/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 It can be shown (see exercises \... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 33

It can be shown (see exercises \(41-45\) ) that an integer is divisible by 3 if, and only if, the sum of its digits is divisible by \(3 .\) An integer is divisible by 9 if, and only if, the sum of its digits is divisible by \(9 .\) An integer is divisible by 5 if, and only if, its right-most digit is a 5 or a \(0 .\) And an integer is divisible by 4 if, and only if, the number formed by its right-most two digits is divisible by 4. Check the following integers for divisibility by \(3,4,5\) and \(9 .\) a. \(637,425,403,705,125\) b. \(12,858,306,120,312\) c. \(517,924,440,926,512\) d. \(14,328,083,360,232\)

Short Answer

Expert verified
a. divisible by 3 and 5 b. divisible by 3 and 4 c. divisible by 4 d. divisible by 4

Step by step solution

01

Add the digits of the integer

Sum all the digits of the given integer.
02

Check divisibility by 3

If the sum from Step 1 is divisible by 3, the integer is divisible by 3. Otherwise, it's not.
03

Check divisibility by 9

If the sum from Step 1 is divisible by 9, the integer is divisible by 9. Otherwise, it's not.
04

Check divisibility by 4

Check the number formed by its right-most two digits. If this number is divisible by 4, the integer is divisible by 4. Otherwise, it's not.
05

Check divisibility by 5

Check the right-most digit of the integer. If it is either a 5 or a 0, the integer is divisible by 5. Otherwise, it's not. Now, let's apply these steps to each of the given integers: a. \(637,425,403,705,125\) Step 1: \(6+3+7+4+2+5+4+0+3+7+0+5+1+2+5=54\) Step 2: \(54\) is divisible by \(3\), so this integer is divisible by \(3\). Step 3: \(54\) is not divisible by \(9\), so this integer is not divisible by \(9\). Step 4: The last two digits are \(25\), which is not divisible by \(4\), so this integer is not divisible by \(4\). Step 5: The last digit is \(5\), so this integer is divisible by \(5\). b. \(12,858,306,120,312\) Step 1: \(1+2+8+5+8+3+0+6+1+2+0+3+1+2=42\) Step 2: \(42\) is divisible by \(3\), so this integer is divisible by \(3\). Step 3: \(42\) is not divisible by \(9\), so this integer is not divisible by \(9\). Step 4: The last two digits are \(12\), which is divisible by \(4\), so this integer is divisible by \(4\). Step 5: The last digit is \(2\), so this integer is not divisible by \(5\). c. \(517,924,440,926,512\) Step 1: \(5+1+7+9+2+4+4+4+0+9+2+6+5+1+2=61\) Step 2: \(61\) is not divisible by \(3\), so this integer is not divisible by \(3\). Step 3: \(61\) is not divisible by \(9\), so this integer is not divisible by \(9\). Step 4: The last two digits are \(12\), which is divisible by \(4\), so this integer is divisible by \(4\). Step 5: The last digit is \(2\), so this integer is not divisible by \(5\). d. \(14,328,083,360,232\) Step 1: \(1+4+3+2+8+0+8+3+3+6+0+2+3+2=43\) Step 2: \(43\) is not divisible by \(3\), so this integer is not divisible by \(3\). Step 3: \(43\) is not divisible by \(9\), so this integer is not divisible by \(9\). Step 4: The last two digits are \(32\), which is divisible by \(4\), so this integer is divisible by \(4\). Step 5: The last digit is \(2\), so this integer is not divisible by \(5\).

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Key Concepts

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

Divisibility by 3
Understanding divisibility by 3 can be very straightforward. If you wonder whether a large number, such as 637,425,403,705,125, can be divided evenly by 3, you don't need to perform the division right away. Instead, simply sum up all its digits. For this example, the sum is 54. Since 54 is a number you can divide by 3 without a remainder (54 梅 3 = 18), the original large number is also divisible by 3.

This rule works because of the way numbers are constructed in our decimal system. Every digit of a number contributes to the total value based on its position. The remarkable trait of the number 3 is that any multiple of 3, when its digits are summed, will result in another multiple of 3. So, if the sum of the digits is a number like 9, 12, or 18, the original number passes the test for divisibility by 3.
Divisibility by 4
To check a number's divisibility by 4 is a bit different from checking for 3. You only need to look at the last two digits of the number. If they form a number that is divisible by 4, the entire number is divisible by 4. Say you have the number 12,858,306,120,312; focus on 12 鈥� the last two digits. Since 12 can be divided by 4 (12 梅 4 = 3), then the whole number can be divided evenly by 4.

