- Which ones digits indicate divisibility by \(5\)? - What ones digit does every multiple of \(10\) have? - Can a number be divisible by \(10\) without also being divisible by \(5\)?

1. A number is divisible by \(5\) when its ones digit is \(0\) or \(5\). This rule identifies \(105,230,600,875,\) and \(990\). 2. A number is divisible by \(10\) when its ones digit is \(0\). This rule identifies \(230,600,\) and \(990\).

a) \(105,230,600,875,990\) b) \(230,600,990\)

- Check the column labeled Ones for each row. - Think about what remains after making as many groups of ten as possible. - Which row shows no leftover ones?

1. The chart shows \(47\) on top and \(50\) on the bottom. 2. \(50\) has \(0\) ones, so it can be divided into groups of \(10\) exactly. 3. The bottom number is divisible by \(10\); \(47\) is not.

The bottom number, \(50\), because its ones digit is \(0\).

- Look at just the last digit of each number. - Which number could be split into two equal groups with nothing left over? - Use the pattern of even numbers to check.

1. Numbers that end in \(0, 2, 4, 6\), or \(8\) are divisible by \(2\). 2. \(46\) ends in \(6\), but \(47\) ends in \(7\).

\(46\) is divisible by \(2\) because it ends in the even digit \(6\). \(47\) is not because it ends in \(7\).

- Look at the last digit of the number. - Which endings appear when you count by fives? - Check the last digit against that pattern.

1. A whole number that ends in \(0\) or \(5\) is divisible by \(5\). 2. \(75\) ends in \(5\), so it is divisible by \(5\).

Yes, because the ones digit is \(5\).

- What does the test let you check instead of dividing the whole number? - Compare the result of your smaller calculation with the threes facts.

1. Add the digits: \(2 + 4 = 6\). 2. Since \(6\) is divisible by \(3\), \(24\) is divisible by \(3\).

Yes. The digit sum is \(6\), a multiple of \(3\).

- Look only at the last digit of each number. - Recall which last digits indicate divisibility by \(2\). - Recall which last digits indicate divisibility by \(5\) and by \(10\).

1. A number is divisible by \(2\) when its last digit is \(0, 2, 4, 6\), or \(8\). It is divisible by \(5\) when its last digit is \(0\) or \(5\). It is divisible by \(10\) when its last digit is \(0\). 2. The numbers divisible by \(2\) are \(240, 252, 260\), and \(278\). 3. The numbers divisible by \(5\) are \(240, 245, 260\), and \(265\). 4. The numbers divisible by \(10\) are \(240\) and \(260\).

Divisible by \(2\): \(240, 252, 260, 278\) Divisible by \(5\): \(240, 245, 260, 265\) Divisible by \(10\): \(240, 260\)

- Which divisibility tests use the sum of a number’s digits? - What must be true of the digit sum for divisibility by \(3\) or \(9\)? - Calculate and test the digit sum of each number separately.

1. Find each digit sum: \(1+1+1=3\), \(2+3+4=9\), \(4+0+5=9\), \(5+6+7=18\), \(8+1+0=9\), and \(9+9+3=21\). 2. A number is divisible by \(3\) when its digit sum is divisible by \(3\). Every digit sum in the list is divisible by \(3\), so all six numbers are divisible by \(3\). 3. A number is divisible by \(9\) when its digit sum is divisible by \(9\). The digit sums \(9\) and \(18\) identify \(234,405,567,\) and \(810\).

a) \(111,234,405,567,810,993\) b) \(234,405,567,810\)

- Which ones digits indicate divisibility by \(2\)? - Which ones digits indicate divisibility by \(5\)? - Which ones digit satisfies both rules? - Check the ones digit of every number in the list.

1. A number is divisible by \(2\) when its ones digit is \(0,2,4,6,\) or \(8\). This identifies \(150,432,60,\) and \(80\). 2. A number is divisible by \(5\) when its ones digit is \(0\) or \(5\). This identifies \(150,275,60,105,\) and \(80\). 3. A number satisfying both rules must end in \(0\). Therefore, the numbers are \(150,60,\) and \(80\).

The numbers are \(150,60,\) and \(80\). Each ends in \(0\), so each is divisible by both \(2\) and \(5\).

