5 \cdot 7^2 + 2 = 247$. If $B$ is equal to $6$, then $2A - 5C = 1$, so $2A - 1 = 5C$ and $2A - 1$ must be divisible by 5. Since $A$ is a base $7$ digit, it follows that $A = 3$ and $C = 1$. This yields the value $n = 361_7 = 3 \cdot 7^2 + 6 \cdot 7 + 1 = 190$. The largest possible value of $n$ in base $10$ is $\boxed{247}$.
human	What is the sum of all two-digit positive integers whose squares end with the digits 01?
gpt	If $n$ is a two-digit number, then we can write $n$ in the form $10a + b$, where $a$ and $b$ are digits. Then the last digit of $n^2$ is the same as the last digit of $b^2$. The last digit of $n^2$ is 1. We know that $b$ is a digit from 0 to 9. Checking these digits, we find that the units digit of $b^2$ is 1 only for $b = 1$ and $b = 9$. If $b = 1$, then $n = 10a + 1$, so \[n^2 = 100a^2 + 20a + 1.\] The last two digits of $100a^2$ are 00, so we want the last two digits of $20a$ to be 00. This occurs only for the digits $a = 0$ and $a = 5$, but we reject $a = 0$ because we want a two-digit number. This leads to the solution $n = 51$. If $b = 9$, then $n = 10a + 9$, so \[n^2 = 100a^2 + 180a + 81 = 100a^2 + 100a + 80a + 81.\] The last two digits of $100a^2 + 100a$ are 00, so we want the last two digits of $80a + 81$ to be 01. In other words, we want the last digit of $8a + 8$ to be 0. This only occurs for the digits $a = 4$ and $a = 9$. This leads to the solutions $n = 49$ and $n = 99$. Therefore, the sum of all two-digit positive integers whose squares end with the digits 01 is $51 + 49 + 99 = \boxed{199}$.
human	Two numbers $90$ and $m$ share exactly three positive divisors. What is the greatest of these three common divisors?
gpt	Recall that the common divisors of two integers are precisely the divisors of the greatest common divisor. So, for two numbers to have exactly three positive divisors in common, those divisors must be $1$, $p$, and $p^2$ such that $p$ is prime. We now look at the prime factorization of $90$: $90=2 \cdot 3^2 \cdot 5$. Since $3^2$ is the only perfect square divisor of $90$, the divisors that $90$ and $m$ share must be $1$, $3$, and $9$. The largest of these three numbers is $\boxed{9}$.
human	The remainders when three positive integers are divided by 5 are 1, 2, and 3. Find the remainder when their product is divided by 5.
gpt	The product of the three integers is equivalent in modulo 5 to the product of the modulo 5 residues of the three integers. We multiply these residues to find the remainder: $$ 1 \cdot 2 \cdot 3 \equiv 6 \equiv \boxed{1} \pmod{5}. $$
human	What is the sum of the last two digits of this portion of the Fibonacci Factorial Series: $1!+1!+2!+3!+5!+8!+13!+21!+34!+55!+89!$?
gpt	This expression $n!$, is the number you get by multiplying $n$ by $(n-1)$ by $(n-2)$ by $(n-3)$ and so on, all the way down to $1$. So $5! = (5)(4)(3)(2)(1) = 120$. Notice that $5!$ ends in a $0$ since it has a factor of $10$ (there is a $5$ and a $2$ in it list of factors) and that $10!$ has to end in two zeroes since it has a factor of $10$, $5$ and $2$ which is really a factor of $100$. Since any factorial greater than $10$ (such as $13!$ or $21!$) includes all of the factors of $10!$, the last two digits of $13!$, $21!$, and so on are zeroes. These terms, therefore will not affect the last two digits of the sum of the Fibonacci factorial series. To find the last two digits, you only need to find the last two digits of each of the terms of $1! + 1! + 2! + 3! + 5! + 8!$. We do not need to calculate $8!$, only to find its last two digits. Starting with $5!$, we can work our way to $8!$, using only the last two digits of each value along the way. We know $5! = 120$, so use $20$ when finding $6!$, which will bring us to $6(20) = 120$ or $20$. Therefore, the last two digits of $7!$ are from $7(20) = 140$ or $40$. Finally $8!$ is $8(40) = 320$ or finally $20$. The last two digits of the entire series will come from $1 + 1 + 2 + 6 + 20 + 20 = 50$. Therefore, the sum of the last two digits is $5 + 0 = \boxed{5}$.
human	Let $A$ be the set of all numbers which can be represented as the sum of three consecutive positive integers. What is the greatest common divisor of all numbers in $A$?
gpt	Each member in the set is of the form $(x-1)+x+(x+1)=3x$. Since $x$ can be any positive integer, the greatest common divisor of all these members is $\boxed{3}$.
human	What is the largest perfect square factor of 1512?
gpt	Let's find the prime factorization of 1512: $1512=2^3\cdot189=2^3\cdot3^3\cdot7$. The only two squares of primes that divide 1512 are $2^2=4$ and $3^2=9$. Therefore, the largest perfect square factor of 1512 is $2^2\cdot3^2=(2\cdot3)^2=\boxed{36}$.
human	Given that $13^{-1} \equiv 29 \pmod{47}$, find $34^{-1} \pmod{47}$, as a residue modulo 47. (Give a number between 0 and 46, inclusive.)
gpt	Note that $34 \equiv -13 \pmod{47}$. Hence, \begin{align} 34^{-1} &\equiv (-13)^{-1} \\ &\equiv (-1)^{-1} \cdot 13^{-1} \\ &\equiv (-1) \cdot 29 \\ &\equiv \boxed{18} \pmod{47}. \end{align}
human	Daniel has exactly $20$ pieces of candy. He has to divide them equally among his $3$ sisters. What is the least number of pieces he should take away so that he could distribute the candy equally?
gpt	When you divide $20$ by $3$, you get $6$ with a remainder of $2$. Therefore, he should take away $\boxed{2 \text{ pieces}}$ so he could give each of his sisters $6$ pieces.
human	Let $f(x) = x^2 - x + 2010$. What is the greatest common divisor of $f(100)$ and $f(101)$?
gpt	We can write $f(x) = x(x-1) + 2010$. From here, it is clear that $f(101) = 101\cdot 100 + 2010$ and $f(100) = 100\cdot 99 + 2010$. We now use the Euclidean algorithm. \begin{align} &\gcd(f(101), f(100)) \\ &= \gcd(101\cdot 100 + 2010, \, \, 100\cdot 99 + 2010) \\ &= \gcd(100\cdot 99 + 2010, \, \, 101\cdot 100 + 2010 - (100\cdot 99 + 2010)) \\ &= \gcd(100\cdot 99 + 2010, \, \, 2\cdot 100) \\ &= \gcd(100\cdot 99 + 2000 + 10, \, \, 2\cdot 100) \\ & = \gcd(100\cdot 99 + 100\cdot 20 + 10, 2\cdot 100) \\ & = \gcd(100\cdot 119 + 10, \, \, 2\cdot 100) \\ & = \gcd(2\cdot 100, \, \, 100\cdot 119 + 10 - 59(2\cdot 100)) \\ & = \gcd(2\cdot 100, \, \, 100\cdot 119 + 10 - 118\cdot 100) \\ & = \gcd(2\cdot 100, \, \, 100 + 10) \\ & = \gcd(200, \, \, 110) \\ & = \gcd(90, \, \, 110) \\ & = \gcd(20, \, \, 90) \\ & = \gcd(20, \, \, 90-4\cdot 20) \\ & = \gcd(20, \, \, 10) \\ &= \boxed{10}. \end{align}
human	In base $b$, there are exactly one hundred three-digit numbers whose digits are all distinct. (That's "one hundred" in the ordinary sense, $100_{10}$.) What is $b$?
gpt	To form a three-digit number in base $b$ whose digits are all distinct, we must choose a first digit, a second digit, and a third digit. We have $b-1$ choices for the first digit ($1,2,3,\ldots,b-2,b-1$). We have $b-1$ choices for the second digit ($0,1,2,\ldots,b-2,b-1$, with the first digit removed from our choices). We have $b-2$ choices for the third digit. So, $$(b-1)^2(b-2) = 100.$$Trial and error is arguably the most reasonable way to solve this equation! Since $100=5\cdot 5\cdot 4$, the answer is $b=\boxed{6}$.
human	A math class has fewer than $40$ students. When the students try to sit in rows of $7$, $3$ students sit in the last row. When the students try to sit in rows of $5$, $1$ student is left in the last row. How many students are in this class?
gpt	Let $n$ be the number of students in the class. When we divide $n$ by $7$, we get a remainder of $3$. Therefore, we need to add multiples of $7$ to $3$ until we get a number which, when we divide by $5$, has a remainder of $1$. When we add $28$ to $3$, we get $31$, a number that gives a remainder of $1$ when dividing by $5$. By the Chinese remainder theorem, the other integers which leave a remainder of 3 when divided by 7 and a remainder of 1 when divided by 5 differ from 31 by a multiple of $7\cdot 5=35$. Therefore, the only integer between 0 and 40 satisfying these conditions is 31, and there are $\boxed{31}$ students in this math class.
human	What is the first digit (from left to right) of the base $8$ representation of $473_{10}$?
gpt	The largest power of $8$ that is still less than $473$ is $8^2 = 64$ and the largest multiple of $64$ that is less than $473$ is $7 \cdot 64 = 448$. Thus, when $473_{10}$ is written in base $8$, its first digit is $\boxed{7}$.
