REMEDIAL MATH PROGRAM Level 1 2016 Copyright Vinay Agarwala, Checked: 6/01/16 This remedial math program applies to students in 5 th grade and above. This program also applies to adults who had trouble with mathematics during their school years. DIAGNOSTIC STEP 1 The first thing to check is the student s ability to divide. Give him the following problem to do. 7715743 38 STEP 2 If the student can do this division comfortably without hesitation then he may move to Remedial Math Level 2. Part of this diagnostic is that the student should be able to show that his answer his correct using a simple calculator. This includes showing the correctness of the remainder. STEP 3 If the student is unable to pass then he continues with Remedial Math Level 1 below. REMEDIAL MATH LEVEL ONE STEP 4 Have the student study Lesson 6, Sections 1 & 2 (see below), and solve all the exercise problems of these sections. Let the student use a Multiplication Table. Do not let him struggle with multiplication.
Lesson 6, Section 1: Basic Division 1. Suppose you have $6 in your pocket, you can give $6 away only one time. There is no money left in your pocket. We write this as follows. 6 6 = 1; because 6 6 = 0 No dollar remaining 2. Suppose there are $8 instead, you can still give $6 away only one time. But there are $2 left in your pocket. We write this as follows. 8 6 = 1 R2; because 8 6 = 2 Two remaining pennies 3. When there are $30 in your pocket, you can give $6 away five times, with no remaining dollars. We write this as follows. 30 6 = 5; because 30 6 6 6 6 6 = 0 Subtract five times No remainder 4. When there are $33 in your pocket, you can still give $6 away five times, but there are $3 left in your pocket. We write this as follows. 33 6 = 5 R3; because 33 6 6 6 6 6 = 3 Subtract five times Remainder 5. The total money in your pocket is the dividend. The money taken out each time is the divisor. 33 6 Dividend Divisor 6. We use the multiplication table to find the maximum number of times the divisor can be taken out of the dividend. This is called the quotient. What is then left is called the remainder. 33 6 = 5 R3 Quotient Remainder 7. The remainder is always less than the divisor; because if the remainder were more than the divisor, you could have taken out the divisor at least one more time. 8. When there is no remainder then we have exact division. 30 6 = 5 (and no remainder) Exact Division 9. When there is a remainder then we have inexact division. 33 6 = 5 R3 Inexact Division
Find the quotient with remainder, if any. Tell if the division is exact or inexact. (a) 9 3 (c) 16 4 (e) 25 5 (g) 12 4 (b) 9 2 (d) 16 5 (f) 23 6 (h) 12 10 Solution: (a) 3 exact (b) 4 R1 inexact (c) 4 exact (d) 3 R1 inexact (e) 5 exact (f) 3 R5 inexact (g) 3 exact (h) 1 R2 inexact Lesson 6, Section 2: Division Facts 10. Division is how many times you can take a divisor out of the dividend. If you want to give away all the dollars in your pocket all together, you may do so only once. Therefore, when the dividend and the divisor are the same the quotient is always 1. 6 6 = 1; 30 30 = 1 11. You can give away $1 only as many times as there are dollars in your pocket. Therefore, when a number is divided by 1, the quotient is the same as the number (dividend). 6 1 = 6; 30 1 = 30 12. When you have no dollars in your pocket, you cannot give away any specific amount. Therefore, when zero is divided by a number, the quotient is zero. 0 6 = 0; 0 30 = 0 13. When you have dollars in your pocket you may give zero dollars away any number of times because the dollars in your pocket never decrease. Therefore, when you divide a number by zero the quotient is unlimited or undefined. 6 0 = undefined; 30 0 = undefined Find the quotients for the following: (a) 9 9 (c) 9 0 (e) 0 23 (g) 24 1 (b) 8 1 (d) 8 8 (f) 28 0 (h) 0 4 Solution: (a) 1 (b) 8 (c) undefined (d) 1 (e) 0 (f) undefined (g) 24 (h) 0 Find the quotient using multiplication tables (a) 27 3 (c) 48 6 (e) 72 9 (g) 36 4 (b) 56 8 (d) 45 5 (f) 16 2 (h) 35 7 Solution: (a) 9 (b) 7 (c) 8 (d) 9 (e) 8 (f) 8 (g) 9 (h) 5