- How do you calculate the digit sum of a number? - What must be true of the digit sum for the original number to be divisible by \(3\)? - Find the digit sum of every number in the list. - Test whether each digit sum is a multiple of \(3\).

1. Compute each digit sum: - For \(214\), \(2+1+4=7\), which is not divisible by \(3\). - For \(333\), \(3+3+3=9\), which is divisible by \(3\). - For \(1002\), \(1+0+0+2=3\), which is divisible by \(3\). - For \(56\), \(5+6=11\), which is not divisible by \(3\). - For \(918\), \(9+1+8=18\), which is divisible by \(3\). 2. Therefore, the numbers divisible by \(3\) are \(333,1002,\) and \(918\).

The numbers divisible by \(3\) are \(333\), with digit sum \(9\); \(1002\), with digit sum \(3\); and \(918\), with digit sum \(18\).

- Add the digits of each number. - Compare each digit sum with the divisibility rules for \(3\) and \(9\). - A number divisible by \(9\) is also divisible by \(3\); use that relationship as a check.

1. The digit sums are \(3,9,13,9,\) and \(6\), respectively. 2. A number is divisible by \(3\) when its digit sum is divisible by \(3\). This identifies \(102,333,1260,\) and \(5001\). 3. A number is divisible by \(9\) when its digit sum is divisible by \(9\). This identifies \(333\) and \(1260\).

<table><tr><th>Number</th><th>Divisible by \(3\)</th><th>Divisible by \(9\)</th></tr><tr><td>\(102\)</td><td>Yes</td><td>No</td></tr><tr><td>\(333\)</td><td>Yes</td><td>Yes</td></tr><tr><td>\(409\)</td><td>No</td><td>No</td></tr><tr><td>\(1260\)</td><td>Yes</td><td>Yes</td></tr><tr><td>\(5001\)</td><td>Yes</td><td>No</td></tr></table>

- Use a divisibility rule when one applies directly. - For another case, nearby known multiples may be quicker than division. - Read carefully whether each statement describes a factor or a multiple.

1. The digit sum of \(2469\) is \(21\), which is divisible by \(3\). Statement a) is true. 2. Since \(15\times8=120\) and \(16\times8=128\), \(124\) is not a multiple of \(8\). Statement b) is false. 3. The digit sum of \(8118\) is \(18\), which is divisible by \(9\). Statement c) is true. 4. The last two digits of \(1010\) form \(10\), which is not divisible by \(4\). Statement d) is false. 5. Since \(12\times12=144\), \(12\) is a factor of \(144\). Statement e) is true.

a) True b) False c) True d) False e) True

- Which ones digits make a number even? - Which ones digits indicate divisibility by \(5\)? - What ones digit does every multiple of \(10\) have? - Is examining the ones digit sufficient for all three tests?

1. A number is divisible by \(2\) when its ones digit is \(0,2,4,6,\) or \(8\). This identifies \(156,480,\) and \(912\). 2. A number is divisible by \(5\) when its ones digit is \(0\) or \(5\). This identifies \(275,480,\) and \(1005\). 3. A number is divisible by \(10\) when its ones digit is \(0\). This identifies only \(480\).

Divisible by \(2\): \(156,480,912\) Divisible by \(5\): \(275,480,1005\) Divisible by \(10\): \(480\)

- Which of the two divisibility conditions is easier to test first? - What must be true of the ones digit for divisibility by \(2\)? - If one required condition can never be met, can the whole problem have a solution?

1. A number divisible by \(2\) must have an even ones digit. 2. The number \(5\square7\) ends in \(7\), so it is odd for every possible tens digit and can never be divisible by \(2\). 3. Because one required condition is impossible, no digit can make the number divisible by both \(2\) and \(3\).

No. The number always ends in the odd digit \(7\), so it can never be divisible by \(2\).

- Use the digit-sum rule for divisibility by \(3\). - List several multiples of \(6\) and examine their ones digits.

1. Statement a is false. The digit sum of \(11{,}111\) is \(1+1+1+1+1=5\), which is not divisible by \(3\). 2. Statement b is false. For example, \(12\div6=2\), so \(12\) is divisible by \(6\) even though its ones digit is \(2\).

a) False b) False

- Think about what happens to place values when dividing by \(100\). - Compare the last two digits of the numbers. - Look for a feature shared by every number that divides evenly by \(100\).