human	What is the first nonzero digit to the right of the decimal point of the fraction $\frac{1}{129}$?
gpt	If we multiply this fraction by 10, we do not change any of the digits; we simply shift all the digits one place to the left. Therefore, if we multiply the fraction by a sufficiently large power of 10, then the first digit obtained when dividing by 129 will be the first nonzero digit to the right of the decimal point of $\frac{1}{129}$. Observing that $100 < 129 < 1000$, we compute $\frac{1000}{129}=7 \frac{97}{129}$. So the first nonzero digit to the right of the decimal point of $\frac{1}{129}$ is $\boxed{7}$.
human	The volume of the box is 240 cubic units where $a,b,$ and $c$ are integers with $1<c<b<a$. What is the largest possible value of $b$? [asy] draw((0,0)--(0,3)--(6,3)--(6,0)--cycle); draw((2,1.5)--(2,4.5)--(8,4.5)--(8,1.5)--cycle); draw((0,0)--(2,1.5)--(2,4.5)--(0,3)--cycle); draw((6,0)--(6,3)--(8,4.5)--(8,1.5)--cycle); label("a",(3,0),S); label("b",(7,0.75),SE); label("c",(8,3.5),E); [/asy]
gpt	The prime factorization of 240 is $2^4\cdot3\cdot5$. We want $c$ to be as small as possible, so $c=2$. Now we have $a\cdot b=2^3\cdot3\cdot5=120$. For the maximum $b$, we seek the minimum $a$ while $b<a$. If $b<a$ then $a\cdot b<a^2$, so $120<a^2$. That means $a$ is at least 11. However, 11 is not a factor of 240. The least factor of 240 greater than 11 is $2^2\cdot3=12$. So the minimum $a$ is 12 and that makes the maximum $b=\frac{2^3\cdot3\cdot5}{2^2\cdot3}=2\cdot5=\boxed{10}$.
human	The least common multiple of two numbers is 3780, and the greatest common divisor is 18. Given that one of the numbers is 180, what is the other number?
gpt	We use the identity $\gcd(a,b) \cdot \mathop{\text{lcm}}[a,b] = ab$ for all positive integers $a$ and $b$. We are told that $\gcd(a,b) = 18$, and $\mathop{\text{lcm}}[a,b] = 3780$, so $ab = 18 \cdot 3780$. If one number is 180, then the other number is $18 \cdot 3780/180 = \boxed{378}$.
human	The sequence 12, 15, 18, 21, 51, 81, $\ldots$ consists of all positive multiples of 3 that contain at least one digit that is a 1. What is the $50^{\mathrm{th}}$ term of the sequence?
gpt	We know that the rule for divisibility by $3$ is that the digits of the number must add up to a multiple of $3$. So, it's clear that there are no other such two-digit numbers beyond the ones listed in the problem. Every number divisible by $3$ between $100$ and $199$ is in the sequence, so that gets us through the $39$th term of the sequence. Using the rule for divisibility by $3$, it is fairly simple to list out the remaining $11$ terms of the sequence: $201, 210, 213, 216, 219, 231, 261, 291, 312, 315, 318$. Thus, the $50$th term is $\boxed{318}$.
human	Each pack of hot dogs contains 10 hot dogs (no buns), but each pack of hot dog buns contains 8 buns. Phil buys a number of these packs for a barbecue. After the barbecue, Phil finds that he has 4 hot dogs left over. What is the SECOND smallest number of packs of hot dogs he could have bought?
gpt	Let $n$ be the number of packs of hot dogs that Phil bought. Then $10n \equiv 4 \pmod{8}$, which reduces to $2n \equiv 4 \pmod{8}$. This congruence tells us that $2n = 8k + 4$ for some integer $k$, or $n = 4k + 2$. The second smallest positive integer of this form is $\boxed{6}$.
human	Abigail, Beatrice, and Carson combine their eggs to sell them at the market. If Abigail has 37 eggs, Beatrice has 49 eggs, and Carson has 14 eggs, and if eggs can only be sold in cartons of 12, how many eggs will be left over if all cartons are sold?
gpt	The remainder when $37+49+14$ is divided by $12$ is the same as the remainder when the congruences of each number mod 12 are summed and then divided by 12. In other words, since we have \begin{align} 37 &\equiv 1\pmod{12}\\ 49 &\equiv 1\pmod{12} \\ 14 &\equiv 2\pmod{12} \end{align}then therefore, $37+49+14 \equiv1+1+2 \equiv \boxed{4}\pmod{12}$.
human	In any month with five Mondays, a school declares the fifth Monday a Marvelous Monday. School begins on Monday, September 4. What is the date of the first Marvelous Monday after school starts? (If you abbreviate, do not use a dot in the abbreviation.)
gpt	September has 30 days. September 4 is a Monday, so September 9 is a Saturday. Since September 30 is exactly 21 days later (or 3 weeks), September 30 is also a Saturday. Then October 1 is a Sunday, and October 2 is a Monday. Then October 2, 9, 16, 23, and 30 are all Mondays, so the first Marvelous Monday is $\boxed{\text{October 30}}$.
human	What is the 308th digit to the right of the decimal point when $\frac{12}{37}$ is expressed as a decimal?
gpt	We use long division to find that the decimal representation of $\frac{12}{37}$ is $0.\overline{324}$. When $308$ is divided by $3$ there is a remainder of $2$ $\left( 308\div 3=102 \ R2\right)$. Therefore the 308th digit to the right of the decimal point is the second digit of $324$, which is $\boxed{2}$.
human	What is the sum of all positive integer divisors of 77?
gpt	Since $77=7\cdot11$, the divisors of 77 are 1, 7, 11, and 77. Their sum is $1+7+11+7\cdot11=\boxed{96}$.
human	How many even perfect square factors does $2^4 \cdot 7^9$ have?
gpt	A positive integer is a factor of $2^4\cdot7^9$ if and only if its prime factorization is of the form $2^a\cdot 7^b$ for exponents $a$ and $b$ satisfying $0\leq a \leq 4$ and $0\leq b\leq 9$. A positive integer is a perfect square if and only if the exponents in its prime factorization are even. A positive integer is even if and only if the exponent of 2 in its prime factorization is at least 1. Therefore, we may choose $a=2$ or $4$ and $b=0,$ $2,$ $4,$ $6,$ or $8.$ Since we have 2 choices for $a$ and 5 choices for $b$, there are $2\times5=\boxed{10}$ ways to make these two decisions.
human	What integer $n$ satisfies $0\le n<9$ and $$-1111\equiv n\pmod 9~?$$
gpt	An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Using this fact, we find that the greatest multiple of $9$ that is less than $-1111$ is $-1116$. We have $-1111 = -1116+5$. That is, $-1111$ is equal to a multiple of $9$ plus $5$. Therefore, $$-1111\equiv \boxed{5}\pmod 9.$$To verify this answer, we can check that $9$ divides $(-1111)-5 = -1116$; it does, so our answer is correct.
human	How many distinct, positive factors does $1100$ have?
gpt	The prime factorization of $1100$ is $2^2\cdot5^2\cdot11$. To find the number of factors, we consider that each factor has a prime factorization of $2^a\cdot5^b\cdot11^c$, where $a$ could be from $0$ to $2$ (3 possible values), $b$ could be from $0$ to $2$ (3 possible values), and $c$ could be $0$ or $1$ (2 possible values). So the number of factors is $3\cdot3\cdot2=\boxed{18}$.
human	What is the units digit of $23^{23}$?
gpt	Let's find the cycle of units digits of $3^n$, starting with $n=1$ (note that the tens digit 2 in 23 has no effect on the units digit): $3, 9, 7, 1, 3, 9, 7, 1,\ldots$ . The cycle of units digits of $23^{n}$ is 4 digits long: 3, 9, 7, 1. Thus, to find the units digit of $23^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 3, $R=2$ corresponds to the units digit 9, etc.) Since $23\div4=5R3$, the units digit of $23^{23}$ is $\boxed{7}$.
human	What is the smallest positive four-digit number divisible by 9 which has two even and two odd digits?
gpt	For a number to be divisible by 9, the sum of its digits must be divisible by 9. But since the number has two even digits and two odd digits, the sum of its digits is even. Thus the sum of its digits must be at least 18. This number will be minimized if its thousands digit is 1 and its hundreds digit is 0. This means the remaining two digits must sum to 17, and are hence 8,9. So we see that the smallest possible integer of the desired form is $\boxed{1089}$.
human	How many positive factors of 72 are perfect cubes?
gpt	Prime factorize $72$ as $2^3\cdot 3^2$. A positive integer is a factor of 72 if and only if the exponents in its prime factorization are less than or equal to the corresponding exponents in the prime factorization of 72. Also, a positive integer is a perfect cube if and only if every exponent is a multiple of 3. Therefore, in forming a perfect cube factor of 72, we have 2 choices for the exponent of $2$ (either 0 or 3) and only 1 choice for the exponent of 3 (namely 0). There are $2\cdot 1=\boxed{2}$ ways to make these choices.
