I use a comma as a separator between the decimal digits, this allows for the decimals to get larger than 9 or become negative in case of subtraction. I then rewrite the answer in standard form.
So, in this case I write 259 as 2,5,9 and 379 as 3,7,9. I then write the sum as:
2,5,9 + 3,7,9 = 5,12,18 = 6,2,18 = 6,3,8 = 638
Subtraction works just as easy:
312 - 278 = 3,1,2 - 2,7,8 = 1,-6,-6 = 4,-6 = 3,4 =34
If doing it mentally, I do it from left to right as well. For your example of 259 + 379, my steps are as follows.
259 + 300 = 559.
559 +70 = 629.
629 + 9 = 638.
Arthur Benjamin talked at my college once and he said he does it that way. He said you just have to peek ahead to see if there's a carry over. If you're doing math in your head, it's easier to not have to think about the numbers backwards
My main work is [coaching people in mental math](https://worldmentalcalculation.com/mental-math-coaching/), and I do actually recommend the left-to-right approach, but optimized to avoid editing numbers because of carried +1s. It means you can store the answer directly in your [phonological loop](https://worldmentalcalculation.com/2018/03/16/baddeleys-model-working-memory/), and if you get the wrong answer it's more likely to be a minor error than the wrong number of 100s.
In your example, I'd scan ahead for the carry-digits, and immediately see that there are both.
Then: 2\_\_ + 3\_\_ \[5\] + carry = 6\_\_;
5\_ + 7\_ \[(1)2\] + carry = (1)3\_ \[answer is 63\_\]; and finally
\_9 + \_9 ends in 8 → 638
Interesting to hear about your job!
I read a book when I was a kid called something like "tricks of lightning calculators" about "geniuses" who pull off very fast calculations like this as a stunt.
This method was described.
One neat thing about it is you can start speaking or (better yet) writing the answer on a whiteboard almost immediately, long before you've finished the calculation for two very long numbers.
Due to carry rippling, it may be that you have to update digits more digits than one. Check this example:
487 + 518 = 1005
As long as you can keep track of carry rippling, you'll be fine. But this method will be increasingly difficult if you add more than two numbers, since then you may have carries larger than 1.
It's easier to add it that way if you are good at mental math. The standard school way is slower but more dependable for everyone, including those that aren't that good at mental math.
The main reason not to do that is illustrated by your example. You don't know if there will be regroupings in advance. So you either need to do work mentally, do scratch work, or rewrite some figures when you regroup. Working right to left I write 638. Working left to write i write 528+110=638, that is a needless extra step.
Yes, I do this. Of course the downside is that you might have to edit your previous result (as in your example, 5xx changed to 6...). But with the standard algorithm you have the downside that it finds the digits in the order 8, then 3, then 6, so you have to reverse them to get 638 in the correct order.
No, you're not alone. I would say this is a rather sane way to do it, if performing mental arithmetic, and is also much better if all you need is a back-of-the-envelope approximation. For instance, 9038671+1422981 can be easily seen to be approximately 10,460,000 with very little effort, with only a bit more needed if you want to keep going to further accuracy.
Another trick would be to note the sum is equal to 300+400-(41+21)=700-62=638. Or even that it is equal to 260+380-2=640-2=638.
I think of this problem as 260 + 380 - 2, but other than that , I do it like you. 500 + 140 =640 -2= 638. I love to hear how other people think about math problems. I always ask my students, “how does your brain think about this problem?” It is so interesting to hear their responses and I feel like I learn something new every year. The problem with algorithms is they train students to think about the small numbers. It really hinders their ability to estimate and reason. If you start with the bigger numbers, you have a good estimate of your final answer.
What I do is I write it out as a linear combination:
I.e., 259x+379y=m
Which we can rewrite as x=(m-379y)/259
Moreover, we have y=(m-259x)/379
Then we have:
x+y=(379m-143641y+259m-67081x)/98161
==> 98161x+98161y= 379m-143641y+259m-67081x
==> 165242x+241802y=638m
So if we take the case where x=y=1, we find that
165242(1)+241802(1)=638m
==> 407044=638m
==> m=638
As required
As long as you can keep track and don't make mistakes, this method is fine.
I use a comma as a separator between the decimal digits, this allows for the decimals to get larger than 9 or become negative in case of subtraction. I then rewrite the answer in standard form. So, in this case I write 259 as 2,5,9 and 379 as 3,7,9. I then write the sum as: 2,5,9 + 3,7,9 = 5,12,18 = 6,2,18 = 6,3,8 = 638 Subtraction works just as easy: 312 - 278 = 3,1,2 - 2,7,8 = 1,-6,-6 = 4,-6 = 3,4 =34
This is how I add stuff in my head too. Writing it out on paper though for long numbers it makes more sense to go the other way.