1. A whole number is divisible by \(100\) when its last two digits are both \(0\). 2. The numbers that meet this condition are \(4800\), \(12{,}000\), and \(30{,}500\). 3. The divisions are \(4800 \div 100 = 48\), \(12{,}000 \div 100 = 120\), and \(30{,}500 \div 100 = 305\).

a) \(4800 \div 100 = 48\) \(12{,}000 \div 100 = 120\) \(30{,}500 \div 100 = 305\) b) A whole number is divisible by \(100\) when it ends in at least two zeros.

- Focus on the number of chips in the ones column. - What happens when a column has enough chips to make one unit of the next place? - Think about the ones digit of a number that divides evenly by \(10\).

1. The columns show \(152{,}007\), including \(7\) ones. 2. To have \(0\) ones after regrouping, add \(3\) chips to make \(10\) ones. 3. Regroup the \(10\) ones as \(1\) ten and \(0\) ones. The new number is \(152{,}010\). 4. It ends in \(0\), so it is divisible by \(10\).

a) \(3\) chips. b) \(152{,}010\). Regrouping leaves \(0\) ones, so the number is divisible by \(10\).

- Think of a hundred or a ten as a group that can be shared in two equal parts. - Look at what is left to decide in the Ones column. - Which row's ones can be paired without a leftover?

1. The chart shows \(236\) and \(237\). 2. Whole hundreds and tens can be split into two equal groups because \(100\) and \(10\) are even. 3. The top row has \(6\) ones, which can also be paired; the bottom has \(7\) ones, with one left over. Only \(236\) is divisible by \(2\).

The top row, \(236\), is divisible by \(2\). Hundreds and tens can already be split into pairs; the \(6\) ones can also be paired, but the \(7\) ones in \(237\) leave one over.

- Break the number into hundreds and the last two digits. - Can every full hundred be split into four equal parts? - Check only the leftover part after the hundreds.

1. Split the number into \(700 + 16\). Since \(100\) is divisible by \(4\), \(700\) is too. 2. The last two digits make \(16\), and \(16 \div 4 = 4\). 3. Both parts divide exactly by \(4\), so \(716\) does as well.

Yes. \(716 = 700+16\), and both parts are divisible by \(4\).

- A number must pass both tests, not just one. - Could a number ending in \(5\) also split evenly into two equal groups?

1. Numbers divisible by \(2\) end in an even digit; numbers divisible by \(5\) end in \(0\) or \(5\). 2. To satisfy both, the last digit must be \(0\). 3. The numbers are \(40\) and \(50\).

\(40\) and \(50\). Both end in \(0\), the only last digit that is both even and either \(0\) or \(5\).

- Add the digits that are already given. - The complete digit sum must be a multiple of nine. - Which one-digit number makes that happen?

1. The known digits add to \(4 + 8 = 12\). 2. To reach a multiple of \(9\) with one more digit, the digit sum must be \(18\). 3. The missing digit is \(18 - 12 = 6\); indeed \(4+6+8=18\).

\(6\), because \(4 + 6 + 8 = 18\), and \(18\) is divisible by \(9\).

- Which final digits indicate divisibility by \(2\) or \(5\)? - How can the digit sum test divisibility by \(3\)? - For divisibility by \(4\), examine the last two digits.

1. A number divisible by \(2\) must be even, so the remaining choices are \(414\), \(424\), and \(450\). 2. For divisibility by \(3\), add the digits. The digit sums of \(414\) and \(450\) are both \(9\), so both are divisible by \(3\). 3. A number divisible by \(4\) has last two digits divisible by \(4\). Since \(14\) is not divisible by \(4\), \(414\) passes this condition. 4. A number divisible by \(5\) ends in \(0\) or \(5\). The number \(414\) ends in \(4\), so it is not divisible by \(5\). 5. Therefore, \(414\) satisfies all four conditions.

b) \(414\)

- A number can reduce a fraction only when it divides both the numerator and denominator. - Use divisibility rules for \(2\), \(3\), \(4\), \(6\), and \(9\). - Check each proposed number separately.

