Do the Chunking Method

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←Older revision		Revision as of 23:48, 18 June 2019
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	===Part One: Basic Instructions===		===Part One: Basic Instructions===
	#Look 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.[[Image:Do the Chunking Method Step 1.jpg|center]]		#Look 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.[[Image:Do the Chunking Method Step 1.jpg|center]]
−	#*This method is also called the "hangman" or the "partial quotients method" — because you are essentially finding the total quotient one part at a time<ref>http://bit.ly/2MV4s4N>. All parts will eventually be added together so that you can find the final, total 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.		#*''Example:'' Use the chunking method to find the quotient of 731 ÷ 5.
	#Know which multiples are easiest to find. The "easy" multiples of your dividend are those that can be quickly calculated in your head.[[Image:Do the Chunking Method Step 2.jpg|center]]		#Know which multiples are easiest to find. The "easy" multiples of your dividend are those that can be quickly calculated in your head.[[Image:Do the Chunking Method Step 2.jpg|center]]
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	#Write your answer, including the remainder. The answer to 931 ÷ 72 is '''12 R67'''.[[Image:Do the Chunking Method Step 22.jpg|center]]		#Write your answer, including the remainder. The answer to 931 ÷ 72 is '''12 R67'''.[[Image:Do the Chunking Method Step 22.jpg|center]]
	===Part Four: Three-Digit Divisors===		===Part Four: Three-Digit Divisors===
−	#Solve 1568 ÷ 112. In this case, short division is less practical as more mental calculations would be required to solve this problem, so using the chunking method can be a practical solution.[[Image:Do the Chunking Method Step 23.jpg|center]]	+	#Solve 1568 ÷ 112. Short division cannot be used to solve this problem, so using the chunking method can be a practical solution.[[Image:Do the Chunking Method Step 23.jpg|center]]
	#Identify the next easiest multiple. The largest easy multiple you can use would be 1120.[[Image:Do the Chunking Method Step 24.jpg|center]]		#Identify the next easiest multiple. The largest easy multiple you can use would be 1120.[[Image:Do the Chunking Method Step 24.jpg|center]]
	#*This multiple is found by multiplying 112 and the easy multiplier 10.		#*This multiple is found by multiplying 112 and the easy multiplier 10.
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	#*You can get 224 by multiplying 112 * 2. In this case, 2 is the easy multiplier used.		#*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.		#*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.
−
	#Subtract 448 – 224. The difference between the two values is 224.[[Image:Do the Chunking Method Step 27.jpg|center]]		#Subtract 448 – 224. The difference between the two values is 224.[[Image:Do the Chunking Method Step 27.jpg|center]]
	#*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.		#*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.
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	#*Note that there is no remainder for this problem.		#*Note that there is no remainder for this problem.
−	== Additional Note ==
−	While chunking usually does not allow for subtracting more than the leftover, it can be adapted into a method that does so — at the expense of incurring some negative numbers in the process.<ref>http://bit.ly/2WRY7vv>
	== References ==		== References ==

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