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What is the greatest common factor of 5!, 10!, and 15!? | Level 1 | Counting & Probability | Note that $5!$ divides $10!$ and $5!$ divides $15!$. Since $5!$ has no factor larger than $5!$, and $5!$ is a factor of all three, the answer is $5!=\boxed{120}$. | [
{
"content": "What is the greatest common factor of 5!, 10!, and 15!?",
"role": "user"
}
] |
To find the greatest common factor (GCF) of 5!, 10!, and 15!, we start by considering the prime factorizations of these factorials.
1. **Prime factorization of 5!**:
- The exponents of the prime factors are:
- \(2^3\) (from \( \left\lfloor \frac{5}{2} \right\rfloor + \left\lfloor \frac{5}{4} \right\rfloor = ... | Okay, so I need to find the greatest common factor (GCF) of 5!, 10!, and 15!. Hmm, factorials, right? Let me remember how factorials work. 5! is 5 times 4 times 3 times 2 times 1, and similarly for 10! and 15!. The GCF of these three numbers would be the largest number that divides all three without leaving a remainder... | 23 | 2,253 | 1,483 |
Determine the number of ways to arrange the letters of the word ELLIPSE. | Level 2 | Counting & Probability | There are two E's, two L's, and seven total letters, so the answer is $\dfrac{7!}{2! \times 2!} = \boxed{1260}$. | [
{
"content": "Determine the number of ways to arrange the letters of the word ELLIPSE.",
"role": "user"
}
] |
To determine the number of ways to arrange the letters of the word "ELLIPSE", we start by noting the letters and their frequencies: E (2), L (2), I (1), P (1), S (1). The word has 7 letters in total with two pairs of repeated letters (E and L).
The formula for permutations of a multiset is used, which is the factori... | Okay, so I need to figure out how many ways there are to arrange the letters of the word "ELLIPSE". Hmm, let's see. The first thing I remember about permutations with repeated elements is that if all the letters were unique, the number of arrangements would just be factorial of the number of letters. But in this case, ... | 21 | 2,166 | 1,953 |
My school's math club has 6 boys and 8 girls. I need to select a team to send to the state math competition. We want 6 people on the team. In how many ways can I select the team to have 3 boys and 3 girls? | Level 2 | Counting & Probability | We are picking 3 boys out of 6, so there are $\binom{6}{3} = 20$ options for the boys on the team. We are picking 3 girls out of 8, so there are $\binom{8}{3} = 56$ options for the girls on the team. This gives a total of $20 \times 56 = \boxed{1120}$ choices. | [
{
"content": "My school's math club has 6 boys and 8 girls. I need to select a team to send to the state math competition. We want 6 people on the team. In how many ways can I select the team to have 3 boys and 3 girls?",
"role": "user"
}
] |
To determine the number of ways to select a team of 6 people (3 boys and 3 girls) from a math club with 6 boys and 8 girls, we use combinations. The formula for combinations is:
\[
\binom{n}{k} = \frac{n!}{k!(n - k)!}
\]
First, we calculate the number of ways to choose 3 boys out of 6:
\[
\binom{6}{3} = \frac{6!}{... | Okay, so I need to figure out how many ways I can select a team of 6 people from my school's math club, which has 6 boys and 8 girls. The requirement is that the team should have 3 boys and 3 girls. Hmm, let's see. I remember something about combinations in math class. Combinations are used when the order doesn't matte... | 64 | 1,707 | 1,478 |
What is the value of $101^{3} - 3 \cdot 101^{2} + 3 \cdot 101 -1$? | Level 2 | Counting & Probability | The given expression is the expansion of $(101-1)^3$. In general, the expansion of $(a-b)^3$ is equal to \[a^3-3\cdot a^2\cdot b+3\cdot a\cdot b^2-b^3\]