STEP 5 Have the student study Lesson 6, Sections 3, and solve all the exercise problems of these sections. On this section the student learns the short form of division. The student also learns how to check the correctness of both quotient and remainder on a simple calculator. Lesson 6, Section 3: Dividing Larger Dividends 14. We divide larger numbers from left to right by breaking them into their place values. 486 2 = (400 + 80 + 6) 2 = 200 + 40 + 3 = 243 15. We may also divide larger numbers by writing them as follows. Hundreds: From 4 we can take 2 out 2 times (no remainder) Tens: From 8 we can take 2 out 4 times (no remainder) Hundreds: From 6 we can take 2 out 3 times (no remainder) Therefore, 486 2 = 243 To check, multiply 243 by 2. You should get back 486. 16. When there is a remainder in a column, we place it to the left of the next digit as a ten for that digit. This is called the short form of division. Hundreds: From 5 we can take 2 out 2 times (remainder is 1) We place the remainder 1 in front of 7 making it 17. Tens: From 17 we can take 2 out 8 times (remainder is 1) We place the remainder 1 in front of 2 making it 12. Hundreds: From 12 we can take 2 out 6 times (no remainder) Therefore, 572 2 = 286 (Check 286 x 2 = 572) 17. When the first digit of the dividend cannot be divided, then start with the first two digits. Make sure you write the digit of the quotient in the correct column. Therefore, 564 6 = 94 (Check 94 x 6 = 564) 18. When a number in a column cannot be divided, put a 0 in the quotient for that column. Therefore, 728 7 = 104
(Check 104 x 7 = 728) 19. We may also write division with quotient below the dividend as shown on the right. This has some advantages as shown later in factoring numbers. In the following exercises use Multiplication Table as an aid, until you can use skip counting with the divisor. Divide the following and check your answers by multiplying back. (a) 396 3 (c) 444 3 (e) 435 5 (g) 450 6 (b) 327 3 (d) 612 6 (f) 315 7 (h) 224 4 Solution: (a) 132 (b) 109 (c) 148 (d) 102 (e) 87 (f) 45 (g) 75 (h) 56 Divide the following and check your answers by multiplying back. (a) 844 4 (c) 732 6 (e) 57 3 (g) 3,174 3 (b) 992 8 (d) 894 6 (f) 847 7 (h) 9,945 5 Solution: (a) 211 (b) 124 (c) 122 (d) 149 (e) 19 (f) 121 (g) 1,058 (h) 1,989 20. When there is a final remainder at the end, you write is next to the quotient. You check your answer by multiplying back and the adding the remainder. Therefore, 764 3 = 254 R2 [Check (254 x 3) + 2 = 764] Divide the following. Check your answer by multiplying back and the adding the remainder. (a) 15 4 (c) 139 6 (e) 578 3 (g) 3,177 2 (b) 92 8 (d) 793 7 (f) 897 5 (h) 9,745 9 Solution: (a) 3 R3 (b) 11 R4 (c) 23 R1 (d) 113 R2 (e) 192 R2 (f) 179 R2 (g) 1588 R1 (h) 1082 R7 STEP 6 Have the student divide a telephone number like 7715743 by single-digit divisors using short form. Teach him how to find and correct his error if the answer is wrong. 7715743 2 = 7715743 3 =
7715743 4 = 7715743 5 = 7715743 6 = 7715743 7 = 7715743 8 = 7715743 9 = STEP 7 Have the student study Lesson 3, Section 1 and 2, and solve all the exercise problems of these sections. Make sure that the student can skip count by single-digit numbers. Lesson 3, Section 1: Addition is Counting More 1. Counting gives us the numbers as follows. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 2. We can count our fingers up to ten because we have ten fingers. Ten has a special role in counting as you shall see later. 3. We can count intervals on a Number Line up to any number. We mark the start of first interval as zero. Then we mark the intervals as we count them. 4. To add two numbers, we start from the first number, and count as many more as the second number. For example, to add 5 + 3, we start from 5 and count 3 more to get the sum of 8. 5. It is perfectly okay to count on fingers. You may also count on a number line by drawing it on paper.
