How to Do the Chunking Method
Four Parts:Basic InstructionsOne-Digit DivisorsTwo-Digit DivisorsThree-Digit Divisors
The chunking method is an alternative to long division. By breaking apart a dividend into easily calculated value chunks, you can solve complex division problems.
Steps
Part 1 Basic Instructions
- 1Look at the problem. When given a division problem that cannot be solved using short division, you can use the chunking method to find the quotient.
- This method is also called the "partial quotients method" because you are essentially finding the total quotient one part at a time. All parts will eventually be added together so that you can find the final, total quotient.
- Example: Use the chunking method to find the quotient of 731 ÷ 5.
- 2Know which multiples are easiest to find. The "easy" multiples of your dividend are those that can be quickly calculated in your head.
- Usually, they will be the multiples calculated when you multiply the dividend by the easy multipliers of 1000, 100, 10, 5, or 2.^{[1]}
- 3Identify the largest easy multiple for the equation. Determine the largest easy multiple you can calculate for the equation. You must multiply the divisor by one of the easy multipliers to come up with a number that is less than the value of the dividend.
- Example: You can multiply the divisor, 5, by the multipliers of 100, 10, 5, and 2 to come up with a product that is less than the value of the dividend, 731. The largest of these multipliers is 100, so you would multiply 5 * 100 to produce an easy multiple of 500.
- 4Subtract the product from the dividend. Subtract the product or partial quotient you just found from the dividend. The difference between the two will be the next value that you work with.^{[2]}
- Example: You'll need to subtract 731 – 500. The answer is 231.
- You'll have to break down the difference, 231, like you broke down the dividend, 731.
- Example: You'll need to subtract 731 – 500. The answer is 231.
- 5Repeat as needed. Identify the next largest easy multiple and subtract it from the difference you just calculated. Repeat this process as needed until the difference between two subtracted numbers is either “0” or a number less than the original divisor.^{[3]}
- Example: The next easy multiple you can work with in this problem is 10, so multiply 5 * 10 to reach a product of 50.
- Subtract 50 from the previous difference, 231, as so: 231 – 50 = 181
- The easy multiple 50 can still be used since it is less than the new difference, 181. As such, you must continue to subtract by 50 until the difference is less than that value: 181 – 50 = 131 – 50 = 81 - 50 = 31
- Identify the next highest easy multiple. The next best multiplier to use would be 5, so your next multiple would be 25 (5 * 5 = 25).
- Subtract 31 – 25, which gives you an answer of 6.
- The divisor, 5, can be subtracted as is from the difference, 6: 6 = 5 = 1
- Since 1 is less than 5 (the original divisor), the calculations end here.
- Example: The next easy multiple you can work with in this problem is 10, so multiply 5 * 10 to reach a product of 50.
- 6Identify any remainder. When you are left with “0” at the end of your calculations, there is no remainder. Any number other than “0” that is smaller than the divisor would be a remainder, though.
- Example: For this problem, there is a remainder of 1.
- 7Add up the multipliers. You'll need to add all of the multipliers you used while chunking the equation to find your final answer. Multipliers used more than once must be added the number of times they were used. Any time you subtracted the actual divisor without multiplying it by a separate multiplier, you must add a 1.^{[4]}
- Example: In this equation, you used the multiplier 100 once, 10 four times, 5 once, and 1 once, so you must add together:
- 100 + 10 + 10 + 10 + 10 + 5 + 1 = 146
- Example: In this equation, you used the multiplier 100 once, 10 four times, 5 once, and 1 once, so you must add together:
- 8Write the final answer. Your final answer is the sum calculated in the previous step, along with any remainder identified in the step before it. The remainder should be proceeded with an “R.”
- Example: The answer to 731 ÷ 5 is 146 R1
Part 2 One-Digit Divisors
- 1Solve 84 ÷ 7. This equation could technically be solved with short division, but if you don't know the answer already, you could still use the chunking method to find the correct answer.
- 2Identify the easiest multiple. The easiest multiple is the largest possible easy multiple of the divisor. In this case, it would be 70.
- You would find the multiple, 70, by multiplying 7 by the easy multiplier of 10.
- Using a lower easy multiplier would give you a value that is smaller than necessary. Using a higher easy multiplier, like 100, would give you a multiple that is larger than the dividend, 84.