1. Both \(126\) and \(162\) are even, so \(2\) works. 2. Each number has digit sum \(9\), so both are divisible by \(3\) and \(9\). 3. Neither number is divisible by \(4\). 4. Both are divisible by \(2\) and \(3\), so both are divisible by \(6\). 5. Therefore, \(2, 3, 6,\) and \(9\) work.

\(2, 3, 6,\) and \(9\)

- For divisibility by \(4\), examine only the last two digits. - For divisibility by \(5\), examine only the ones digit. - In part c, identify the numbers that appear in both earlier lists.

1. For divisibility by \(4\), test the last two digits. The endings \(40,60,00,\) and \(24\) are divisible by \(4\), so the numbers are \(140,360,500,\) and \(624\). 2. For divisibility by \(5\), the ones digit must be \(0\) or \(5\). The numbers are \(140,255,360,500,\) and \(750\). 3. The numbers appearing in both lists are \(140,360,\) and \(500\).

a) \(140,360,500,624\) b) \(140,255,360,500,750\) c) \(140,360,500\)

- Focus on the part of each number that the divisibility-by-\(4\) rule uses. - List possible endings that match the fixed digit pattern. - In a part with no solution, explain why the fixed digits prevent every choice.

1. A number is divisible by \(4\) when the number formed by its last two digits is divisible by \(4\). 2. In part a), the possible endings are \(20,24,28\), so \(\ast\) can be \(0,4,8\). 3. In part b), the possible endings are \(16,36,56,76,96\), so \(\ast\) can be \(1,3,5,7,9\). 4. In part c), the possible endings are \(12,32,52,72,92\), so \(\ast\) can be \(1,3,5,7,9\). 5. In part d), the ending is always \(18\), which is not divisible by \(4\), so no digit works.

a) \(0,4,8\) b) \(1,3,5,7,9\) c) \(1,3,5,7,9\) d) No digit works because \(18\) is not divisible by \(4\).

- What must be true about every number divisible by \(4\)? - Can a number ending in \(7\) be a multiple of \(4\)? - Which even final digit could create a last-two-digit multiple of \(4\)?

1. Every number divisible by \(4\) is even. The number \(3\square7\) ends in \(7\), so it is odd regardless of the tens digit. Therefore, no digit can make it divisible by \(4\). 2. Replace \(7\) with an even digit, such as \(6\). 3. For \(3\square6\), choose \(\square=1\). The last two digits are \(16\), which is divisible by \(4\), so \(316\) is divisible by \(4\).

No digit works in \(3\square7\) because the number is odd. One possible change is \(3\square6\). Setting \(\square=1\) gives \(316\), which is divisible by \(4\).

- Which divisibility test uses the digit sum? - How does each possible box digit change the digit sum? - Could more than one digit work?

1. A number is divisible by \(3\) when its digit sum is divisible by \(3\). 2. The known digits have sum \(8+2=10\). 3. The sums \(10+2=12\), \(10+5=15\), and \(10+8=18\) are divisible by \(3\). 4. Therefore, the possible digits are \(2,5,\) and \(8\).

\(2,5,\) and \(8\)

- Recall the divisibility rule for \(9\). - First add the digits that are already known. - How much is needed to reach the next multiple of \(9\)?

1. A number is divisible by \(9\) when its digit sum is divisible by \(9\). 2. The known digits have sum \(3+7+5=15\). 3. The next multiple of \(9\) after \(15\) is \(18\), so the missing digit is \(18-15=3\). 4. The next possible digit sum, \(27\), would require a missing value of \(12\), which is not a digit. Thus \(3\) is the only possibility.

The missing digit is \(3\).

- Which possible ones digits make a number divisible by \(2\)? - Recall the divisibility rule for \(3\). - Test only the digits that satisfy the first condition.

1. For divisibility by \(2\), the missing ones digit must be even: \(0,2,4,6,\) or \(8\). 2. For divisibility by \(3\), the digit sum must be divisible by \(3\). The known digits have sum \(7+4=11\). 3. Testing the even digits gives sums \(11,13,15,17,\) and \(19\). Only \(15\) is divisible by \(3\), which occurs when the missing digit is \(4\). 4. Therefore, the only possible digit is \(4\).

\(4\)

- Does the leading digit affect divisibility by \(2\) when the ones digit is fixed? - Add the digits that are already known. - Which leading digits make the total a multiple of \(3\)? - Exclude \(0\) because the box is the leading digit.