In this case, $a=101,b=1$. Thus $101^3-3\cdot 101^2+3\cdot 101-1=(101-1)^3$; we can easily calculate $100^3=\boxed{1000000}$. | [
{
"content": "What is the value of $101^{3} - 3 \\cdot 101^{2} + 3 \\cdot 101 -1$?",
"role": "user"
}
] |
The given expression is \(101^3 - 3 \cdot 101^2 + 3 \cdot 101 - 1\). Recognizing this as the expansion of \((a - b)^3\) where \(a = 101\) and \(b = 1\), we can simplify the expression using the binomial theorem:
\[
(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3
\]
Substituting \(a = 101\) and \(b = 1\), we get:
\[
(101 - 1... | Okay, let's see. I need to figure out the value of this expression: 101 cubed minus three times 101 squared plus three times 101 minus one. Hmm. That sounds a bit complicated, but maybe there's a trick or a formula that can simplify things. Let me think... Wait, the expression is 101³ - 3·101² + 3·101 - 1. Hmm. That st... | 36 | 1,235 | 920 |
What is the coefficient of $xy$ in the expansion of $(3x+(2y+1))^2$? | Level 3 | Counting & Probability | Using the Binomial Theorem, we can expand this to get $(3x)^2+2(3x)(2y+1)+(2y+1)^2$. The only $xy$ comes from the middle term $2(3x)(2y+1)=12xy+6x$, hence the coefficient is $\boxed{12}$. | [
{
"content": "What is the coefficient of $xy$ in the expansion of $(3x+(2y+1))^2$?",
"role": "user"
}
] |
To find the coefficient of \(xy\) in the expansion of \((3x + (2y + 1))^2\), we start by expanding the expression using the binomial theorem \((A + B)^2 = A^2 + 2AB + B^2\), where \(A = 3x\) and \(B = (2y + 1)\).
1. Expand \(A^2\):
\[
(3x)^2 = 9x^2
\]
2. Expand \(2AB\):
\[
2 \cdot 3x \cdot (2y + 1) =... | Alright, let's see. I need to find the coefficient of the xy term in the expansion of (3x + (2y + 1))². Hmm, okay. Let me start by recalling how to expand expressions like this. When you have something squared, like (a + b)², the formula is a² + 2ab + b². So maybe I can use that here.
First, let me simplify the expres... | 30 | 3,065 | 2,746 |
How many ways are there to put 5 balls in 2 boxes if the balls are not distinguishable but the boxes are? | Level 3 | Counting & Probability | Since the balls are indistinguishable, we need only count the number of balls in the distinguishable boxes. We can put 5, 4, 3, 2, 1, or 0 balls in Box 1 (and the rest go in Box 2). So there are $\boxed{6}$ different arrangements. | [
{
"content": "How many ways are there to put 5 balls in 2 boxes if the balls are not distinguishable but the boxes are?",
"role": "user"
}
] |
To determine the number of ways to put 5 indistinguishable balls into 2 distinguishable boxes, we consider the different distributions of balls between the boxes. Since the boxes are distinguishable, each distribution where the number of balls in each box differs is unique.
We can list the possible distributions:
1.... | Okay, so I need to figure out how many ways there are to put 5 indistinguishable balls into 2 distinguishable boxes. Hmm, let's think about this step by step.
First, since the balls are not distinguishable, the order in which we place them doesn't matter. What matters is how many balls end up in each box. The boxes a... | 30 | 2,044 | 1,724 |
What is the remainder when $301^4$ is divided by 10,000? | Level 3 | Counting & Probability | By the Binomial Theorem, we have \begin{align*}
301^4 &= (3(100) + 1)^4\\
&= \binom40 \cdot 3^4 \cdot 100^4 \cdot 1^0 + \binom41 \cdot 3^3 \cdot 100^3 \cdot 1^1 \\
&\qquad\qquad+ \binom42 \cdot 3^2 \cdot 100^2 \cdot 1^2+ \binom43 \cdot 3^1 \cdot 100^1 \cdot 1^3 \\
&\qquad\qquad+ \binom44 \cdot 3^0 \cdot 100^0 \cdot 1^... | [
{
"content": "What is the remainder when $301^4$ is divided by 10,000?",
"role": "user"
}
] |
To find the remainder when \(301^4\) is divided by 10,000, we can use the binomial theorem to expand \((300 + 1)^4\) and simplify modulo 10,000.
