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Chapter 4: Problem 17

Find the smallest positive integer \(n\) for which the product \(1260 \times n\) is a perfect cube.

Short Answer

Expert verified
The smallest positive integer \(n\) for which \(1260 \times n\) is a perfect cube is \(2 \times 3 \times 5^2 \times 7^2 = 1470\).

Step by step solution

01

Find prime factors of 1260

The prime factorization of 1260 can be computed by dividing the number by prime numbers, starting from 2. The process yields the prime factors: 2, 3, 5 and 7. Their exponents are: \(2^2\), \(3^2\), \(5^1\) and \(7^1\).
02

Make all exponents multiples of 3

To make a number a perfect cube, we need all prime factor exponents to be multiples of 3. For 2 and 3 (since they currently have an exponent of 2), we need one more of each to make the exponent 3. Now, for 5 and 7 (since they have an exponent of 1), we need two more of each to make the exponent 3.
03

Determine smallest positive integer \(n\)

The required values to make \(1260 \times n\) a perfect cube are \(2^1\), \(3^1\), \(5^2\) and \(7^2\). The smallest positive integer \(n\) is the product of these values: \(n = 2 \times 3 \times 5^2 \times 7^2\).

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Most popular questions from this chapter

For any \(n \in \mathbf{Z}^{*}\), what are \(\operatorname{gcd}(n, n+1)\) and \(\operatorname{lcm}(n, n+1) ?\)

Let \(a, b, c, d, m, n, s, t \in \mathbf{Z}^{+}\)with \(a d-b c=1, s=\) \(a m+b n\), and \(t=c m+d n\). a) Solve for \(m, n\) in terms of \(s, t .\) b) Prove that \(\operatorname{gcd}(s, t)=\operatorname{gcd}(m, n)\).

For \(a, b, c, d \in \mathbf{Z}^{+}\), prove that if \(d=a+b c\), then \(\operatorname{gcd}(b, d)=\operatorname{gcd}(a, b)\).

If a machine stores integers by the two's complement method, what are the largest and smallest integers that it can store if it uses bit patterns of (a) 4 bits? (b) 8 bits? (c) 16 bits? (d) 32 bits? (e) \(2^{n}\) bits, \(n \in \mathbf{Z}^{+}\)?

Prove that \(\sqrt{p}\) is irrational for any prime \(p\).
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