1. The number ends in \(2\), so it is divisible by \(2\) for every possible leading digit. 2. For divisibility by \(3\), the digit sum must be divisible by \(3\). The known digits have sum \(1+5+2=8\). 3. Among the possible leading digits \(1\) through \(9\), the sums \(8+1=9\), \(8+4=12\), and \(8+7=15\) are divisible by \(3\). 4. Therefore, the possible digits are \(1,4,\) and \(7\).

\(1,4,\) and \(7\)

- First determine which tens digits make the final two digits divisible by \(4\). - Then test which of those digits make the full digit sum divisible by \(3\). - Which digits satisfy both conditions?

1. For divisibility by \(4\), the last two digits \(\square2\) must form a multiple of \(4\). This occurs for \(12,32,52,72,\) and \(92\), so the missing digit could be \(1,3,5,7,\) or \(9\). 2. For divisibility by \(3\), the digit sum \(2+5+\square+2=9+\square\) must be divisible by \(3\). This occurs when the missing digit is \(0,3,6,\) or \(9\). 3. The digits in both sets are \(3\) and \(9\).

\(3\) and \(9\)

- What must be true of the digit sum for divisibility by \(9\)? - What must be true of the final two digits for divisibility by \(4\)? - Does the digit that passes the first test also pass the second?

1. For divisibility by \(9\), the digit sum \(1+\square+2=3+\square\) must be divisible by \(9\). The only possible digit is \(6\), which gives a digit sum of \(9\). 2. With \(\square=6\), the number is \(162\). Its last two digits form \(62\), which is not divisible by \(4\). 3. Therefore, no digit satisfies both conditions.

No digit works. The only digit that makes the number divisible by \(9\) is \(6\), but \(162\) is not divisible by \(4\).

- What does the final digit tell you about divisibility by \(5\)? - How are the divisibility rules for \(3\) and \(9\) related? - Add the known digits and determine what is needed to reach a multiple of \(9\).

1. The number ends in \(0\), so it is divisible by \(5\) for every possible box digit. 2. For divisibility by \(9\), the digit sum \(4+\square+3+0=7+\square\) must be divisible by \(9\). This occurs only when \(\square=2\), giving a digit sum of \(9\). 3. Every number divisible by \(9\) is also divisible by \(3\). Therefore, \(2\) satisfies all three conditions.

\(2\)

- How can the digit sum test divisibility by \(3\)? - Which two digits determine divisibility by \(4\)? - Make one list for each rule and find the numbers common to both.

1. Test divisibility by \(3\) using digit sums: the sums are \(6,15,7,9,15,\) and \(9\), respectively. Every number except \(412\) is divisible by \(3\). 2. Test divisibility by \(4\) using the final two digits: \(32\), \(12\), \(44\), and \(08\) are divisible by \(4\), so \(132,412,744,\) and \(1008\) pass this test. 3. The numbers in both lists are \(132,744,\) and \(1008\).

\(132,744,\) and \(1008\)

- Examine the ones digit for divisibility by \(2\), \(5\), and \(10\). - For divisibility by \(25\), examine the last two digits. - Can a number be divisible by \(10\) without also being divisible by \(5\)?

1. A number is divisible by \(2\) when its ones digit is even. This identifies \(250,1020,\) and \(884\). 2. A number is divisible by \(5\) when its ones digit is \(0\) or \(5\). This identifies \(250,475,1020,\) and \(915\). 3. A number is divisible by \(10\) when its ones digit is \(0\). This identifies \(250\) and \(1020\). 4. A number is divisible by \(25\) when its last two digits are \(00,25,50,\) or \(75\). This identifies \(250\) and \(475\). 5. Organize the results by number.

<table> <tr><th>Number</th><th>Divisible by</th></tr> <tr><td>\(250\)</td><td>\(2,5,10,25\)</td></tr> <tr><td>\(475\)</td><td>\(5,25\)</td></tr> <tr><td>\(1020\)</td><td>\(2,5,10\)</td></tr> <tr><td>\(884\)</td><td>\(2\)</td></tr> <tr><td>\(915\)</td><td>\(5\)</td></tr> </table>

- Can you find a number ending in \(3\) that is not a multiple of \(3\)? - How are multiples of \(9\) related to multiples of \(3\)? - Which two divisibility conditions together guarantee divisibility by \(6\)?