human	The prime factorization of 2007 is $3^{2}\times223$. How many ordered pairs of positive integers $(x,y)$ satisfy the equation $xy=2007$?
gpt	$x$ must be a divisor of 2007. To count the number of divisors, we note that a divisor can have zero factors of 3, one factor, or two factors, and (independently) it can have zero factors of 223 or one factor. The total number of divisors is then $3\cdot2=6$ (3 choices for the factors of 3 and 2 choices for the factors of 223). Since each divisor is unique, each creates a unique ordered pair. So there are $\boxed{6}$ ordered pairs.
human	What is the smallest integer $k$ such that $k>1$ and $k$ has remainder $1$ when divided by any of $17,$ $6,$ and $2?$
gpt	We can re-write the problem as the following three equations: $$k = 17a+1\\k = 6b+1 \\ k = 2c + 1$$Therefore, $k-1$ is divisible by $17,$ $6,$ and $2.$ The smallest positive value of $k-1$ is thus $$\text{lcm}[17,6,2] = \text{lcm}[17,6] = 17\cdot 6 = 102,$$and so the smallest possible value of $k$ is $k = 102+1 = \boxed{103}.$
human	Let $a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}$. What is the remainder when $a$ is divided by $11$?
gpt	One way to do this is to find each inverse explicitly: \begin{align} (3^{-1}+5^{-1}+7^{-1})^{-1} &\equiv (4+9+8)^{-1} \pmod{11} \\ &\equiv 21^{-1} \pmod{11} \\ &\equiv 10^{-1} \pmod{11} \\ &\equiv \boxed{10}\pmod{11}. \end{align} Another way to do this is through manipulation: \begin{align} & (3^{-1}+5^{-1}+7^{-1})^{-1}\\ \equiv~ & (3\cdot 5\cdot 7)(3\cdot 5\cdot 7)^{-1}(3^{-1}+5^{-1}+7^{-1})^{-1}\\ \equiv~ & (3\cdot 5\cdot 7)(3\cdot 5+5\cdot 7+ 7\cdot 3)^{-1}\\ \equiv~ & 6\cdot(15+35+21)^{-1}\\ \equiv~ & 6\cdot 5^{-1}\\ \equiv~ & 6\cdot 9\\ \equiv~ & \boxed{10} \pmod{11} \end{align}
human	What is the remainder of $8^6 + 7^7 + 6^8$ divided by 5?
gpt	First of all, $6$ has a remainder of $1$ when divided by $5,$ thus any power of $6$ will have a remainder of $1$ when divided by $5.$ As for $7,$ it has a remainder of $2$ when divided by $5$, so let us look at the powers of $2$: \begin{align} 2^1 &\equiv 2 \pmod{5} \\ 2^2 &\equiv 4 \pmod{5} \\ 2^3 &\equiv 3 \pmod{5} \\ 2^4 &\equiv 1 \pmod{5}. \end{align}Since $2^4 \equiv 1 \pmod{5},$ we see that $2^{7} \equiv 2^3 \cdot 2^4 \equiv 3 \pmod{5},$ hence the remainder of $7^4$ when divided by $5$ is $3.$ Now, there is a bit of a shortcut we can use for $8.$ Since $8 \equiv -2 \pmod{5},$ we can see that $8^6 \equiv (-2)^6 \equiv 2^6 \equiv 2^2 \cdot 2^4 \equiv 4 \pmod {5},$ hence our desired remainder is $4.$ Adding them up, we have $4 + 3 + 1 \equiv \boxed{3} \pmod{5}.$
human	What is the remainder of $5^{2010}$ when it is divided by 7?
gpt	We start by writing out some powers of five modulo 7. \begin{align} 5^1 &\equiv 5 \pmod{7} \\ 5^2 &\equiv 4 \pmod{7} \\ 5^3 &\equiv 6 \pmod{7} \\ 5^4 &\equiv 2 \pmod{7} \\ 5^5 &\equiv 3 \pmod{7} \\ 5^6 &\equiv 1 \pmod{7} \end{align}Therefore, we have that $5^6 \equiv 1$ modulo 7. Thus, $5^{2010} \equiv (5^6)^{335} \equiv 1^{335} \equiv \boxed{1}$ modulo 7.
human	Find the largest prime factor of $15^3+10^4-5^5$.
gpt	The greatest common divisor of the three terms is $5^3$. Factoring $5^3$ out of each term and using the distributive property gives: \begin{align} 15^3+10^4-5^5 &= 5^3\cdot3^3 + 5^3\cdot5\cdot2^4-5^3\cdot5^2 \\ &= 5^3(3^3+5\cdot2^4-5^2)\\ & = 5^3(27+80-25) \\ &= 5^3(82)=2\cdot5^3\cdot41. \end{align}So the largest prime factor is $\boxed{41}$.
human	What is the sum of the six positive integer factors of 18?
gpt	$18=1\cdot18=2\cdot9=3\cdot6$. The sum is $1+18+2+9+3+6=\boxed{39}$.
human	How many integers between $500$ and $1000$ contain both the digits $3$ and $4$?
gpt	First we notice that the hundreds digit of any integer between $500$ and $1000$ cannot be $3$ or $4$, so $3$ and $4$ must be the units and tens digits (in either the order $34$ or $43$). Since the integer must be between $500$ and $1000$, there are $5$ choices for the hundreds digit of this integer ($5$, $6$, $7$, $8$, or $9$). Thus, there are $2 \times 5 = \boxed{10}$ ways to form such a number.
human	The product of the positive integer divisors of a positive integer $n$ is 729. Find $n$.
gpt	Suppose that $n$ has $d$ divisors and that $d$ is even. Since the divisors come in pairs whose product is $n$, the product of the divisors of $n$ is $n^{d/2}$. For example, if $n=12$ then the product of the divisors is $(1\cdot 12)(2\cdot 6)(3\cdot 4)=12^3$. If $d$ is odd, then there are $(d-1)/2$ pairs which give a product of $n$, as well as the divisor $\sqrt{n}$. So again the product of the divisors is $n^{(d-1)/2}n^{1/2}=n^{d/2}$. For example, if $n=16$, then the product of the divisors of $n$ is $(1\cdot 16)(2\cdot 8)(4)=16^{5/2}$. To summarize, we have found that the product of the positive integers divisors of a positive integer $n$ is $n^{d/2}$, where $d$ is the number of divisors of $n$. Writing 729 as $3^6$, then, we have $n^{d/2}=3^6$, which implies $n^d=3^{12}$. So the possible values of $(n,d)$ are $(3,12)$, $(9,6)$, $(27,4)$, $(81,3)$ and $(729,2)$. We find that only in the third case $(n,d)=(27,4)$ is $d$ the number of divisors of $n$, so $n=\boxed{27}$.
human	Grady distributed $x$ pieces of candy evenly among nine Halloween bags such that every bag received the greatest possible number of whole pieces of candy, but some candy was left over. What is the greatest possible number of pieces that could have been left over?
gpt	Grady distributes his candy in groups of 9 until he no longer has any groups left. The largest number of possible pieces left is $\boxed{8},$ since if he has more than 8 he can distribute another group of 9.
human	When the greatest common divisor and least common multiple of two integers are multiplied, their product is 200. How many different values could be the greatest common divisor of the two integers?
gpt	We know that $\gcd(a,b) \cdot \mathop{\text{lcm}}[a,b] = ab$ for all positive integers $a$ and $b$. Hence, in this case, $ab = 200$. The prime factorization of 200 is $2^3 \cdot 5^2$, so $a = 2^p \cdot 5^q$ and $b = 2^r \cdot 5^s$ for some nonnegative integers $p$, $q$, $r$, and $s$. Then $ab = 2^{p + r} \cdot 5^{q + s}$. But $ab = 200 = 2^3 \cdot 5^2$, so $p + r = 3$ and $q + s = 2$. We know that $\gcd(a,b) = 2^{\min\{p,r\}} \cdot 5^{\min\{q,s\}}$. The possible pairs $(p,r)$ are $(0,3)$, $(1,2)$, $(2,1)$, and $(3,0)$, so the possible values of $\min\{p,r\}$ are 0 and 1. The possible pairs $(q,s)$ are $(0,2)$, $(1,1)$, and $(2,0)$, so the possible values of $\min\{q,s\}$ are 0 and 1. Therefore, the possible values of $\gcd(a,b)$ are $2^0 \cdot 5^0 = 1$, $2^1 \cdot 5^0 = 2$, $2^0 \cdot 5^1 = 5$, and $2^1 \cdot 5^1 = 10$, for a total of $\boxed{4}$ possible values.
human	The greatest common divisor of two positive integers is $(x+5)$ and their least common multiple is $x(x+5)$, where $x$ is a positive integer. If one of the integers is 50, what is the smallest possible value of the other one?
gpt	We know that $\gcd(m,n) \cdot \mathop{\text{lcm}}[m,n] = mn$ for all positive integers $m$ and $n$. Hence, in this case, the other number is \[\frac{(x + 5) \cdot x(x + 5)}{50} = \frac{x(x + 5)^2}{50}.\]To minimize this number, we minimize $x$. We are told that the greatest common divisor is $x + 5$, so $x + 5$ divides 50. The divisors of 50 are 1, 2, 5, 10, 25, and 50. Since $x$ is a positive integer, the smallest possible value of $x$ is 5. When $x = 5$, the other number is $5 \cdot 10^2/50 = 10$. Note that that the greatest common divisor of 10 and 50 is 10, and $x + 5 = 5 + 5 = 10$. The least common multiple is 50, and $x(x + 5) = 5 \cdot (5 + 5) = 50$, so $x = 5$ is a possible value. Therefore, the smallest possible value for the other number is $\boxed{10}$.
human	What is the sum of all positive integers less than 100 that are squares of perfect squares?
gpt	Squares of perfect squares are fourth powers. $1^4=1$, $2^4=16$, and $3^4=81$ are the only fourth powers less than 100. Their sum is $1+16+81=\boxed{98}$.