Add the following by counting more (a) 2 + 1 (c) 3 + 2 (e) 5 + 2 (g) 4 + 3 (i) 6 + 3 (b) 4 + 5 (d) 1 + 6 (f) 2 + 6 (h) 2 + 7 (j) 3 + 5 Answer: (a) 3 (b) 9 (c) 5 (d) 7 (e) 7 (f) 8 (g) 7 (h) 9 (i) 9 (j) 8 Lesson 3, Section 2: Skip Counting 6. Skip counting is counting by a number greater than one by adding it repeatedly. You may skip count by 2 on your fingers by thinking of each finger as 2. 7. When you skip count by 5, you are adding 5 repeatedly. The skip counts of a number are called multiples of that number. 8. Skip counting by 10 is the same as the counting by 1, and putting a 0 to the right of each multiple. 9. The following table shows the skip counts of numbers up to 10. This is called a Multiplication Table. Exercise: Practice making the above multiplication table, by skip counting, at least three times.
STEP 8 (a) Have the student divide different phone numbers by 2 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (b) Have the student divide different phone numbers by 3 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (c) Have the student divide different phone numbers by 4 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (d) Have the student divide different phone numbers by 5 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. STEP 9 Have the student study Lesson 3, Section 3 & 4, and solve all the exercise problems of these sections. Make sure that the student can skip count by single-digit numbers. Lesson 3, Section 3: Adding to 10 and Beyond 10. When we add to 0, we get the same number that we added. 0 + 3 = 3; 0 + 6 = 6; 0 + 9 = 9 11. It is easy to add to 10 because 0 in ten changes to the number you are adding. 10 + 3 = 13; 10 + 6 = 16; 10 + 9 = 19 12. When the count goes beyond 10, count up to 10 first, and then add the rest to 10 mentally. Example 1: Add 9 + 5. (a) We start from 9 and count forward. (b) It takes one count to get 10. We have 4 counts left. (c) We add them mentally as 10 + 4 = 14. Add the following by counting up to 10, and then the rest mentally (a) 9 + 3 (b) 9 + 6 (c) 9 + 4 (d) 9 + 8 (e) 9 + 7 Answer: (a) 12 (b) 15 (c) 13 (d) 17 (e) 16
Example 2: Add 8 + 7. (a) We start from 8 and count forward. (b) It takes 2 counts to get 10. We have 5 counts left. (c) We add them mentally as 10 + 5 = 15. Add the following by counting up to 10, and then the rest mentally (a) 8 + 4 (b) 8 + 6 (c) 8 + 5 (d) 8 + 3 (e) 8 + 8 Answer: (a) 12 (b) 14 (c) 13 (d) 11 (e) 16 Example 3: Add 7 + 5. (a) We start from 7 and count forward. (b) It takes 3 counts to get 10. We have 2 counts left. (c) We add them mentally as 10 + 2 = 12. 