1. Statement a is false. Divisibility by \(3\) depends on the digit sum, not only the ones digit. For example, \(13\) ends in \(3\), but its digit sum is \(1+3=4\), so it is not divisible by \(3\). 2. Statement b is true. Since \(9=3\times3\), every multiple of \(9\) is also a multiple of \(3\). 3. Statement c is true. A number is divisible by \(6\) exactly when it is divisible by both \(2\) and \(3\); being even satisfies the first condition, and having a digit sum divisible by \(3\) satisfies the second.

a) False. One counterexample is \(13\). b) True. c) True.

- Add the digits of each number for part a. - Examine only the tens and ones digits for part b. - Compare your two lists for part c.

1. The digit sums are \(6,9,10,9,\) and \(9\), respectively. Therefore, \(225,450,\) and \(504\) are divisible by \(9\). 2. The final two digits are \(32,25,16,50,\) and \(04\). The endings \(32,16,\) and \(04\) are divisible by \(4\), so \(132,316,\) and \(504\) are divisible by \(4\). 3. The only number in both lists is \(504\).

a) \(225,450,504\) b) \(132,316,504\) c) \(504\)

- Use the digit-sum rule for divisibility by \(3\). - Which ones digits make a number divisible by \(5\)? - When all four digits are equal, test the possible repeated digits in increasing order.

1. In part a, the number has the form \(aaaa\), where \(a\) is a nonzero digit. Its digit sum is \(4a\). Testing digits in increasing order, \(4\times1=4\), \(4\times2=8\), and \(4\times3=12\). The first digit sum divisible by \(3\) occurs at \(a=3\), so the number is \(3333\). 2. In part b, the number has the form \(a50b\). To maximize the number, choose \(a=9\). For divisibility by \(5\), the ones digit must be \(0\) or \(5\); choose \(b=5\) to maximize the number. The result is \(9505\).

a) \(3333\) b) \(9505\)

- List the whole numbers that round to \(640\) to the nearest ten. - Recall the divisibility rule for \(5\). - Test the numbers in your rounding range against that rule.

1. The whole numbers from \(635\) through \(644\) round to \(640\) to the nearest ten. 2. A whole number is divisible by \(5\) when its ones digit is \(0\) or \(5\). 3. Among the numbers from \(635\) through \(644\), only \(635\) and \(640\) meet that rule.

\(635\) and \(640\)

- List several multiples of \(25\) and examine their last two digits. - What pattern do the possible endings follow? - Which endings match the form \(\triangle0\)?

1. A number is divisible by \(25\) when its last two digits are \(00,25,50,\) or \(75\). 2. The last two digits of \(8\triangle0\) have the form \(\triangle0\). 3. Of the possible endings, only \(00\) and \(50\) match this form. 4. Therefore, \(\triangle=0\) or \(\triangle=5\).

\(0\) and \(5\)

- Which two divisibility conditions together guarantee divisibility by \(6\)? - What does the final digit \(2\) tell you? - Write the digit sum, remembering that the same missing digit appears twice. - Which digits make that sum a multiple of \(3\)?

1. A number is divisible by \(6\) when it is divisible by both \(2\) and \(3\). 2. The number \(1\ast\ast2\) ends in \(2\), so it is divisible by \(2\) for every possible digit. 3. Its digit sum is \(1+\ast+\ast+2=3+2\times\ast\). 4. For \(\ast=0,3,6,\) and \(9\), the digit sums are \(3,9,15,\) and \(21\), respectively, all divisible by \(3\). No other digit gives a digit sum divisible by \(3\). 5. Therefore, the possible digits are \(0,3,6,\) and \(9\).

\(0,3,6,\) and \(9\)

- Split the number into hundreds, tens, and ones. - Use the two given facts to separate parts that are divisible by three. - Compare the small leftover parts with the original digits.

1. Split the number by place value: \(132 = 100 + 30 + 2\). 2. Rewrite it as \(99 + 1 + 27 + 3 + 2\), or \(99 + 27 + 6\). 3. Both \(99\) and \(27\) are divisible by \(3\), and the leftover digit sum \(1+3+2=6\) is divisible by \(3\). So \(132\) is divisible by \(3\).