Expanding \((300 + 1)^4\) using the binomial theorem:
\[
(300 + 1)^4 = \sum_{k=0}^{4} \binom{4}{k} \cdot 300^{4-k} \cdot 1^k
\]
Breaking down each term:
- \(\binom{4}{0} \... | Okay, so I have this problem: What is the remainder when 301^4 is divided by 10,000? Hmm, let's see. I need to find 301⁴ mod 10,000. I remember that when dealing with large exponents modulo a number, there are some tricks like Euler's theorem or binomial expansion that can simplify things. Let me think about how to app... | 23 | 2,962 | 2,562 |
Given that $a$ and $b$ are real numbers such that $-3\leq a\leq1$ and $-2\leq b\leq 4$, and values for $a$ and $b$ are chosen at random, what is the probability that the product $a\cdot b$ is positive? Express your answer as a common fraction. | Level 5 | Counting & Probability | The probability that both $a$ and $b$ are positive is $\left(\frac{1}{4}\right)\left(\frac{2}{3}\right)=\frac{1}{6}$. The probability that $a$ and $b$ are both negative is $\left(\frac{3}{4}\right)\left(\frac{1}{3}\right)=\frac{1}{4}$. Since $ab$ is positive if and only if one of these two events occur, the probabilit... | [
{
"content": "Given that $a$ and $b$ are real numbers such that $-3\\leq a\\leq1$ and $-2\\leq b\\leq 4$, and values for $a$ and $b$ are chosen at random, what is the probability that the product $a\\cdot b$ is positive? Express your answer as a common fraction.",
"role": "user"
}
] |
Given that \(a\) and \(b\) are real numbers such that \(-3 \leq a \leq 1\) and \(-2 \leq b \leq 4\), we need to find the probability that the product \(a \cdot b\) is positive when values for \(a\) and \(b\) are chosen at random.
First, we calculate the total area of the rectangle formed by the ranges of \(a\) and \... | Okay, so I need to find the probability that the product of a and b is positive, given that a is between -3 and 1, and b is between -2 and 4. Both a and b are chosen at random, so I assume this means uniformly at random. Therefore, the probability would be the area of the region where a*b is positive divided by the tot... | 74 | 2,790 | 2,363 |
A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$, it moves at random to one of the points $(a-1,b)$, $(a,b-1)$, or $(a-1,b-1)$, each with probability $\frac{1}{3}$, independently of its previous moves. The prob... | Level 5 | Counting & Probability | One could recursively compute the probabilities of reaching $(0,0)$ as the first axes point from any point $(x,y)$ as\[P(x,y) = \frac{1}{3} P(x-1,y) + \frac{1}{3} P(x,y-1) + \frac{1}{3} P(x-1,y-1)\]for $x,y \geq 1,$ and the base cases are $P(0,0) = 1, P(x,0) = P(y,0) = 0$ for any $x,y$ not equal to zero. We then recurs... | [
{
"content": "A moving particle starts at the point $(4,4)$ and moves until it hits one of the coordinate axes for the first time. When the particle is at the point $(a,b)$, it moves at random to one of the points $(a-1,b)$, $(a,b-1)$, or $(a-1,b-1)$, each with probability $\\frac{1}{3}$, independently of its p... |
To solve the problem, we need to determine the probability that a particle starting at \((4,4)\) will hit the coordinate axes at \((0,0)\) for the first time. The particle can move to \((a-1,b)\), \((a,b-1)\), or \((a-1,b-1)\) each with probability \(\frac{1}{3}\). We use a dynamic programming approach to compute thi... | Okay, so I have this problem here where a particle starts at (4,4) and moves until it hits one of the coordinate axes for the first time. At each point (a,b), it can move to (a-1,b), (a,b-1), or (a-1,b-1), each with probability 1/3. I need to find the probability that it hits the axes at (0,0), and express it as m/3^n ... | 141 | 5,232 | 3,803 |
Ice-cream-o-rama is eager to advertise how many flavors it has. But it really only has three basic flavors: chocolate, vanilla, and strawberry. However, they can make "new" flavors by taking four scoops of ice cream of those basic flavors and blending them together. Different proportions of the basic flavors give diffe... | Level 5 | Counting & Probability | We can think of the $3$ basic flavors as $3$ distinguishable boxes, and the $4$ scoops as $4$ indistinguishable balls. For every ball we put in the chocolate box, for instance, we put a scoop of chocolate ice cream in the blending machine. In this way we can correlate each new flavor with an arrangement of balls in box... | [
{