7 + 5 = 7, 8, 9, 10 and 2 more = 10 + 2 = 12 Add the following by counting up to 10, and then the rest mentally (a) 7 + 4 (b) 7 + 6 (c) 7 + 5 (d) 7 + 7 (e) 7 + 8 Answer: (a) 11 (b) 13 (c) 12 (d) 14 (e) 15 Add the following (a) 0 + 4 (e) 0 + 3 (i) 0 + 7 (m) 0 + 8 (q) 0 + 9 (b) 10 + 4 (f) 10 + 9 (j) 10 + 2 (n) 10 + 8 (r) 10 + 6 (c) 9 + 4 (g) 9 + 6 (k) 9 + 3 (o) 9 + 8 (s) 9 + 9 (d) 8 + 5 (h) 8 + 7 (l) 8 + 4 (p) 8 + 6 (t) 8 + 3 Answer: (a) 4 (b) 14 (c) 13 (d) 13 (e) 3 (f) 19 (g) 15 (h) 15 (i) 7 (j) 12 (k) 12 (l) 12 (m) 8 (n) 18 (o) 17 (p) 14 (q) 9 (r) 16 (s) 18 (t) 11 Add the following: (Note: You may add two numbers in any order.) (a) 8 + 5 (e) 3 + 9 (i) 5 + 9 (m) 6 + 8 (q) 7 + 9 (b) 9 + 6 (f) 7 + 6 (j) 8 + 6 (n) 6 + 5 (r) 8 + 4 (c) 3 + 8 (g) 5 + 8 (k) 9 + 4 (o) 7 + 4 (s) 6 + 6 (d) 7 + 8 (h 9 + 5 (l) 9 + 8 (p) 7 + 5 (t) 2 + 9 Answer: (a) 13 (b) 15 (c) 11 (d) 15 (e) 12 (f) 13 (g) 13 (h) 14 (i) 14 (j) 14 (k) 13 (l) 17 (m) 14 (n) 11 (o) 11 (p) 12 (q) 16 (r) 12 (s) 12 (t) 11 Lesson 3, Section 4: Adding beyond 20, 30, 40, etc. 13. It is actually easy to add to multiples of 10 because 0 in ten changes to the number you are adding. 20 + 3 = 23; 30 + 6 = 36; 40 + 9 = 49
14. When the count goes beyond a multiple of 10, count up to the multiple of 10 first, and then add the rest to it mentally. Example 1: Add 39 + 5. (a) We start from 39 and count forward. (b) It takes one count to get 40. We have 4 counts left. (c) We add them mentally as 40 + 4 = 44. Add the following by counting up to 10, and then the rest mentally (a) 29 + 3 (b) 69 + 6 (c) 49 + 4 (d) 89 + 8 (e) 39 + 7 Answer: (a) 32 (b) 75 (c) 53 (d) 97 (e) 46 Example 2: Add 66 + 8. (a) We start from 66 and count forward. (b) It takes 4 counts to get 70. We have 4 counts left. (c) We add them mentally as 70 + 4 = 74. Add the following by counting up to 10, and then the rest mentally (a) 26 + 8 (b) 36 + 8 (c) 46 + 8 (d) 86 + 8 (e) 56 + 8 Answer: (a) 34 (b) 44 (c) 54 (d) 94 (e) 64 Example 3: Add 84 + 9. (d) We start from 84 and count forward. (e) It takes 6 counts to get 90. We have 3 counts left. (f) We add them mentally as 90 + 3 = 93. That is the answer to this addition. 84 + 9 = 84, 85, 86, 87, 88, 89, 90 and 3 = 90 + 3 = 93 Add the following by counting up to 10, and then the rest mentally (a) 24 + 7 (b) 34 + 8 (c) 44 + 9 (d) 87 + 7 (e) 56 + 7 Answer: (a) 31 (b) 42 (c) 53 (d) 94 (e) 63 15. Please note that there are other ways to add mentally. For example, you may add as follows. You may find you own ways to add mentally. 84 + 9 = 83 + 1 + 9 = 83 + 10 = 93 16. The key point to remember is that all these different ways of mental addition involve the intermediate sum of ten (10, 20, 30, 40, etc.).