\(132 = 99 + 27 + 6\). The first two parts are multiples of \(3\), and the digit sum \(1 + 3 + 2 = 6\) is too.

- Which numbers from \(30\) through \(39\) are multiples of \(4\)? - How does the missing digit affect the digit sum? - Test the digits from \(0\) through \(9\) for each of the first two conditions. - Which digit appears in both lists?

1. For divisibility by \(4\), the last two digits \(3\square\) must form a multiple of \(4\). The possibilities from \(30\) through \(39\) are \(32\) and \(36\), so the missing digit can be \(2\) or \(6\). 2. For divisibility by \(3\), the digit sum \(7+3+\square=10+\square\) must be divisible by \(3\). This occurs for digit sums \(12,15,\) and \(18\), so the missing digit can be \(2,5,\) or \(8\). 3. The only digit in both sets is \(2\), so it is the only digit that makes the number divisible by both \(4\) and \(3\).

a) \(2\) or \(6\) b) \(2\), \(5\), or \(8\) c) \(2\)

- Which last-two-digit combinations make a number divisible by \(4\)? - How can the digit sum test divisibility by \(3\)? - Compare your two lists of possible digits.

1. For divisibility by \(4\), the last two digits \(2\square\) must form a multiple of \(4\). The possibilities are \(20,24,\) and \(28\), so the missing digit can be \(0,4,\) or \(8\). 2. For divisibility by \(3\), the digit sum \(2+0+2+\square=4+\square\) must be divisible by \(3\). This occurs when the missing digit is \(2,5,\) or \(8\). 3. The only digit in both sets is \(8\).

a) \(0\), \(4\), or \(8\) b) \(2\), \(5\), or \(8\) c) Yes, \(8\).

- Find a number that satisfies the stated condition but not the conclusion. - Check small two-digit numbers. - Recall the actual divisibility rules for \(5\) and \(6\). - Only one counterexample is needed for each statement.

1. For part a, choose a number ending in \(6\) that is not divisible by \(6\). The number \(16\) works because \(16\div6=2\) remainder \(4\). 2. For part b, choose a number whose digit sum is divisible by \(5\) but whose ones digit is not \(0\) or \(5\). The number \(14\) has digit sum \(1+4=5\), but \(14\) is not divisible by \(5\).

a) One counterexample is \(16\). b) One counterexample is \(14\).

- What kind of numbers are not divisible by \(2\)? - Look for a multiple of \(4\) whose ones digit is not \(4\) or \(8\). - Add two simple odd numbers. - Is one counterexample enough to disprove a universal claim?

1. For part a, \(12\) is divisible by \(4\) because \(12\div4=3\), but its ones digit is \(2\). This disproves the statement. 2. For part b, choose two odd numbers. For example, \(3+5=8\). Both addends are not divisible by \(2\), but their sum is divisible by \(2\), so the statement is false.

a) One counterexample is \(12\). b) One counterexample is \(3+5=8\).

- Is every multiple of \(3\) also a multiple of \(9\)? - Can a tens digit alone determine divisibility by \(3\)? - Look for a familiar multiple of \(3\) with a different tens digit. - Write the first several multiples of \(3\) and test the claims.

1. For part a, \(6\) is divisible by \(3\), but it is not divisible by \(9\). This counterexample disproves the statement. 2. For part b, \(12\) is divisible by \(3\) because \(12\div3=4\), but its tens digit is \(1\). This counterexample disproves the statement.

a) One counterexample is \(6\). b) One counterexample is \(12\).

- Can two factors each contribute part of the factorization needed for divisibility by \(4\)? - Begin with small whole-number factors.

Choose two factors that are not divisible by \(4\) but whose product is divisible by \(4\). The numbers \(2\) and \(2\) work because \(2\times2=4\). Neither factor is divisible by \(4\), but the product is.

One counterexample is \(2\times2=4\). The product is divisible by \(4\), but neither factor is divisible by \(4\).

- How do you compute a digit sum when one digit is unknown? - Which two-digit numbers ending in \(2\) are divisible by \(4\)? - Does one digit satisfy more than one condition?