human

What is the greatest whole number that must be a divisor of the product of any three consecutive positive integers?

gpt

We know that at least one of our three consecutive positive integers must be a multiple of $2$ since every other integer in a list of consecutive integers is divisible by $2$. Similarly, one of our three consecutive integers must also be divisible by $3$. Thus, the product of our three integers must be divisible by $2 \cdot 3 = 6$. By choosing the example where our three consecutive integers are $1$, $2$, and $3$ and their product is $6$, we see that $\boxed{6}$ is indeed the greatest whole number that must be a factor of the product of any three consecutive positive integers.

human

If $A,B$ and $C$ are non-zero distinct digits in base $6$ such that $\overline{ABC}_6 + \overline{BCA}_6+ \overline{CAB}_6 = \overline{AAA0}_6$, find $B+C$ in base $6$.

gpt

By definition of a base, it follows that $\overline{ABC}_6 = 6^2 \cdot A + 6 \cdot B + C$. Noting that every digit appears in every possible slot once, it follows that $\overline{ABC}_6 + \overline{BCA}_6+ \overline{CAB}_6 = (6^2 + 6 + 1)(A + B + C).$ The value is equal to the sum, $\overline{AAA0}_6 = 6^3 \cdot A + 6^2 \cdot A + 6 \cdot A = (6^2 + 6 + 1) \cdot (6 \cdot A)$. Setting them equal, $$(6^2 + 6 + 1)(A + B + C) = (6^2 + 6 + 1) \cdot (6 \cdot A) \Longrightarrow B+C = 5 \cdot A.$$Since $B,C < 6$, then $B+C < 2 \cdot 6$, so $A = 1,2$. However, there do not exist distinct base $6$ digits such that $B + C = 2 \cdot 5$, so it follows that $A = 1_6$, and $B+C = \boxed{5}_6$.

human

How many distinct, natural-number factors does $4^3 \cdot 5^4 \cdot 6^2$ have?

gpt

We prime factorize the given number as $2^8\cdot 3^2\cdot 5^4$. A factor of this number takes the form $2^a3^b5^c$ for integers $a$ between 0 and 8, $b$ between 0 and 2, and $c$ between 0 and 4. There are $9$ ways to choose $a$, 3 ways to choose $b$, and 5 ways to choose $c$. In total, there are $9\cdot3\cdot5 = \boxed{135}$ factors.

human

Find the units digit of the following within the indicated number base: $52_7 + 62_7$

gpt

Since $2_7 + 2_7 = 4_7$, the units digit is $\boxed{4}$.

human

Divide the product of the first five positive composite integers by the product of the next five composite integers. Express your answer as a common fraction.

gpt

The first ten positive composite integers are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18. The desired quotient is $\frac{4\cdot6\cdot8\cdot9\cdot10}{12\cdot14\cdot15\cdot16\cdot18}$. After cancellation, we get $\frac{1}{14\cdot3}=\boxed{\frac{1}{42}}$.

human

How many of the positive divisors of 3240 are multiples of 3?