Add the following (a) 18 + 5 (e) 73 + 9 (i) 75 + 9 (m) 16 + 8 (q) 87 + 9 (b) 33 + 8 (f) 25 + 8 (j) 59 + 4 (n) 37 + 4 (r) 66 + 6 (c) 54 + 9 (g) 34 + 8 (k) 46 + 9 (o) 48 + 9 (s) 54 + 7 (d) 67 + 8 (h) 69 + 5 (l) 29 + 8 (p) 67 + 5 (t) 32 + 9 Answer: (a) 23 (b) 41 (c) 63 (d) 75 (e) 82 (f) 33 (g) 42 (h) 74 (i) 84 (j) 63 (k) 55 (l) 37 (m) 24 (n) 41 (o) 57 (p) 72 (q) 96 (r) 72 (s) 61 (t) 41 STEP 10 (a) Have the student divide different phone numbers by 6 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (b) Have the student divide different phone numbers by 7 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (c) Have the student divide different phone numbers by 8 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. (d) Have the student divide different phone numbers by 9 by skip counting only, until he can do so comfortably. He may use the multiplication table only to reinforce his skip counting. STEP 11 Have the student study Lesson 3, Sections 5 and 6, and solve all the exercise problems of these sections. Make sure that the student can generate skip count for numbers 11 to 20. Lesson 3, Section 5: Adding Double-digit Numbers 17. A double-digit number is made up of a ten and a one as follows. 26 = 20 + 6; 33 = 30 + 3; 57 = 50 + 7 18. We add double-digit numbers by adding tens and ones separately. 26 + 33 = (20 + 30) + (6 + 3) = 50 + 9 = 59 33 + 57 = (30 + 50) + (3 + 7) = 80 + 10 = 90 57 + 26 = (50 + 20) + (7 + 6) = 70 + 13 = 83 Add the following (a) 18 + 35 (f) 73 + 29 (k) 75 + 39 (p) 16 + 78 (u) 87 + 19 (b) 33 + 28 (g) 25 + 38 (l) 59 + 24 (q) 37 + 44 (v) 66 + 26 (c) 54 + 19 (h) 34 + 58 (m) 46 + 29 (r) 48 + 19 (w) 54 + 37 (d) 67 + 28 (i) 69 + 15 (n) 29 + 68 (s) 67 + 15 (x) 32 + 49 (e) 89 + 47 (j) 28 + 77 (o) 15 + 76 (t) 78 + 13 (y) 28 + 58 Answer: (a) 53 (b) 61 (c) 73 (d) 95 (e) 136 (f) 102 (g) 63 (h) 92 (i) 84 (j) 105 (k) 114 (l) 83 (m) 75 (n) 97 (o) 91 (p) 94 (q) 81 (r) 67 (s) 82 (t) 91 (u) 106 (v) 92 (w) 91 (x) 81 (y) 86
Lesson 3, Section 6: Skip Counting by Double-digit Numbers 19. The skip counting by 20, is done by putting zero in the skip count of 2. 1 2 3 4 5 6 7 8 9 10 2 4 6 8 10 12 14 16 18 20 20 40 60 80 100 120 140 160 180 200 Similarly, skip counting may be done for 30, 40, 50, etc. 1 2 3 4 5 6 7 8 9 10 3 6 9 12 15 18 21 24 27 30 30 60 90 120 150 180 210 240 270 300 20. The skip counting by 12 is done by adding the skip counts of 10 and 2. 1 2 3 4 5 6 7 8 9 10 10 20 30 40 50 60 70 80 90 100 2 4 6 8 10 12 14 16 18 20 12 24 36 48 60 72 84 96 108 120 The skip counting by 16, is done by adding the skip counting of 10 and 6. 1 2 3 4 5 6 7 8 9 10 10 20 30 40 50 60 70 80 90 100 6 12 18 24 30 36 42 48 54 60 16 32 48 64 80 96 112 128 144 160 Similarly, skip counting by a double-digit number is done by adding the skip counts of the tens and ones. For example, skip counting by 23 is as follows. 1 2 3 4 5 6 7 8 9 10 20 40 60 80 100 120 140 160 180 200 3 6 9 12 15 18 21 24 27 30 23 46 69 92 115 138 161 184 207 230 Exercise: Practice making the multiplication table from 11 to 20, by skip counting, at least once. STEP 12 Have the student study Lesson 6, Section 4, and solve all the exercise problems of these sections. The long form of division is used to get the remainder using subtraction. Make sure that the student understands this.
