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Chapter 4: Problem 15

Determine the smallest perfect square that is divisible by \(7 !\)

Short Answer

Expert verified
The smallest perfect square divisible by 7! is 2,646,720.

Step by step solution

01

Calculation of 7!

To calculate 7 factorial (7!), simply take the product of all positive integers from 7 down to 1, i.e.,\(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 5040\). This is because by definition, n! (n factorial) is the product of all positive integers from n down to 1.
02

Determine Perfect Squares

Analyze the prime factors of 5040. It factors to \(2^4 * 3^2 * 5 * 7\). In order for a number to be a perfect square, each factor should be to an even power. Our number is missing two 2's, one 3, one 5 and one 7 to become a perfect square. So, we need to multiply these missing factors to our 7-factorial (5040).
03

Find the smallest perfect square

Multiply the missing factors to 7-factorial (5040): \(5040 * 2^2 * 3 * 5 * 7 = 2,646,720\). This is the smallest perfect square divisible by 7!.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Factorial Calculation
Understanding factorial calculation is vital for solving numerous mathematical problems, including combinatorics, algebra, and calculus. In essence, the factorial of a positive integer, represented by the symbol '!', signifies the product of all the positive integers up to that number.

For example, the factorial of 7, written as '7!', is computed by multiplying every whole number from 7 down to 1. The calculation goes as follows:
Prime Factorization
To appreciate the relevance of prime factorization, it's beneficial to start with the basics: prime numbers. Prime numbers are integers greater than 1 that have no positive divisors other than 1 and themselves. When you break down a composite number into the product of prime numbers, that process is known as prime factorization.

It's a foundational concept in mathematics because it's the method you use to find the prime numbers that, when multiplied together, result in the original number. Understanding prime factorization enables students to simplify fractions, find the greatest common divisors, and solve problems involving perfect squares.
Perfect Squares
A perfect square is an integer that can be expressed as the square of another integer. For example, 16 is a perfect square because it is the result of squaring 4 (4 x 4). The property that characterizes a perfect square in terms of prime factorization is having even powers for all its prime factors.

When a question involves figuring out the smallest perfect square that a number is divisible by, what's generally required is adjusting the powers of the prime factors so that they are all even. That often involves multiplying the original number by missing factors to balance the prime factorization into even exponents.

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Most popular questions from this chapter

Write each of the following integers in two's complement representation. Here the results are eight-bit patterns. a) 15 b) \(-15\) c) 100 d) \(-65\) e) 127 f) \(-128\)

Consider the set \(\\{1,2,3\\}\). Here we may write \(\\{1,2,3\\}=\) \(\\{1,2\\} \cup\\{3\\}\), where \(1+2=3\). For the set \(\\{1,2,3,4\\}\) we find that \(\\{1,2,3,4\\}=\\{1,4\\} \cup\\{2,3\\}\), where \(1+4=2+3\). However, things change when we examine the set \(\\{1,2,3,4,5\\} .\) In this case, if \(C \subseteq\\{1,2,3,4,5\\}\) and we let \(s_{C}\) denote the sum of the elements in \(C\), then we find that there is no way to write \(\\{1,2,3,4,5\\}=A \cup B\), with \(A \cap B=\emptyset\) and \(s_{A}=s_{B}\) a) For which \(n \in \mathbf{Z}^{+}, n \geq 3\), can we write \(\\{1,2,3, \ldots\), \(n\\}=A \cup B\), with \(A \cap B=\emptyset\) and \(s_{A}=s_{B} ?\) (As above, \(s_{A}\) and \(s_{B}\) denote the sums of the elements in \(A\) and \(B\), respectively.) b) Let \(n \in \mathbf{Z}^{+}\)with \(n \geq 3\). If we can write \(\\{1,2,3, \ldots\), \(n\\}=A \cup B\) with \(A \cap B=\emptyset\) and \(s_{A}=s_{B}\), describe how such sets \(A\) and \(B\) can be determined.

Determine those integers \(n\) for which \(\frac{5 n-4}{6}\) and \(\frac{7 n+1}{4}\) are also integers.

When does a positive integer \(n\) have exactly a) two positive divisors? b) three positive divisors? c) four positive divisors? d) five positive divisors?

Let \(a, b \in \mathbf{Z}^{+}\). a) Prove that if \(a^{2} \mid b^{2}\) then \(a \mid b\). b) Is it true that if \(a^{2} \mid b^{3}\) then \(a \mid b\) ?
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