Subtract 389,002-76,151 using the standard algorithm.
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- this is more confusing than addition(13 votes)
- I must admit, yes at times it is more confusing than addition but if you can get the methods and get used to it, it is really really easy(2 votes)
- is there any more methods(5 votes)
- Yes! There’s a Vedic (Indian) method of subtraction.
We subtract one column at a time. When the bottom number is bigger we subtract in the reverse order but put a bar over the digit in the answer. Bar digits can be thought of as negative digits!
We then convert from bar digits to a normal number using the following steps:
1) For each group of one or more 0’s (if any) immediately to the left of a group of one or more bar digits, we also put bars on those 0’s.
2) For each group of bar digits after the previous step:
a) We subtract “all from 9 except last from 10”.
b) We take away 1 from the digit immediately to the left of the group of bar digits.
This might seem hard to understand at first, but it should become easier to understand when you read the example below.
Example: 676,354,508 - 146,378,168.
We could go right to left, or left to right. Let’s go left to right.
From left to right:
6-1 = 5.
7-4 = 3.
6-6 = 0.
3-3 = 0.
5-7 = 2bar (because 7-5 = 2).
4-8 = 4bar (because 8-4 = 4).
5-1 = 4.
0-6 = 6bar (because 6-0 = 6).
8-8 = 0.
So we have 5 3 0 0 2bar 4bar 4 6bar 0.
We now need to convert to a normal number.
1) We have a group of two 0’s immediately to the left of the group of bar digits 2bar 4bar. So we put bars on these two 0’s.
So we now have 5 3 0bar 0bar 2bar 4bar 4 6bar 0.
2) We have the two groups of bar digits 0bar 0bar 2bar 4bar, and 6bar.
For the group 0bar 0bar 2bar 4bar:
a) 9-0 = 9.
9-0 = 9.
9-2 = 7.
10-4 = 6.
b) The 3 immediately to the left of this group becomes a 2.
For the group 6bar:
a) 10-6 = 4.
b) The 4 immediately to the left of this group becomes a 3.
Note that the 5 on the far left and the 0 on the far right both stay as is.
The final answer is 529,976,340.
Have a blessed, wonderful day!(16 votes)
Is there a faster method to solve substraction problems?
The distribution of digits - in a long number - to the right 10^x category is helpful, but is there a faster way to solve these kind of problems? It takes to long to regroup them, even thou I get to the solution im unhappy with the timing.(3 votes)
- is there any more methods(2 votes)
- Is there any more methods(2 votes)
- I like this man's videos they are helpful!:)(2 votes)
- [Instructor] What we're gonna do in this video is figure out what 389,002 minus 76,151 is. Like always, I encourage you to pause the video and try to figure it out on your own. That's the best way to really, even if you're not able to figure out, or if you get a different answer, then when I work through it with you it will really stick in your brain that much more. Alright, now let's work through it together. The way I'm gonna do it is sometimes called the standard method or the standard algorithm, algorithm being a fancy word for a method. What I'm gonna do is first write the 389,002. 389,002. And I'm subtracting 76,151. You notice the first thing that I did is I aligned the digits to the appropriate place value. I put the ones below the ones. The 10s below the 10s, the 100s below the 100s, the 1,000s below the 1,000s, the 10,000s below the 10,000s, so on and so forth. And now we're ready to subtract. So the first thing we might do is well, let's look at the ones place. Here I have two ones, and I'm gonna take away one one. So I'm gonna be left with one one. That was pretty straightforward. But then things get a little bit more difficult when we get to the 10s place. How do I take five 10s from zero 10s? So let me just not think about that for a second, but I have the same problem in the 100s place. How do I take away one 100 from zero 100s? Now when I go to the 1,000s place, I can take away six 1,000s from nine 1,000s, but before I do that what I want to do is regroup so that I don't have zeros here so that I can take away from the 100s and the 10s place. And so what I can do is I can rewrite nine 1,000s, so I'm gonna take away one of those 1,000s, so I'm gonna have eight 1,000s. And I'm gonna regroup it as 10 100s. So this can be that 1,000 would be 10 100s. Now that solves a problem, except for the 10s place. But what I can then do is I could take away one of those 100s so I only have nine 100s now, and I could regroup that extra 100 as 10 10s. So as 10 10s. And now I can keep subtracting. So in the 10s place, 10 10s minus five 10s, is five 10s. I go to the 100s place, nine 100s minus 100 is 800. I go to the 1,000s place. 8,000 minus 6,000 is 2,000. And then I can go to the 10,000s place. This is essentially eight 10,000s, or 80,000 minus 70,000 is going to be 10,000. One 10,000. And then last but not least, I have my three 100,000s. So there you go. We're done. This is 312,851. This is the standard method. I started at the ones place. Sometimes it's good to just do a check to make sure every digit on top in the appropriate place is at least equal to the digit that you're subtracting from it. And so you can do the regrouping ahead of time. But either way, you will end up with a similar process.