Lesson 6, Section 4: Division with Large Divisors by Skip Counting 21. When you divide by 10, the quotient is same as the dividend without the last digit. The last digit appears as the remainder. Check this out. 25 10 = 2 R5 764 10 = 76 R4 7715743 10 = 771574 R3 22. When the divisor is between 10 and 20, write down the skip count for it per Section 6 of Lesson 3: Addition. Divide, 10853 16 Write down the multiplication table for 16 using repeated addition. 1 2 3 4 5 6 7 8 9 10 10 20 30 40 50 60 70 80 90 100 6 12 18 24 30 36 42 48 54 60 16 32 48 64 80 96 112 128 144 160 Then we can divide as before. Note that it is easier to find the remainder with long form. The steps are: (a) 16 cannot be taken out of the first two digits. Try taking it out the first 3 digits. (b) 16 can be taken out of 108 six times (16 x 6 = 96). Place 6 in the column above 8. Subtract 96 from 108. The remainder is 12. Bring down the next digit 5. We get 125. (c) 16 can be taken out of 125 seven times (16 x 7 = 112). Place 7 in the column above 5. Subtract 112 from 125. The remainder is 13. Bring down the next digit 3. We get 133. (d) 16 can be taken out of 133 eight times (16 x 8 = 128). Place 8 in the column above 3. Subtract 128 from 133. The remainder is 5. There are no more digits. (e) The quotient is 678, and there is a remainder of 5. Therefore, 10853 16 = 678 R5 [Check (678 x 16) + 5 = 10853]
Divide by first writing down the multiples of the divisor (a) 108 12 (d) 3225 12 (g) 976 11 (b) 911 11 (e) 1111 11 (h) 322 13 (c) 432 14 (f) 4555 15 (i) 3567 18 Solution: (a) 9 (b) 82 R9 (c) 30 R12 (d) 268 R9 (e) 101 (f) 303 R10 (g) 88 R8 (h) 24 R10 (i) 198 R3 STEP 13 Have the student divide different phone numbers by divisors 10 to 20 by skip counting, until he can do so comfortably. STEP 14 Have the student study Lesson 5, Sections 1, 2 and 3, and solve all the exercise problems of these sections. This will help the student generate skip counts more rapidly. Lesson 5, Section 1: Repeated Addition as Multiplication 1. We write repeated addition of skip counting as multiplication as follows. The word MULTIPLY means many layers. Multiplication is a shortcut for writing many additions for a number. 2. We may write the skip counting of 2 with multiplication as follows. 2 x 1 = 2 2 x 6 = 12 2 x 2 = 4 2 x 7 = 14 2 x 3 = 6 2 x 8 = 16 2 x 4 = 8 2 x 9 = 18 2 x 5 = 10 2 x 10 = 20 3. This may be shown in a table as follows.
4. The product of 3 x 5 can be found at the intersection of the row and column marked by 3 and 5. We find that 3 x 5 = 5 x 3 = 15 This is because 3 rows of 5 each are equal to 5 columns of 3 each. Multiply using the table. You may multiply these numbers in any order. (a) 7 x 6 (c) 9 x 5 (e) 5 x 8 (g) 8 x 3 (b) 5 x 7 (d) 9 x 8 (f) 7 x 4 (h) 8 x 7 Solution: (a) 42 (b) 35 (c) 45 (d) 72 (e) 40 (f) 28 (g) 24 (h) 56 Lesson 5, Section 2: Mental Math for Multiplication 5. Zero means no count. You may multiply it any number of times you still have 0. 0 x 6 = 0 0 x 75 = 0 6. One means a single item. 1 multiplied a number of times produces that number.