gpt

$$ 3240 = 2^3 \cdot 3^4 \cdot 5^1 $$A positive divisor of 3240 is a multiple of 3 when it has a prime factorization in the form $2^a \cdot 3^b \cdot 5^c$ where $0 \le a \le 3$, $1 \le b \le 4$, and $0 \le c \le 1$. There are $4 \cdot 4 \cdot 2 = \boxed{32}$ choices for $a$, $b$, and $c$, giving the number of positive divisors of 3240 that are multiples of 3.

human

Convert $\rm{A}03_{16}$ to a base 10 integer, where the 'digits' A through F represent the values 10, 11, 12, 13, 14, and 15 in order.

gpt

In base 16, $A = 10$, $B = 11$, $C = 12$, $D = 13$, $E = 14$, and $F = 15$. So $A03 = 10\cdot16^2 + 0\cdot16 + 3 = \boxed{2563}.$

human

Find the integer $n$, $0 \le n \le 5$, such that \[n \equiv -3736 \pmod{6}.\]

gpt

Since $-3736 \equiv 2 \pmod{6}$, the integer $n$ we seek is $n = \boxed{2}$.

human

When the set of natural numbers is listed in ascending order, what is the smallest prime number that occurs after a sequence of five consecutive positive integers all of which are nonprime?

gpt

Consider the differences between consecutive prime numbers and look for the first difference of 6 or greater. The first several prime numbers are \[ 2,3,5,7,11,13,17,19,23,29,31, 37,\ldots, \] and the differences between consecutive terms of this sequence are \[ 1,2,2,4,2,4,2,4,6,2,\ldots. \] The first appearance of a difference of 6 or greater occurs between 23 and $\boxed{29}$.

human

Suppose that $n$ is a positive integer such that in base $7$, then $n$ can be expressed as $\overline{ABC}_7$, and in base $11$, then $n$ can be expressed as $\overline{CBA}_{11}$. Find the largest possible value of $n$ in base $10$.

gpt

We convert $n$ to base $10$. The base $7$ expression implies that $n = 49A + 7B + C$, and the base $11$ expression implies that $n = 121C + 11B + A$. Setting the two expressions equal to each other yields that $$n = 49A + 7B + C = 121C + 11B + A \Longrightarrow 48A - 4B - 120C = 0.$$Isolating $B$, we get $$B = \frac{48A - 120C}{4} = 12A - 30C = 6(2A - 5C).$$It follows that $B$ is divisible by $6$, and since $B$ is a base $7$ digit, then $B$ is either $0$ or $6$. If $B$ is equal to $0$, then $2A - 5C = 0 \Longrightarrow 2A = 5C$, so $A$ must be divisible by $5$ and hence must be either $0$ or $5$. Since $n$ is a three-digit number in base $7$, then $A \neq 0$, so $A = 5$ and $C = 2$. Thus, $n = 502_7 = 5 \cdot 7^2 + 2 = 247$. If $B$ is equal to $6$, then $2A - 5C = 1$, so $2A - 1 = 5C$ and $2A - 1$ must be divisible by 5. Since $A$ is a base $7$ digit, it follows that $A = 3$ and $C = 1$. This yields the value $n = 361_7 = 3 \cdot 7^2 + 6 \cdot 7 + 1 = 190$. The largest possible value of $n$ in base $10$ is $\boxed{247}$.

human

What is the sum of all two-digit positive integers whose squares end with the digits 01?

gpt

If $n$ is a two-digit number, then we can write $n$ in the form $10a + b$, where $a$ and $b$ are digits. Then the last digit of $n^2$ is the same as the last digit of $b^2$. The last digit of $n^2$ is 1. We know that $b$ is a digit from 0 to 9. Checking these digits, we find that the units digit of $b^2$ is 1 only for $b = 1$ and $b = 9$. If $b = 1$, then $n = 10a + 1$, so \[n^2 = 100a^2 + 20a + 1.\] The last two digits of $100a^2$ are 00, so we want the last two digits of $20a$ to be 00. This occurs only for the digits $a = 0$ and $a = 5$, but we reject $a = 0$ because we want a two-digit number. This leads to the solution $n = 51$. If $b = 9$, then $n = 10a + 9$, so \[n^2 = 100a^2 + 180a + 81 = 100a^2 + 100a + 80a + 81.\] The last two digits of $100a^2 + 100a$ are 00, so we want the last two digits of $80a + 81$ to be 01. In other words, we want the last digit of $8a + 8$ to be 0. This only occurs for the digits $a = 4$ and $a = 9$. This leads to the solutions $n = 49$ and $n = 99$. Therefore, the sum of all two-digit positive integers whose squares end with the digits 01 is $51 + 49 + 99 = \boxed{199}$.

human

Two numbers $90$ and $m$ share exactly three positive divisors. What is the greatest of these three common divisors?

gpt

Recall that the common divisors of two integers are precisely the divisors of the greatest common divisor. So, for two numbers to have exactly three positive divisors in common, those divisors must be $1$, $p$, and $p^2$ such that $p$ is prime. We now look at the prime factorization of $90$: $90=2 \cdot 3^2 \cdot 5$. Since $3^2$ is the only perfect square divisor of $90$, the divisors that $90$ and $m$ share must be $1$, $3$, and $9$. The largest of these three numbers is $\boxed{9}$.

human

The remainders when three positive integers are divided by 5 are 1, 2, and 3. Find the remainder when their product is divided by 5.

gpt

The product of the three integers is equivalent in modulo 5 to the product of the modulo 5 residues of the three integers. We multiply these residues to find the remainder: $$ 1 \cdot 2 \cdot 3 \equiv 6 \equiv \boxed{1} \pmod{5}. $$

human

What is the sum of the last two digits of this portion of the Fibonacci Factorial Series: $1!+1!+2!+3!+5!+8!+13!+21!+34!+55!+89!$?

gpt

This expression $n!$, is the number you get by multiplying $n$ by $(n-1)$ by $(n-2)$ by $(n-3)$ and so on, all the way down to $1$. So $5! = (5)(4)(3)(2)(1) = 120$. Notice that $5!$ ends in a $0$ since it has a factor of $10$ (there is a $5$ and a $2$ in it list of factors) and that $10!$ has to end in two zeroes since it has a factor of $10$, $5$ and $2$ which is really a factor of $100$. Since any factorial greater than $10$ (such as $13!$ or $21!$) includes all of the factors of $10!$, the last two digits of $13!$, $21!$, and so on are zeroes. These terms, therefore will not affect the last two digits of the sum of the Fibonacci factorial series. To find the last two digits, you only need to find the last two digits of each of the terms of $1! + 1! + 2! + 3! + 5! + 8!$. We do not need to calculate $8!$, only to find its last two digits. Starting with $5!$, we can work our way to $8!$, using only the last two digits of each value along the way. We know $5! = 120$, so use $20$ when finding $6!$, which will bring us to $6(20) = 120$ or $20$. Therefore, the last two digits of $7!$ are from $7(20) = 140$ or $40$. Finally $8!$ is $8(40) = 320$ or finally $20$. The last two digits of the entire series will come from $1 + 1 + 2 + 6 + 20 + 20 = 50$. Therefore, the sum of the last two digits is $5 + 0 = \boxed{5}$.

human

Let $A$ be the set of all numbers which can be represented as the sum of three consecutive positive integers. What is the greatest common divisor of all numbers in $A$?

gpt

Each member in the set is of the form $(x-1)+x+(x+1)=3x$. Since $x$ can be any positive integer, the greatest common divisor of all these members is $\boxed{3}$.

human

What is the largest perfect square factor of 1512?

gpt

Let's find the prime factorization of 1512: $1512=2^3\cdot189=2^3\cdot3^3\cdot7$. The only two squares of primes that divide 1512 are $2^2=4$ and $3^2=9$. Therefore, the largest perfect square factor of 1512 is $2^2\cdot3^2=(2\cdot3)^2=\boxed{36}$.