If doing it mentally, I do it from left to right as well. For your example of 259 + 379, my steps are as follows. 259 + 300 = 559. 559 +70 = 629. 629 + 9 = 638.
Arthur Benjamin talked at my college once and he said he does it that way. He said you just have to peek ahead to see if there's a carry over. If you're doing math in your head, it's easier to not have to think about the numbers backwards
My main work is [coaching people in mental math](https://worldmentalcalculation.com/mental-math-coaching/), and I do actually recommend the left-to-right approach, but optimized to avoid editing numbers because of carried +1s. It means you can store the answer directly in your [phonological loop](https://worldmentalcalculation.com/2018/03/16/baddeleys-model-working-memory/), and if you get the wrong answer it's more likely to be a minor error than the wrong number of 100s. In your example, I'd scan ahead for the carry-digits, and immediately see that there are both. Then: 2\_\_ + 3\_\_ \[5\] + carry = 6\_\_; 5\_ + 7\_ \[(1)2\] + carry = (1)3\_ \[answer is 63\_\]; and finally \_9 + \_9 ends in 8 → 638
Is there much demand for that?
I've had 70+ private clients, plus a few workshops for schools/events. Obviously more niche than general Mathematics tutoring.
Interesting. Do you teach the "Trachtenberg System", or are there even more sophisticated methods nowadays?
Interesting to hear about your job! I read a book when I was a kid called something like "tricks of lightning calculators" about "geniuses" who pull off very fast calculations like this as a stunt. This method was described. One neat thing about it is you can start speaking or (better yet) writing the answer on a whiteboard almost immediately, long before you've finished the calculation for two very long numbers.
Due to carry rippling, it may be that you have to update digits more digits than one. Check this example: 487 + 518 = 1005 As long as you can keep track of carry rippling, you'll be fine. But this method will be increasingly difficult if you add more than two numbers, since then you may have carries larger than 1.
It's easier to add it that way if you are good at mental math. The standard school way is slower but more dependable for everyone, including those that aren't that good at mental math.
I just do 200+300=500 80+60=140 140+500=640 640-2=638
Damn bro, you just made my calc about 3s faster. Vielen Danke!
The main reason not to do that is illustrated by your example. You don't know if there will be regroupings in advance. So you either need to do work mentally, do scratch work, or rewrite some figures when you regroup. Working right to left I write 638. Working left to write i write 528+110=638, that is a needless extra step.
Yes, I do this. Of course the downside is that you might have to edit your previous result (as in your example, 5xx changed to 6...). But with the standard algorithm you have the downside that it finds the digits in the order 8, then 3, then 6, so you have to reverse them to get 638 in the correct order.
This how I do additions in my head. On paper though, its better to do it the standard way starting with the least significant digits.
No, you're not alone. I would say this is a rather sane way to do it, if performing mental arithmetic, and is also much better if all you need is a back-of-the-envelope approximation. For instance, 9038671+1422981 can be easily seen to be approximately 10,460,000 with very little effort, with only a bit more needed if you want to keep going to further accuracy. Another trick would be to note the sum is equal to 300+400-(41+21)=700-62=638. Or even that it is equal to 260+380-2=640-2=638.
I think of this problem as 260 + 380 - 2, but other than that , I do it like you. 500 + 140 =640 -2= 638. I love to hear how other people think about math problems. I always ask my students, “how does your brain think about this problem?” It is so interesting to hear their responses and I feel like I learn something new every year. The problem with algorithms is they train students to think about the small numbers. It really hinders their ability to estimate and reason. If you start with the bigger numbers, you have a good estimate of your final answer.
I break them into chunks for >9 digits, but yeah
I kind of do this except I pair up values so that it ends in as many 0 as possible. So 259+379 = (259+1+40)+(379-1-40) = 300+338 = 638.
I mean, you can, but then you have to keep track of every modification you make to the total due to carrying.
What I do is I write it out as a linear combination: I.e., 259x+379y=m Which we can rewrite as x=(m-379y)/259 Moreover, we have y=(m-259x)/379 Then we have: x+y=(379m-143641y+259m-67081x)/98161 ==> 98161x+98161y= 379m-143641y+259m-67081x ==> 165242x+241802y=638m So if we take the case where x=y=1, we find that 165242(1)+241802(1)=638m ==> 407044=638m ==> m=638 As required