1 x 6 = 6 1 x 75 = 75 7. Ten is the base of the number. 10 multiplied a number of times produces that many tens (the number with a 0 on the right) 10 x 6 = 60 10 x 75 = 750 8. 100 multiplied by a number is always the same number followed by a 00. 1000 multiplied by a number is always the same number followed by a 000 100 x 6 = 600 1000 x 75 = 75000 (NOTE: Place as many 0 s after the number as there are 0 s in 10, 100, 1000, etc.) 9. When multiplying numbers with trailing 0 s, multiply the numbers without the trailing 0 s, and then place that many trailing 0 s after the product. 7 x 30 = (7 x 3) and one trailing 0 = 210 40 x 60 = (4 x 6) and two trailing 0 s = 2,400 300 x 2000 = (3 x 2) and five trailing 0 s = 600,000 Multiply the following numbers using mental math (You may use paper and pencil to aid you) (a) 3 x 10 (d) 3 x 1000 (g) 9 x 1 (j) 16,437 x 0 (b) 14 x 100 (e) 231 x 1 (h) 32,862 x 10 (k) 446 x 1000 (c) 6 x 30 (f) 50 x 300 (i) 500 x 20 (l) 3,000 x 4,000 Solution: (a) 30 (b) 1,400 (c) 180 (d) 3,000 (e) 231 (f) 15,000 (g) 9 (h) 328,620 (i) 10,000 (j) 0 (k) 446,000 (l) 12,000,000 Lesson 5, Section 3: Skip Counting by Multiplication 10. When multiplying numbers greater than 10, we may multiply TENS and ONES separately. 23 x 2 = (20 + 3) x 2 = 40 + 6 = 46 35 x 5 = (30 + 5) x 5 = 150 + 25 = 175 47 x 8 = (40 + 7) x 8 = 320 + 56 = 376
Multiply the following numbers separating them by TENS and ONES (You may use paper and pencil to aid you) (a) 14 x 3 (c) 23 x 7 (e) 83 x 2 (g) 34 x 6 (b) 38 x 3 (d) 58 x 7 (f) 93 x 4 (h) 47 x 9 Solution: (a) 42 (b) 114 (c) 161 (d) 406 (e) 166 (f) 372 (g) 204 (h) 423 11. You may create skip counts for a double-digit number like 37 as follows: 37 x 1 = (30 + 7) x 1 = 30 + 7 = 37 37 x 2 = (30 + 7) x 2 = 60 + 14 = 74 37 x 3 = (30 + 7) x 3 = 90 + 21 = 111 37 x 4 = (30 + 7) x 4 = 120 + 28 = 148 37 x 5 = (30 + 7) x 5 = 150 + 35 = 185 37 x 6 = (30 + 7) x 6 = 180 + 42 = 222 37 x 7 = (30 + 7) x 7 = 210 + 49 = 259 37 x 8 = (30 + 7) x 8 = 240 + 56 = 296 37 x 9 = (30 + 7) x 9 = 270 + 63 = 333 Create skip counts for the following: (a) 23 (b) 31 (c) 42 (d) 55 (e) 73 STEP 15 Have the student divide different phone numbers by divisors between 21 and 30 by skip counting, until he can do so comfortably. STEP 16 Have the student study Lesson 6, Sections 5, and solve all the exercise problems of these sections. Make sure that the student learns to guess the quotient as shown in this section. For large divisors this is faster than skip counting.
Lesson 5, Section 5: Division with Large Divisors by Guessing 23. When the divisor is larger, find the quotient by guessing. Divide, 563 62 Round to tens as follows: 563 / 62 is about 560 / 60; the tens are 56 / 6. We can take 6 out of 56 nine times. Therefore, 62 may be taken out of 563 about 9 times. Compute 62 x 9 = 558. This gives us, 563 62 = 9 R5 Divide, 396 44 Round to tens as follows: 396 / 44 is about 400 / 40; the tens are 40 / 4. We can take 4 out of 40 ten times. Therefore, 44 may be taken out of 396 about 10 times. Compute 44 x 10 = 440 (more than 396) Compute 44 x 9 = 396. This gives us, 396 44 = 9 Divide, 611 87 Round to tens as follows: 611 / 87 is about 610 / 90; the tens are 61 / 9. We can take 9 out of 61 six times. Therefore, 87 may be taken out of 611 about 6 times. Compute 87 x 6 = 522 (much smaller than 611) Compute 87 x 7 = 609. This gives us, 611 87 = 7 R2 Divide 630526 37 The steps are
(a) For 63 37 round to 60 and 40. The TENS are 6 and 4. You can take 4 from 6 at most 1 time. Check 37 x 1 = 37; 37 x 2 = 74. This gives us 63 37 = 1 R 26. (b) Place 1 above 63, and 26 before 0. (c) For 260 37 round to 260 and 40. The TENS are 26 and 4. You can take 4 from 26 at most 6 times. Check 37 x 6 = 222, 37 x 7= 259. This gives us 260 37 = 7 R1. (d) Place 7 above 0, and 1 before 5. (e) For 15 37 you obviously get 0 R15. (f) Place 0 above 5, and 15 before 2. (g) For 152 37 round to 150 and 40. The TENS are 15 and 4. You can take 4 from 15 at most 3 times. Check 37 x 3 = 111, 37 x 4 = 148. This gives us 152 37 = 4 R4. (h) Place 4 above 2, and the remainder 4 before 6. (i) We get 46 37 = 1 R9. Place 1 above 6. There are no more digits, so the final remainder is 9. Therefore, 630526 37 = 17041 with remainder of 9 [Check (17041 x 37) + 9 = 630526] Divide by approximating the quotient. (a) 144 16 (d) 217 45 (g) 7488 61 (b) 123 23 (e) 318 53 (h) 82593 71 (c) 259 37 (f) 419 59 (i) 994720 89 STEP 17 Have the student divide different phone numbers by divisors between 31 and 50 by guessing, until he can do so comfortably. STEP 18 Have the student study Lesson 4, Sections 1 and 2, and solve all the exercise problems of these sections. Make sure that the student learns to subtract rapidly. The method of subtraction by reverse addition works faster with large numbers.