human

Given that $13^{-1} \equiv 29 \pmod{47}$, find $34^{-1} \pmod{47}$, as a residue modulo 47. (Give a number between 0 and 46, inclusive.)

gpt

Note that $34 \equiv -13 \pmod{47}$. Hence, \begin{align*} 34^{-1} &\equiv (-13)^{-1} \\ &\equiv (-1)^{-1} \cdot 13^{-1} \\ &\equiv (-1) \cdot 29 \\ &\equiv \boxed{18} \pmod{47}. \end{align*}

human

Daniel has exactly $20$ pieces of candy. He has to divide them equally among his $3$ sisters. What is the least number of pieces he should take away so that he could distribute the candy equally?

gpt

When you divide $20$ by $3$, you get $6$ with a remainder of $2$. Therefore, he should take away $\boxed{2 \text{ pieces}}$ so he could give each of his sisters $6$ pieces.

human

Let $f(x) = x^2 - x + 2010$. What is the greatest common divisor of $f(100)$ and $f(101)$?

gpt

We can write $f(x) = x(x-1) + 2010$. From here, it is clear that $f(101) = 101\cdot 100 + 2010$ and $f(100) = 100\cdot 99 + 2010$. We now use the Euclidean algorithm. \begin{align*} &\gcd(f(101), f(100)) \\ &= \gcd(101\cdot 100 + 2010, \, \, 100\cdot 99 + 2010) \\ &= \gcd(100\cdot 99 + 2010, \, \, 101\cdot 100 + 2010 - (100\cdot 99 + 2010)) \\ &= \gcd(100\cdot 99 + 2010, \, \, 2\cdot 100) \\ &= \gcd(100\cdot 99 + 2000 + 10, \, \, 2\cdot 100) \\ & = \gcd(100\cdot 99 + 100\cdot 20 + 10, 2\cdot 100) \\ & = \gcd(100\cdot 119 + 10, \, \, 2\cdot 100) \\ & = \gcd(2\cdot 100, \, \, 100\cdot 119 + 10 - 59(2\cdot 100)) \\ & = \gcd(2\cdot 100, \, \, 100\cdot 119 + 10 - 118\cdot 100) \\ & = \gcd(2\cdot 100, \, \, 100 + 10) \\ & = \gcd(200, \, \, 110) \\ & = \gcd(90, \, \, 110) \\ & = \gcd(20, \, \, 90) \\ & = \gcd(20, \, \, 90-4\cdot 20) \\ & = \gcd(20, \, \, 10) \\ &= \boxed{10}. \end{align*}

human

In base $b$, there are exactly one hundred three-digit numbers whose digits are all distinct. (That's "one hundred" in the ordinary sense, $100_{10}$.) What is $b$?

gpt

To form a three-digit number in base $b$ whose digits are all distinct, we must choose a first digit, a second digit, and a third digit. We have $b-1$ choices for the first digit ($1,2,3,\ldots,b-2,b-1$). We have $b-1$ choices for the second digit ($0,1,2,\ldots,b-2,b-1$, with the first digit removed from our choices). We have $b-2$ choices for the third digit. So, $$(b-1)^2(b-2) = 100.$$Trial and error is arguably the most reasonable way to solve this equation! Since $100=5\cdot 5\cdot 4$, the answer is $b=\boxed{6}$.

human

A math class has fewer than $40$ students. When the students try to sit in rows of $7$, $3$ students sit in the last row. When the students try to sit in rows of $5$, $1$ student is left in the last row. How many students are in this class?

gpt

Let $n$ be the number of students in the class. When we divide $n$ by $7$, we get a remainder of $3$. Therefore, we need to add multiples of $7$ to $3$ until we get a number which, when we divide by $5$, has a remainder of $1$. When we add $28$ to $3$, we get $31$, a number that gives a remainder of $1$ when dividing by $5$. By the Chinese remainder theorem, the other integers which leave a remainder of 3 when divided by 7 and a remainder of 1 when divided by 5 differ from 31 by a multiple of $7\cdot 5=35$. Therefore, the only integer between 0 and 40 satisfying these conditions is 31, and there are $\boxed{31}$ students in this math class.

human

What is the first digit (from left to right) of the base $8$ representation of $473_{10}$?

gpt

The largest power of $8$ that is still less than $473$ is $8^2 = 64$ and the largest multiple of $64$ that is less than $473$ is $7 \cdot 64 = 448$. Thus, when $473_{10}$ is written in base $8$, its first digit is $\boxed{7}$.

human

What is the first nonzero digit to the right of the decimal point of the fraction $\frac{1}{129}$?

gpt

If we multiply this fraction by 10, we do not change any of the digits; we simply shift all the digits one place to the left. Therefore, if we multiply the fraction by a sufficiently large power of 10, then the first digit obtained when dividing by 129 will be the first nonzero digit to the right of the decimal point of $\frac{1}{129}$. Observing that $100 < 129 < 1000$, we compute $\frac{1000}{129}=7 \frac{97}{129}$. So the first nonzero digit to the right of the decimal point of $\frac{1}{129}$ is $\boxed{7}$.

human

The volume of the box is 240 cubic units where $a,b,$ and $c$ are integers with $1<c<b<a$. What is the largest possible value of $b$? [asy] draw((0,0)--(0,3)--(6,3)--(6,0)--cycle); draw((2,1.5)--(2,4.5)--(8,4.5)--(8,1.5)--cycle); draw((0,0)--(2,1.5)--(2,4.5)--(0,3)--cycle); draw((6,0)--(6,3)--(8,4.5)--(8,1.5)--cycle); label("a",(3,0),S); label("b",(7,0.75),SE); label("c",(8,3.5),E); [/asy]

gpt

The prime factorization of 240 is $2^4\cdot3\cdot5$. We want $c$ to be as small as possible, so $c=2$. Now we have $a\cdot b=2^3\cdot3\cdot5=120$. For the maximum $b$, we seek the minimum $a$ while $b<a$. If $b<a$ then $a\cdot b<a^2$, so $120<a^2$. That means $a$ is at least 11. However, 11 is not a factor of 240. The least factor of 240 greater than 11 is $2^2\cdot3=12$. So the minimum $a$ is 12 and that makes the maximum $b=\frac{2^3\cdot3\cdot5}{2^2\cdot3}=2\cdot5=\boxed{10}$.

human

The least common multiple of two numbers is 3780, and the greatest common divisor is 18. Given that one of the numbers is 180, what is the other number?

gpt

We use the identity $\gcd(a,b) \cdot \mathop{\text{lcm}}[a,b] = ab$ for all positive integers $a$ and $b$. We are told that $\gcd(a,b) = 18$, and $\mathop{\text{lcm}}[a,b] = 3780$, so $ab = 18 \cdot 3780$. If one number is 180, then the other number is $18 \cdot 3780/180 = \boxed{378}$.

human

The sequence 12, 15, 18, 21, 51, 81, $\ldots$ consists of all positive multiples of 3 that contain at least one digit that is a 1. What is the $50^{\mathrm{th}}$ term of the sequence?

gpt

We know that the rule for divisibility by $3$ is that the digits of the number must add up to a multiple of $3$. So, it's clear that there are no other such two-digit numbers beyond the ones listed in the problem. Every number divisible by $3$ between $100$ and $199$ is in the sequence, so that gets us through the $39$th term of the sequence. Using the rule for divisibility by $3$, it is fairly simple to list out the remaining $11$ terms of the sequence: $201, 210, 213, 216, 219, 231, 261, 291, 312, 315, 318$. Thus, the $50$th term is $\boxed{318}$.

human

Each pack of hot dogs contains 10 hot dogs (no buns), but each pack of hot dog buns contains 8 buns. Phil buys a number of these packs for a barbecue. After the barbecue, Phil finds that he has 4 hot dogs left over. What is the SECOND smallest number of packs of hot dogs he could have bought?