1. For divisibility by \(3\), the digit sum \(4+7+\square+2=13+\square\) must be divisible by \(3\). This occurs for \(\square=2,5,\) or \(8\). 2. For divisibility by \(4\), the last two digits \(\square2\) must form a multiple of \(4\). The possibilities are \(12,32,52,72,\) and \(92\), so \(\square=1,3,5,7,\) or \(9\). 3. For divisibility by \(9\), \(13+\square\) must be a multiple of \(9\). Only \(\square=5\) gives \(18\). 4. For divisibility by both \(3\) and \(4\), the digit must appear in both sets from steps 1 and 2. The only common digit is \(5\).

a) \(2\), \(5\), or \(8\) b) \(1\), \(3\), \(5\), \(7\), or \(9\) c) \(5\) d) \(5\)

- Which ones digits make a number divisible by \(2\)? - How can you use the digit sum when one digit is missing? - Which two divisibility conditions together guarantee divisibility by \(6\)?

1. For divisibility by \(2\), the ones digit must be even. Thus the possible digits are \(0,2,4,6,\) and \(8\). 2. For divisibility by \(3\), the digit sum \(2+5+\square=7+\square\) must be divisible by \(3\). This occurs for \(\square=2,5,\) or \(8\). 3. A number is divisible by \(6\) when it is divisible by both \(2\) and \(3\). The common digits from the first two sets are \(2\) and \(8\), giving \(252\) and \(258\).

a) \(0,2,4,6,8\) b) \(2,5,8\) c) \(2\) and \(8\), because the number must be divisible by both \(2\) and \(3\).

- Which endings make a number divisible by \(25\)? - What is the least possible leading digit of the five-digit number? - How can the remaining digits be chosen to keep the number as small as possible?

1. A number divisible by \(25\) ends in \(00,25,50,\) or \(75\). The greatest four-digit number with one of these endings is \(9975\). 2. The least five-digit number using only \(0\) and \(1\) must begin with \(1\), and all remaining digits should be \(0\). It is \(10{,}000\). 3. Add: \(9975+10{,}000=19{,}975\).

\(19{,}975\)

- First add the digits from \(0\) through \(9\). - Compare that total with the digit sums of your results. - Think about how one regrouping changes a digit sum.

1. Three valid examples are \(10{,}234 + 98{,}765 = 108{,}999\), with digit sum \(36\); \(50{,}246 + 13{,}789 = 64{,}035\), with digit sum \(18\); and \(24{,}680 + 13{,}579 = 38{,}259\), with digit sum \(27\). 2. Each result has a digit sum that is divisible by \(9\). 3. The sum of the digits \(0\) through \(9\) is \(45\), a multiple of \(9\). During addition, each regrouping replaces \(10\) in one place with \(1\) in the next place, decreasing the digit sum by \(9\). Therefore, regrouping can change the digit sum only by multiples of \(9\), so the result still has a digit sum divisible by \(9\).

a) One possible set is: \(10{,}234 + 98{,}765 = 108{,}999\), digit sum \(36\) \(50{,}246 + 13{,}789 = 64{,}035\), digit sum \(18\) \(24{,}680 + 13{,}579 = 38{,}259\), digit sum \(27\) b) Each digit sum is a multiple of \(9\). c) The digits used have a total of \(45\), and each regrouping changes the digit sum by \(9\). Therefore, the result's digit sum remains a multiple of \(9\).

- How often do even numbers occur among consecutive whole numbers? - How often do multiples of \(4\) occur? - Find a group of four consecutive whole numbers containing no multiple of \(5\). - What kind of factor would guarantee divisibility by \(5\)?

1. Among four consecutive whole numbers, two are even. Therefore, at least one factor is divisible by \(2\), so the whole product is divisible by \(2\). 2. Among any four consecutive whole numbers, exactly one is a multiple of \(4\). Because that number is a factor of the product, the product is divisible by \(4\). 3. The product is not always divisible by \(5\). For example, \(1\times2\times3\times4=24\), which is not divisible by \(5\). Four consecutive whole numbers do not always include a multiple of \(5\).

1) The product is always divisible by \(2\) because it includes even factors. 2) The product is always divisible by \(4\) because the four whole numbers include a multiple of \(4\). 3) No. A counterexample is \(1\times2\times3\times4=24\).