Lesson 4, Section 1: Subtraction is Reverse Addition 1. In subtraction we take a smaller number away from a larger number and find the remainder. 11 6 = 5 2. The remainder is the gap between the two numbers. Therefore, we may subtract by counting the gap on the number line as follows. This gives us 5 again for 11 6. 3. This shows that subtraction is reverse addition. To find 11 6 = what? we ask, 6 + what? = 11 From counting more we know that 6 + 5 = 11 Therefore, what? = 5; and, 11 6 = 5 (the gap) 4. We may, therefore subtract by reverse addition as follows. 16 7 = what 7 + what = 16 what = 9 (by counting from 7 to 16) Subtract by reverse addition (a) 3 + what = 9 (c) 35 + what = 39 (e) 68 + what = 71 (b) 8 + what = 11 (d) 49 + what = 53 (f) 88 + what = 96 Answer: (a) 6 (b) 3 (c) 4 (d) 4 (e) 3 (f) 8 Lesson 4, Section 2: Subtraction by Column 5. We may subtract larger numbers, such as, 88 57 = what? by reverse addition also. In this case we get the help by using columns. We place the smaller number at the top, the larger number at the bottom. Then we write the gap between them one column at a time by starting from the right.
Rightmost column, 7 + what = 8; what = 1; write 1 in the middle row. Next column, 5 + what = 8; what = 3; write 3 in the middle row Subtraction: 88 57 = 31. 6. When the bottom digit is smaller we increase it by 10. Then we carry the ten over to the next column as in addition. Rightmost column: Since bottom digit 3 is smaller than 7, we increase it to 13. 7 + what = 13; what = 6; write 6 in the middle row Next left column: 8 and 1 carried over is 9; increase bottom digit 2 to 12 9 + what = 12; what = 3; write 3 in the middle row Next left column: 3 and 1 carried over is 4; bottom digit 6 is larger. 4 + what = 6; what = 2; write 2 in the middle row We get 623 387 = 236 (you may check this on calculator) 7. Check out the following problem of subtraction by reverse addition. Therefore, 70,001 23,854 = 46,147 Subtract the following by column using reverse addition (a) 465 243 = (c) 407 389 = (e) 311 179 = (b) 396 254 = (d) 300 289 = (f) 254 169 = Answer: (a) 222 (b) 142 (c) 18 (d) 11 (e) 132 (f) 85 8. You may still subtract by reverse addition when numbers are written in the conventional order. For example, when 325 148 is written in the conventional order as follows. We simply treat the larger number at the top to be the sum, and the smaller number in the middle being added to the remainder at the bottom. The carry overs are applied to the smaller number in the middle.
Write the following subtractions in the conventional order, but subtract by reverse addition (a) 352 238 = (c) 523 376 = (e) 645 453 = (b) 444 288 = (d) 416 237 = (f) 911 199 = Answer: (a) 114 (b) 156 (c) 147 (d) 179 (e) 192 (f) 712 STEP 19 Have the student divide different phone numbers by double-digit divisors, until he can do so comfortably. STEP 20 Have the student review Lessons 3 through 6. STEP 21 Have the student review Lessons 1 and 2. Let the student take the Diagnostic Test in STEP 1 again. If he passes the test he can move forward to Remedial Math Level 2.