gpt

Let $n$ be the number of packs of hot dogs that Phil bought. Then $10n \equiv 4 \pmod{8}$, which reduces to $2n \equiv 4 \pmod{8}$. This congruence tells us that $2n = 8k + 4$ for some integer $k$, or $n = 4k + 2$. The second smallest positive integer of this form is $\boxed{6}$.

human

Abigail, Beatrice, and Carson combine their eggs to sell them at the market. If Abigail has 37 eggs, Beatrice has 49 eggs, and Carson has 14 eggs, and if eggs can only be sold in cartons of 12, how many eggs will be left over if all cartons are sold?

gpt

The remainder when $37+49+14$ is divided by $12$ is the same as the remainder when the congruences of each number mod 12 are summed and then divided by 12. In other words, since we have \begin{align*} 37 &\equiv 1\pmod{12}\\ 49 &\equiv 1\pmod{12} \\ 14 &\equiv 2\pmod{12} \end{align*}then therefore, $37+49+14 \equiv1+1+2 \equiv \boxed{4}\pmod{12}$.

human

In any month with five Mondays, a school declares the fifth Monday a Marvelous Monday. School begins on Monday, September 4. What is the date of the first Marvelous Monday after school starts? (If you abbreviate, do not use a dot in the abbreviation.)

gpt

September has 30 days. September 4 is a Monday, so September 9 is a Saturday. Since September 30 is exactly 21 days later (or 3 weeks), September 30 is also a Saturday. Then October 1 is a Sunday, and October 2 is a Monday. Then October 2, 9, 16, 23, and 30 are all Mondays, so the first Marvelous Monday is $\boxed{\text{October 30}}$.

human

What is the 308th digit to the right of the decimal point when $\frac{12}{37}$ is expressed as a decimal?

gpt

We use long division to find that the decimal representation of $\frac{12}{37}$ is $0.\overline{324}$. When $308$ is divided by $3$ there is a remainder of $2$ $\left( 308\div 3=102 \ R2\right)$. Therefore the 308th digit to the right of the decimal point is the second digit of $324$, which is $\boxed{2}$.

human

What is the sum of all positive integer divisors of 77?

gpt

Since $77=7\cdot11$, the divisors of 77 are 1, 7, 11, and 77. Their sum is $1+7+11+7\cdot11=\boxed{96}$.

human

How many even perfect square factors does $2^4 \cdot 7^9$ have?

gpt

A positive integer is a factor of $2^4\cdot7^9$ if and only if its prime factorization is of the form $2^a\cdot 7^b$ for exponents $a$ and $b$ satisfying $0\leq a \leq 4$ and $0\leq b\leq 9$. A positive integer is a perfect square if and only if the exponents in its prime factorization are even. A positive integer is even if and only if the exponent of 2 in its prime factorization is at least 1. Therefore, we may choose $a=2$ or $4$ and $b=0,$ $2,$ $4,$ $6,$ or $8.$ Since we have 2 choices for $a$ and 5 choices for $b$, there are $2\times5=\boxed{10}$ ways to make these two decisions.

human

What integer $n$ satisfies $0\le n<9$ and $$-1111\equiv n\pmod 9~?$$

gpt

An integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. Using this fact, we find that the greatest multiple of $9$ that is less than $-1111$ is $-1116$. We have $-1111 = -1116+5$. That is, $-1111$ is equal to a multiple of $9$ plus $5$. Therefore, $$-1111\equiv \boxed{5}\pmod 9.$$To verify this answer, we can check that $9$ divides $(-1111)-5 = -1116$; it does, so our answer is correct.

human

How many distinct, positive factors does $1100$ have?

gpt

The prime factorization of $1100$ is $2^2\cdot5^2\cdot11$. To find the number of factors, we consider that each factor has a prime factorization of $2^a\cdot5^b\cdot11^c$, where $a$ could be from $0$ to $2$ (3 possible values), $b$ could be from $0$ to $2$ (3 possible values), and $c$ could be $0$ or $1$ (2 possible values). So the number of factors is $3\cdot3\cdot2=\boxed{18}$.

human

What is the units digit of $23^{23}$?

gpt

Let's find the cycle of units digits of $3^n$, starting with $n=1$ (note that the tens digit 2 in 23 has no effect on the units digit): $3, 9, 7, 1, 3, 9, 7, 1,\ldots$ . The cycle of units digits of $23^{n}$ is 4 digits long: 3, 9, 7, 1. Thus, to find the units digit of $23^n$ for any positive $n$, we must find the remainder, $R$, when $n$ is divided by 4 ($R=1$ corresponds to the units digit 3, $R=2$ corresponds to the units digit 9, etc.) Since $23\div4=5R3$, the units digit of $23^{23}$ is $\boxed{7}$.

human

What is the smallest positive four-digit number divisible by 9 which has two even and two odd digits?

gpt

For a number to be divisible by 9, the sum of its digits must be divisible by 9. But since the number has two even digits and two odd digits, the sum of its digits is even. Thus the sum of its digits must be at least 18. This number will be minimized if its thousands digit is 1 and its hundreds digit is 0. This means the remaining two digits must sum to 17, and are hence 8,9. So we see that the smallest possible integer of the desired form is $\boxed{1089}$.

human

How many positive factors of 72 are perfect cubes?

gpt

Prime factorize $72$ as $2^3\cdot 3^2$. A positive integer is a factor of 72 if and only if the exponents in its prime factorization are less than or equal to the corresponding exponents in the prime factorization of 72. Also, a positive integer is a perfect cube if and only if every exponent is a multiple of 3. Therefore, in forming a perfect cube factor of 72, we have 2 choices for the exponent of $2$ (either 0 or 3) and only 1 choice for the exponent of 3 (namely 0). There are $2\cdot 1=\boxed{2}$ ways to make these choices.

human

The prime factorization of 2007 is $3^{2}\times223$. How many ordered pairs of positive integers $(x,y)$ satisfy the equation $xy=2007$?

gpt

$x$ must be a divisor of 2007. To count the number of divisors, we note that a divisor can have zero factors of 3, one factor, or two factors, and (independently) it can have zero factors of 223 or one factor. The total number of divisors is then $3\cdot2=6$ (3 choices for the factors of 3 and 2 choices for the factors of 223). Since each divisor is unique, each creates a unique ordered pair. So there are $\boxed{6}$ ordered pairs.

human

What is the smallest integer $k$ such that $k>1$ and $k$ has remainder $1$ when divided by any of $17,$ $6,$ and $2?$

gpt

We can re-write the problem as the following three equations: $$k = 17a+1\\k = 6b+1 \\ k = 2c + 1$$Therefore, $k-1$ is divisible by $17,$ $6,$ and $2.$ The smallest positive value of $k-1$ is thus $$\text{lcm}[17,6,2] = \text{lcm}[17,6] = 17\cdot 6 = 102,$$and so the smallest possible value of $k$ is $k = 102+1 = \boxed{103}.$

human

Let $a\equiv (3^{-1}+5^{-1}+7^{-1})^{-1}\pmod{11}$. What is the remainder when $a$ is divided by $11$?

gpt

One way to do this is to find each inverse explicitly: \begin{align*} (3^{-1}+5^{-1}+7^{-1})^{-1} &\equiv (4+9+8)^{-1} \pmod{11} \\ &\equiv 21^{-1} \pmod{11} \\ &\equiv 10^{-1} \pmod{11} \\ &\equiv \boxed{10}\pmod{11}. \end{align*} Another way to do this is through manipulation: \begin{align*} & (3^{-1}+5^{-1}+7^{-1})^{-1}\\ \equiv~ & (3\cdot 5\cdot 7)(3\cdot 5\cdot 7)^{-1}(3^{-1}+5^{-1}+7^{-1})^{-1}\\ \equiv~ & (3\cdot 5\cdot 7)(3\cdot 5+5\cdot 7+ 7\cdot 3)^{-1}\\ \equiv~ & 6\cdot(15+35+21)^{-1}\\ \equiv~ & 6\cdot 5^{-1}\\ \equiv~ & 6\cdot 9\\ \equiv~ & \boxed{10} \pmod{11} \end{align*}

human

What is the remainder of $8^6 + 7^7 + 6^8$ divided by 5?

