Example Question - counting
Here are examples of questions we've helped users solve.
Finding Natural Numbers with Four Positive Divisors
<p>To find natural numbers with exactly four positive divisors less than 65, we must identify the form of such numbers. A natural number can have exactly four positive divisors if it is in one of the following forms:</p> <p>1. The square of a prime number: \( p^2 \)</p> <p>2. The product of two distinct prime numbers: \( p \times q \)</p> <p>Now, we will find numbers less than 65 for each case.</p> <p>1. For \( p^2 \): The prime numbers less than 8 (since \( 8^2 = 64 \)) are 2, 3, 5, and 7.</p> <p>- \( 2^2 = 4 \)</p> <p>- \( 3^2 = 9 \)</p> <p>- \( 5^2 = 25 \)</p> <p>- \( 7^2 = 49 \)</p> <p>The valid numbers are 4, 9, 25, and 49.</p> <p>2. For \( p \times q \): We consider pairs of primes less than 65:</p> <p>- \( 2 \times 3 = 6 \)</p> <p>- \( 2 \times 5 = 10 \)</p> <p>- \( 2 \times 7 = 14 \)</p> <p>- \( 2 \times 11 = 22 \)</p> <p>- \( 2 \times 13 = 26 \)</p> <p>- \( 2 \times 17 = 34 \)</p> <p>- \( 2 \times 19 = 38 \)</p> <p>- \( 2 \times 23 = 46 \)</p> <p>- \( 2 \times 29 = 58 \)</p> <p>- \( 3 \times 5 = 15 \)</p> <p>- \( 3 \times 7 = 21 \)</p> <p>- \( 3 \times 11 = 33 \)</p> <p>- \( 3 \times 13 = 39 \)</p> <p>- \( 3 \times 17 = 51 \)</p> <p>- \( 3 \times 19 = 57 \)</p> <p>- \( 5 \times 7 = 35 \)</p> <p>- \( 5 \times 11 = 55 \)</p> <p>- \( 5 \times 13 = 65 \) (not valid)</p> <p>- \( 7 \times 11 = 77 \) (not valid)</p> <p>The valid products are 6, 10, 14, 15, 21, 22, 26, 34, 35, 38, 39, 46, 49, 51, 55, 57, and 58.</p> <p>The complete list of natural numbers less than 65 with exactly four positive divisors is: 4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 34, 35, 38, 39, 49, 51, 55, 57, and 58.</p> <p>Counting these, we have a total of 18 numbers.</p> <p>Thus, the answer is 18.</p>
Counting Palindromic Numbers with Specific Conditions
<p>To find the number of different palindromic keys that can be generated using four-digit numbers with the specified conditions (even, odd, and four digits), consider the following:</p> <p>A four-digit palindromic number has the form:</p> <p>\[ abba \]</p> <p>Where \(a\) is the first and last digit, and \(b\) is the second and third digit.</p> <p>1. \(a\) must be an odd digit (1, 3, 5, 7, 9) to satisfy the condition of being odd. Therefore, there are 5 choices for \(a\).</p> <p>2. \(b\) can be any digit from 0 to 9, giving us 10 choices for \(b\).</p> <p>Thus, the total number of different palindromic four-digit keys that can be generated is:</p> <p>\[ 5 \times 10 = 50 \]</p> <p>The answer is 50.</p>
Analysis of a Mathematical Expansion and Counting Problem
<p>The question you've asked appears to be two separate math problems:</p> <p>1. For the number of words each of 3 vowels and 2 consonants that can be formed from "UTE" - This seems like a permutations problem, but the question seems incomplete as it does not state how many vowels and consonants are available to choose from. Without further information, this part of the question cannot be solved.</p> <p>2. For finding the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) - This can be solved using the binomial theorem which states that \(n\)th term from the end is given by \(T_{(r+1)} = ^nC_r \cdot a^{(n-r)} \cdot b^r\) for the expansion of \((a + b)^n\), where \(n\) is the power and \(r\) is \(n-k\) if you're looking for the \(k\)th term from the end.</p> <p>We're looking for the 3rd term from the end (which is the same as the 5th term from the beginning since there are 7 terms in total), so \(r=7-3=4\). Using this, the term is:</p> <p>\[T_{(4+1)} = ^7C_4 \cdot \left(\frac{3}{x^2}\right)^{(7-4)} \cdot (x^3)^4\]</p> <p>\[T_5 = ^7C_4 \cdot \frac{3^3}{x^6} \cdot x^{12}\]</p> <p>\[T_5 = 35 \cdot \frac{27}{x^6} \cdot x^{12}\]</p> <p>\[T_5 = 35 \cdot 27 \cdot x^6\]</p> <p>\[T_5 = 945x^6\]</p> <p>So the 3rd term from the end in the expansion of \(\left(\frac{3}{x^2} - x^3\right)^7\) is \(945x^6\).</p>
Matrix Method for Counting Battery Production
<p>The question asks to find the number P of batteries in a certain box out of a total Q using a matrix method. This is a mathematical problem involving systems of equations that can be expressed in matrix form.</p> <p>Let the number of batteries on Monday be M and on Tuesday be T. We have:</p> <p>M = 960</p> <p>T = 1016</p> <p>Assuming there is a linear relationship between the days and the number of batteries, we can express this as a system of equations, with x representing the constant difference in daily production, and P being the production on an unknown day (using a 0-based index for days where Monday is day 0):</p> <p>M = P + 0*x</p> <p>T = P + 1*x</p> <p>We can then solve for P and x using matrix operations:</p> <p>\[ \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix} * \begin{bmatrix}P\\x\end{bmatrix} = \begin{bmatrix}960\\1016\end{bmatrix} \]</p> <p>To find the values of P and x, we can use matrix inversion or other methods:</p> <p>\[ \begin{bmatrix}P\\x\end{bmatrix} = \begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}^{-1} * \begin{bmatrix}960\\1016\end{bmatrix} \]</p> <p>By finding the inverse of the coefficient matrix and multiplying it by the production matrix, we can find P. However, the question lacks sufficient information to calculate an exact answer without additional context or assumptions about the pattern of battery production.</p>
Counting Divisible Numbers in a Sequence
\[ \text{令} S \text{为序列} 10^1, 10^2, 10^3, \ldots \text{。序列中的第} n \text{个数可以表示为} 10^{n} \text{。} \] \[ \text{如果} 10^n \text{能被} 10^2 \text{整除，那么} n \text{必须大于或等于} 2 \text{。} \] \[ \text{在} 2018 \text{个数中，第一个数} 10^1 \text{不能被} 10^2 \text{整除，其余} 2017 \text{个数可以。} \] \[ \text{因此，} 2018 \text{个数中有} 2017 \text{个数能被} 10^2 \text{整除。} \] \[ \text{答案是} (D)2017 \text{。} \]
Counting People in Rows
<p>首先我们观察图像中的排列，有两排人。</p> <p>第一排有5个女生，第二排有7个男生。</p> <p>题目问的是女生和男生的比例，所以我们用女生的数量除以男生的数量来得到答案：</p> <p>\[ \frac{女生人数}{男生人数} = \frac{5}{7} \]</p> <p>答案是女生和男生的比例是\(\frac{5}{7}\)。</p>
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