"content": "Ice-cream-o-rama is eager to advertise how many flavors it has. But it really only has three basic flavors: chocolate, vanilla, and strawberry. However, they can make \"new\" flavors by taking four scoops of ice cream of those basic flavors and blending them together. Different proportions of the ... |
To determine the number of total flavors Ice-cream-o-rama can create by combining four scoops of three basic flavors (chocolate, vanilla, and strawberry), we need to count the different proportions of these basic flavors. This is a combinatorial problem where we distribute four indistinguishable scoops into three dis... | Okay, so Ice-cream-o-rama wants to advertise their number of flavors. They have three basic ones: chocolate, vanilla, and strawberry. But they can create new flavors by blending four scoops of these basic flavors. The key here is that different proportions of the basic flavors count as different new flavors. Even if it... | 123 | 2,374 | 2,122 |
An integer between $1000$ and $9999$, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?
| Level 5 | Counting & Probability | If the common sum of the first two and last two digits is $n$, such that $1 \leq n \leq 9$, there are $n$ choices for the first two digits and $n + 1$ choices for the second two digits (since zero may not be the first digit). This gives $\sum_{n = 1}^9 n(n + 1) = 330$ balanced numbers. If the common sum of the first tw... | [
{
"content": "An integer between $1000$ and $9999$, inclusive, is called balanced if the sum of its two leftmost digits equals the sum of its two rightmost digits. How many balanced integers are there?\n",
"role": "user"
}
] |
To determine the number of balanced integers between 1000 and 9999, we need to count four-digit numbers where the sum of the first two digits equals the sum of the last two digits.
A four-digit number \(ABCD\) is balanced if \(A + B = C + D\). We need to compute the number of such numbers for each possible sum \(S\... | Okay, so I need to figure out how many balanced four-digit integers there are between 1000 and 9999. A balanced integer is one where the sum of the first two digits equals the sum of the last two digits. Hmm, let me think about how to approach this.
First, let me understand what exactly constitutes a balanced number. ... | 48 | 4,774 | 4,210 |
The first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \begin{tabular}{ccccccccccc}
&&&&&1&&&&&\\
&&&&1&&1&&&&\\
&&&1... | Level 4 | Counting & Probability | Each number in Pascal's triangle is the sum of the two numbers above it. If we use $0\text{'s}$ and $1\text{'s}$ to stand for "even" and "odd", then by using the rules $0+0=0,$ $0+1=1,$ and $1+1=0,$ we can efficiently compute the parity (even- or oddness) of the entries without computing the entries themselves: \begin{... | [
{
"content": "The first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \\begin{tabular}{ccccccccccc}\n&&&&&1&&&... |
To determine how many of the first 20 rows (excluding row 0 and row 1) of Pascal's triangle have the property that all numbers except the 1s at each end are even, we need to analyze the parity of the binomial coefficients.
Key observations:
1. A row \( n \) in Pascal's triangle corresponds to the binomial coefficien... | Alright, so I need to figure out how many of the first 20 rows in Pascal's triangle (excluding row 0 and row 1) have the property that all the numbers in the row, except for the 1s at each end, are even. From the example given, rows 2 and 4 have this property. The problem mentions row 4 as having only even numbers exce... | 142 | 4,123 | 3,733 |
Two standard 6-sided dice are tossed. What is the probability that the sum of the numbers shown on the dice is a prime number? Express your answer as a common fraction. | Level 4 | Counting & Probability | There are $6\cdot6=36$ possible outcomes when two dice are tossed. The greatest possible sum is 12 and the prime numbers less than 12 are 2, 3, 5, 7, 11. There is exactly one way to get a sum of 2 (1+1), exactly two ways to get 3 (1+2, 2+1), exactly four ways to get 5 (1+4, 4+1, 2+3, 3+2), exactly six ways to get 7 (1+... | [
{
"content": "Two standard 6-sided dice are tossed. What is the probability that the sum of the numbers shown on the dice is a prime number? Express your answer as a common fraction.",
"role": "user"
}
] |
To find the probability that the sum of two standard 6-sided dice is a prime number, we first identify the prime numbers between 2 and 12, which are 2, 3, 5, 7, and 11.