This rule is based on the concept of place value. The last two digits in a number represent the part of the number that is critical for divisibility by 4, given that 100 is divisible by 4 and any multiple of 100 will also be divisible by 4. As long as the digits in the tens and ones place form a number that is a multiple of 4, the entire number maintains this property.
Divisibility by 5
Another easy rule to remember is for divisibility by 5. It's even simpler than the one for 3 or 4. For a number to be divisible by 5, its last digit 鈥� the rightmost one 鈥� must be either 0 or 5. That's it! Consider the numbers 403,705,125 or 517,924,440,926,512. The first ends in 5, and the second ends in 2. Without considering anything else about these numbers, we can conclude that the first is divisible by 5 and the second isn't.

The reasoning behind this rule is due to the base-10 structure of our number system. Since 5 is a factor of the base (10 = 5 x 2), any number with a last digit of 0 or 5 can be divided by 5 evenly. This is because the number will end in a 5 or 0 regardless of the other digits.
Divisibility by 9
The divisibility rule for 9 is somewhat akin to that for 3, but with a stronger condition. To check if a number is divisible by 9, add up all of its digits. If the resulting sum is itself divisible by 9, then so is the original number. Take the sum 54 from earlier; it is not divisible by 9 outright (since 54 梅 9 = 6 with no remainder), so the large number we started with, 637,425,403,705,125, isn't divisible by 9 either.

This rule holds because, mathematically, 9 is the largest single-digit number, and its influence on the property of divisibility within the decimal system is robust. A number that is divisible by 9 has all of its components 鈥� the sum of its digits 鈥� sharing that same property. This principle is often used to spot errors in calculations or entries, as numbers that are expected to be multiples of 9 must adhere to this pattern.

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Most popular questions from this chapter

a. Rewrite the following theorem in the form \(\forall \longrightarrow\) if then b. Fill in the blanks in the proof, Theorem: The sum of any even integer and any odd integer is odd. Proof: Suppose \(m\) is any even integer and \(n\) is (a) By definition of even, \(m=2 r\) for some \((b)\), and by definition of odd, \(n=2 s+1\) for some integer \(s\). By substitution and algebra, \(m+n=\underline{(c)}=2(r+s)+1 .\) Since \(r\) and \(s\) are both integers, so is their sum \(r+s\). Hence \(m+n\) has the form 2 . (some integer) \(+1\), and \(s o\) (d) by definition of odd.

a. Fermat's last theorem says that for all integers \(n>2\), the equation \(x^{n}+y^{n}=z^{n}\) has no positive integer solution (solution for which \(x, y\), and \(z\) are positive integers). Prove the following: If for all prime numbers \(p>2\), \(x^{p}+y^{p}=z^{p}\) has no positive integer solution, then for any integer \(n>2\) that is not a power of \(2, x^{n}+y^{n}=z^{n}\) has no positive integer solution. b. Fermat proved that there are no integers \(x, y\), and \(z\) such that \(x^{4}+y^{4}=z^{4}\). Use this result to remove the restriction in part (a) that \(n\) not be a power of 2 . That is, prove that if \(n\) is a power of 2 and \(n>4\), then \(x^{n}+y^{n}=z^{n}\) has no positive integer solution.

When expressions of the form \((x-r)(x-s)\) are multiplied out, a quadratic polynomial is obtained. For instance, \((x-2)(x-(-7))=(x-2)(x+7)=x^{2}+5 x-14 .\) \(H\) a. What can be said about the coefficients of the polynomial obtained by multiplying out \((x-r)(x-s)\) when both \(r\) and \(s\) are odd integers? when both \(r\) and \(s\) are even integers? when one of \(r\) and \(s\) is even and the other is odd? b. It follows from part (a) that \(x^{3}-1253 x+255\) cannot be written as a product of two polynomials with integer coefficients. Explain why this is so.

Assume that \(k\) is a particular integer. a. Is \(-17\) an odd integer? b. Is 0 an even integer? c. Is \(2 k-1\) odd?

Prove that there is at most one real number \(b\) with the property that \(b r=r\) for all real numbers \(r\). (Such a number is called a multiplicative identity.)
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