- 3Subtract 84 – 70. The difference is 14.
- Since 14 is still larger than 7, you need to continue you calculations further.
- 4Identify the next easiest multiple. If you've memorized your multiplication tables, you will already know that 7 * 2 = 14. Since the product of 7 and 2 is no larger than the difference calculated in the previous step, this product (14) is your next easiest multiple.
- Note that the multiplier used here is 2, which happens to be one of the standard easy multiples.
- 5Subtract 14 – 14. The difference between these values is 0.
- When you reach a difference of 0, you have found all of the partial quotients you can find. The chunking portion of your calculations is complete.
- 6Add the multipliers together. In this case, you would need to add 10 + 2, giving you an answer of 12.
- You used the multiplier 10 once.
- You used the multiplier 2 once.
- 7Write your answer. The answer to 84 ÷ 7 is 12.
- Note that there was no remainder in this problem.
Part 3 Two-Digit Divisors
- 1Solve 931 ÷ 72. Since this equation cannot be solved simply by using short division, it makes sense to use the chunking method of division to find the quotient.
- 2Identify the easiest multiple. The largest possible easy multiple of your divisor, 72, would be 720.
- This multiple is found by multiplying 72 by the easy multiplier 10.
- A larger easy multiplier, like 100, would produce a multiple that is too large for the equation (7200) since the multiple must be smaller than the dividend, 931.
- 3Subtract 931 – 720. The difference between the dividend and the multiple is 211.
- Since 211 is larger than 72, you must continue chunking to find the final answer.
- Note that 211 is smaller than 720, so you will need to find a new multiple to use.
- 4Identify the next easiest multiple. The next easiest multiple you can use would 144.
- You need to use an easy multiplier smaller than the previous multiplier, 10.
- The next highest easy multiplier is 5, but 72 * 5 = 360. Since 360 is larger than 211, this multiple cannot be used.
- The next highest easy multiplier after that is 2, and 72 * 2 = 144. Since 144 is smaller than 211, this is the multiple you should use.
- 5Subtract 211 – 144. The difference between the two values is 67.
- Since 67 is smaller than the original divisor, 72, your chunking calculations must stop here.
- 6Add the multipliers together. You'll need to add together 10 + 2, giving you an answer of 12.
- Note, however, that there is also a remainder value for this equation: 67
- The remainder must be included when you write your final answer.
- 7Write your answer, including the remainder. The answer to 931 ÷ 72 is 12 R67.
Part 4 Three-Digit Divisors
- 1Solve 1568 ÷ 112. Short division cannot be used to solve this problem, so using the chunking method can be a practical solution.
- 2Identify the next easiest multiple. The largest easy multiple you can use would be 1120.
- This multiple is found by multiplying 112 and the easy multiplier 10.
- A larger easy multiplier, like 100, would create a product that is larger than the quotient, so it cannot be used. A smaller easy multiplier would not be as practical, even though it could technically be used.
- 3Subtract 1568 – 1120. The difference between the quotient and the multiple is 448.
- Since 448 is larger than 112, you will need to continue chunking the equation.
- Since 1120 is larger than the difference, 112, you can no longer use that multiple.
- 4Identify the next easiest multiple. The largest easy multiple you can use at this point would be 224.
- You can get 224 by multiplying 112 * 2. In this case, 2 is the easy multiplier used.
- Even though the multiplier 5 is smaller than the multiplier 10 and larger than the multiplier 2, 112 * 5 = 560. Since 560 is larger than 224, it cannot serve as an easy multiple in this problem.
- 5Subtract 448 – 224. The difference between the two values is 224.
- Notice that 224 is the same value as your chosen multiple. As such, you will continue to use 224 as your chosen multiple and subtract it from the difference.
- 6Subtract 224 – 224. The answer is 0.
- Since you've reached 0, there can be no further chunking for this problem.
- 7Add the multipliers. You must add 10 + 2 + 2, giving you an answer of 14.
- You used the multiplier 10 only once.
- You used the multiplier 2 a total of two times.
- 8Write your answer. The answer to 1568 ÷ 112 is 14.
- Note that there is no remainder for this problem.
Sources and Citations
Article Info
Categories: Mathematics | Algebra