gpt

First of all, $6$ has a remainder of $1$ when divided by $5,$ thus any power of $6$ will have a remainder of $1$ when divided by $5.$ As for $7,$ it has a remainder of $2$ when divided by $5$, so let us look at the powers of $2$: \begin{align*} 2^1 &\equiv 2 \pmod{5} \\ 2^2 &\equiv 4 \pmod{5} \\ 2^3 &\equiv 3 \pmod{5} \\ 2^4 &\equiv 1 \pmod{5}. \end{align*}Since $2^4 \equiv 1 \pmod{5},$ we see that $2^{7} \equiv 2^3 \cdot 2^4 \equiv 3 \pmod{5},$ hence the remainder of $7^4$ when divided by $5$ is $3.$ Now, there is a bit of a shortcut we can use for $8.$ Since $8 \equiv -2 \pmod{5},$ we can see that $8^6 \equiv (-2)^6 \equiv 2^6 \equiv 2^2 \cdot 2^4 \equiv 4 \pmod {5},$ hence our desired remainder is $4.$ Adding them up, we have $4 + 3 + 1 \equiv \boxed{3} \pmod{5}.$

human

What is the remainder of $5^{2010}$ when it is divided by 7?

gpt

We start by writing out some powers of five modulo 7. \begin{align*} 5^1 &\equiv 5 \pmod{7} \\ 5^2 &\equiv 4 \pmod{7} \\ 5^3 &\equiv 6 \pmod{7} \\ 5^4 &\equiv 2 \pmod{7} \\ 5^5 &\equiv 3 \pmod{7} \\ 5^6 &\equiv 1 \pmod{7} \end{align*}Therefore, we have that $5^6 \equiv 1$ modulo 7. Thus, $5^{2010} \equiv (5^6)^{335} \equiv 1^{335} \equiv \boxed{1}$ modulo 7.

human

Find the largest prime factor of $15^3+10^4-5^5$.

gpt

The greatest common divisor of the three terms is $5^3$. Factoring $5^3$ out of each term and using the distributive property gives: \begin{align*} 15^3+10^4-5^5 &= 5^3\cdot3^3 + 5^3\cdot5\cdot2^4-5^3\cdot5^2 \\ &= 5^3(3^3+5\cdot2^4-5^2)\\ & = 5^3(27+80-25) \\ &= 5^3(82)=2\cdot5^3\cdot41. \end{align*}So the largest prime factor is $\boxed{41}$.

human

What is the sum of the six positive integer factors of 18?

gpt

$18=1\cdot18=2\cdot9=3\cdot6$. The sum is $1+18+2+9+3+6=\boxed{39}$.

human

How many integers between $500$ and $1000$ contain both the digits $3$ and $4$?

gpt

First we notice that the hundreds digit of any integer between $500$ and $1000$ cannot be $3$ or $4$, so $3$ and $4$ must be the units and tens digits (in either the order $34$ or $43$). Since the integer must be between $500$ and $1000$, there are $5$ choices for the hundreds digit of this integer ($5$, $6$, $7$, $8$, or $9$). Thus, there are $2 \times 5 = \boxed{10}$ ways to form such a number.

human

The product of the positive integer divisors of a positive integer $n$ is 729. Find $n$.

gpt

Suppose that $n$ has $d$ divisors and that $d$ is even. Since the divisors come in pairs whose product is $n$, the product of the divisors of $n$ is $n^{d/2}$. For example, if $n=12$ then the product of the divisors is $(1\cdot 12)(2\cdot 6)(3\cdot 4)=12^3$. If $d$ is odd, then there are $(d-1)/2$ pairs which give a product of $n$, as well as the divisor $\sqrt{n}$. So again the product of the divisors is $n^{(d-1)/2}n^{1/2}=n^{d/2}$. For example, if $n=16$, then the product of the divisors of $n$ is $(1\cdot 16)(2\cdot 8)(4)=16^{5/2}$. To summarize, we have found that the product of the positive integers divisors of a positive integer $n$ is $n^{d/2}$, where $d$ is the number of divisors of $n$. Writing 729 as $3^6$, then, we have $n^{d/2}=3^6$, which implies $n^d=3^{12}$. So the possible values of $(n,d)$ are $(3,12)$, $(9,6)$, $(27,4)$, $(81,3)$ and $(729,2)$. We find that only in the third case $(n,d)=(27,4)$ is $d$ the number of divisors of $n$, so $n=\boxed{27}$.

human

Grady distributed $x$ pieces of candy evenly among nine Halloween bags such that every bag received the greatest possible number of whole pieces of candy, but some candy was left over. What is the greatest possible number of pieces that could have been left over?

gpt

Grady distributes his candy in groups of 9 until he no longer has any groups left. The largest number of possible pieces left is $\boxed{8},$ since if he has more than 8 he can distribute another group of 9.

human

When the greatest common divisor and least common multiple of two integers are multiplied, their product is 200. How many different values could be the greatest common divisor of the two integers?

gpt

We know that $\gcd(a,b) \cdot \mathop{\text{lcm}}[a,b] = ab$ for all positive integers $a$ and $b$. Hence, in this case, $ab = 200$. The prime factorization of 200 is $2^3 \cdot 5^2$, so $a = 2^p \cdot 5^q$ and $b = 2^r \cdot 5^s$ for some nonnegative integers $p$, $q$, $r$, and $s$. Then $ab = 2^{p + r} \cdot 5^{q + s}$. But $ab = 200 = 2^3 \cdot 5^2$, so $p + r = 3$ and $q + s = 2$. We know that $\gcd(a,b) = 2^{\min\{p,r\}} \cdot 5^{\min\{q,s\}}$. The possible pairs $(p,r)$ are $(0,3)$, $(1,2)$, $(2,1)$, and $(3,0)$, so the possible values of $\min\{p,r\}$ are 0 and 1. The possible pairs $(q,s)$ are $(0,2)$, $(1,1)$, and $(2,0)$, so the possible values of $\min\{q,s\}$ are 0 and 1. Therefore, the possible values of $\gcd(a,b)$ are $2^0 \cdot 5^0 = 1$, $2^1 \cdot 5^0 = 2$, $2^0 \cdot 5^1 = 5$, and $2^1 \cdot 5^1 = 10$, for a total of $\boxed{4}$ possible values.

human

The greatest common divisor of two positive integers is $(x+5)$ and their least common multiple is $x(x+5)$, where $x$ is a positive integer. If one of the integers is 50, what is the smallest possible value of the other one?

gpt

We know that $\gcd(m,n) \cdot \mathop{\text{lcm}}[m,n] = mn$ for all positive integers $m$ and $n$. Hence, in this case, the other number is \[\frac{(x + 5) \cdot x(x + 5)}{50} = \frac{x(x + 5)^2}{50}.\]To minimize this number, we minimize $x$. We are told that the greatest common divisor is $x + 5$, so $x + 5$ divides 50. The divisors of 50 are 1, 2, 5, 10, 25, and 50. Since $x$ is a positive integer, the smallest possible value of $x$ is 5. When $x = 5$, the other number is $5 \cdot 10^2/50 = 10$. Note that that the greatest common divisor of 10 and 50 is 10, and $x + 5 = 5 + 5 = 10$. The least common multiple is 50, and $x(x + 5) = 5 \cdot (5 + 5) = 50$, so $x = 5$ is a possible value. Therefore, the smallest possible value for the other number is $\boxed{10}$.

human

What is the sum of all positive integers less than 100 that are squares of perfect squares?

gpt

Squares of perfect squares are fourth powers. $1^4=1$, $2^4=16$, and $3^4=81$ are the only fourth powers less than 100. Their sum is $1+16+81=\boxed{98}$.