Next, we count the number of ways to achieve each of these prime sums:
- Sum of 2: Only (1,1), which is 1 way.
- Sum of 3: (1,2) and (2,1), which a... | Okay, so I need to find the probability that the sum of two standard 6-sided dice is a prime number. Let me think step by step. First, I know that when you toss two dice, the possible sums range from 2 to 12, right? Because the smallest each die can show is 1, so 1+1=2, and the largest is 6+6=12. Now, prime numbers bet... | 41 | 1,498 | 1,190 |
Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a $\diamondsuit$ and the second card is an ace? | Level 4 | Counting & Probability | We have two cases because if the first card is a $\diamondsuit$, it could be an ace or not be an ace.
There is a $\dfrac{1}{52}$ chance that the ace of $\diamondsuit$ is drawn first, and a $\dfrac{3}{51} = \dfrac{1}{17}$ chance that the second card drawn is one of the three remaining aces, which gives a probability of... | [
{
"content": "Two cards are dealt at random from a standard deck of 52 cards. What is the probability that the first card is a $\\diamondsuit$ and the second card is an ace?",
"role": "user"
}
] |
To find the probability that the first card is a diamond and the second card is an ace when two cards are dealt from a standard deck of 52 cards, we consider the following steps:
1. **Case 1: The first card is a diamond but not an ace.**
- There are 12 diamonds that are not aces.
- Probability of drawing a dia... | Okay, so I need to find the probability that when two cards are dealt from a standard deck, the first one is a diamond and the second one is an ace. Hmm, let's see. Let me remember how probability works when dealing with dependent events since the two cards are dealt without replacement, right?
First, a standard deck ... | 45 | 4,671 | 4,286 |
Let $f(x) : \mathbb{R} \to \mathbb{R}$ be a function such that
\[\frac{f(x) f(y) - f(xy)}{3} = x + y + 2\]for all $x,$ $y \in \mathbb{R}.$ Find $f(x).$ | Level 4 | Intermediate Algebra | We write the functional equation as
\[f(x)f(y) - f(xy) = 3x + 3y + 6.\]Setting $x = y = 0,$ we get
\[f(0)^2 - f(0) = 6.\]Then $f(0)^2 - f(0) - 6 = 0,$ which factors as $(f(0) - 3)(f(0) + 2) = 0.$ Hence, $f(0) = 3$ or $f(0) = -2.$
Setting $y = 0,$ we get
\[f(0) f(x) - f(0) = 3x + 6.\]Then
\[f(x) - 1 = \frac{3x + 6}{f(... | [
{
"content": "Let $f(x) : \\mathbb{R} \\to \\mathbb{R}$ be a function such that\n\\[\\frac{f(x) f(y) - f(xy)}{3} = x + y + 2\\]for all $x,$ $y \\in \\mathbb{R}.$ Find $f(x).$",
"role": "user"
}
] |
Given the function \( f(x) : \mathbb{R} \to \mathbb{R} \) such that
\[
\frac{f(x) f(y) - f(xy)}{3} = x + y + 2
\]
for all \( x, y \in \mathbb{R} \), we need to find \( f(x) \).
First, we multiply both sides by 3 to simplify the equation:
\[
f(x)f(y) - f(xy) = 3x + 3y + 6.
\]
Setting \( x = 0 \) in the equation, we ... | Okay, let me try to figure out this problem. So, we have a function f from the real numbers to the real numbers, and it satisfies this equation: (f(x)f(y) - f(xy))/3 = x + y + 2 for all real numbers x and y. We need to find f(x). Hmm.
First, maybe I can rewrite the equation to make it a bit clearer. Let me multiply bo... | 71 | 3,072 | 2,486 |
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