Computing Digit Sums

Lesson 1.1, Part 2

In Part One I talked about what Digit Sums were, and why they are useful. Now I’ll show you some examples where I compute Digit Sums.

Example: 27
27 —-> (2 + 7) —-> 9
Therefore the Digit Sum of 27 is 9.

Example: 435
435 —-> (4 + 3 + 5) —-> 12

Since ’12′ contains two digits, and were trying to get down to one…we add them again.
12 —-> (1 + 2) —-> 3. Therefore the Digit Sum of 435 is 3.

Note: Notice that I didn’t use an ‘=’ sign because 435 does not equal 3, only the Digit Sum does.

Example: 102372
102372 —-> (1 + 0 + 2 + 3 + 7 + 2) —-> 15
15 —-> (1 + 5) —-> 6.
Therefore the Digit Sum of 102372 is 6.

Try to solve the problems below on your own first, whether mentally or by writing it down, before clicking on the link to the solutions.

PRACTICE PROBLEMS
1. 29
2. 32
3. 57
4. 354
5. 271
6. 10253
7. 27361
8. 56381
9. 1029301
10. 1425361

Answers to Practice Problems.

Other Parts of Lesson 1.1:
Part One – What are Digit Sums?
Part Three – The Nine-Point Circle
Part Four – Casting Out the 9′s

What are Digit Sums?

Lesson 1.1, Part 1

So what exactly are Digit Sums?
The Digit Sum of a number is the number you get when you add all the digits together, then add those digits together all the way until you have a single digit left between 1 and 9. You may be wondering about ’0′, it has some special properties which we will talk about in a future lesson. As you will come to see, there are patterns involving digit sums and shortcuts to make computing Digit Sums easier.

Why are they useful?
Digit Sums can find use in many things, a couple examples being: checking your answers, checking divisibility, and for dividing by nine.

How do you get them?
In order to get a Digit Sum all you have to do is add the digits. If the number you get is a single digit then your done, if not then you add them again until your left with a single digit. Take the number 26 for example; the Digit Sum would be 8 because 2 + 6 = 8.

Other Parts of Lesson 1.1
Part Two – Computing Digit Sums
Part Three – The Nine-Point Ciricle
Part Four – Casting Out the 9′s

New Page Added

Just wanted to let you guys know that I added a Vedic Math page to the blog. You can access it either at the top of the blog, or from the side-bar on the right.

At the moment some of the pages are password-protected and/or hidden. This is because they are not finished yet, once they’re done I’ll open them up for everyone.

Hope you enjoy it! Let me know if you catch any errors, or have any suggestions.

UPDATE: Just opened up the “Table of Contents” & “Section 1″ pages. Right now none of the links in Section 1 work because I haven’t written up the lessons yet. However, looking at it will give you a rough outline of whats to come. Yes the topics look basic, but you have to build a foundation first before you start doing anything else otherwise you will just get lost.

Update

I just wanted to let everyone know that I am aware the posts have been kind of long, and that they can be intimidating. For this reason, I’ve decided I’m going to go back and repost some of the stuff I’ve already posted but in shorter posts. These posts will be shorter, contain more practice problems and will have a link to the the solutions for the practice problems.

Right now I’m working on re-writing the posts. Before I can post them though, I need to come up with the order in which I want to post them since most topics build upon other topics.

If you have any questions, or suggestions you can email me at liberius1776@gmail.com

Vinculum Numbers

In Vedic Mathematics there exists a way to write higher numbers (6, 7, 8, 9) in terms of lower numbers (0, 1, 2, 3, 4). Why is this important? It’s important because it makes “difficult” looking multiplication problems fairly simple. This is because it is much easier to work with the lower numbers, especially in multiplication. Using this method we only need to know multiplication table up to 5 x 5. This method of writing numbers is called vinculum numbers, and results in numbers containing both positive and negative digits.

How to get vinculum numbers
The secret to vinculum numbers lies in the “Nikhilam Navatashcaramam Dashatah” sutra, which translates as “All from 9 the Last from 10” and the “Ekadhikena Purvena” sutra which translates as “By one more than the previous one.” These sutra is used to obtain the vinculum numbers as I will show you below. Let us begin with an example…

Example #1: 26
Here we wish to convert the number ’26′ into a number that only consists of a mixture of (0, 1, 2, 3, 4, 5).

1) For this first step we will be using the “All from 9 the Last from 10” sutra. The first thing we do is identify the last digit that happens to be a higher number (6, 7, 8, 9). In this case that number is ’6′, the last digit.

2) We see that ’6′ is 4 less than 10 (we use ’10′ because of the sutra…”All from 9 the Last from 10”), so we write this as 4. (Side Note: In books on Vedic math, the line will be on top of the numbers, but since I’m typing this on a computer for simplicity’s sake its easier to write it below.)

3) Now we use the “By one more than the previous one.” The previous one is ’2′ in this case, and by one more means ’3′. Our number is now…

26 = 34

You can think of 34 as (30 – 4) which gives you 26.

Example #2: 183

183 = 223

1) Since 8 is the last “higher number”, we take it from 10….10-8 = 2. We write it as 2.
2) We then add one to the previous one…1 + 1 = 2.
3) The ’3′ is unaffected so it stays the same.

This should be thought of as 203 – 20.

Example #3: 169

169 = 231

1) The last in this example is the ’9′, so we take it from 10….10 – 9 = 1. We write it as 1.

2) In this example we have a second digit that is a “higher number”…’6′. For this we use the first part of the Nikhilam sutra…”All from 9.” So we take ’6′ from 9….9 – 6 = 3. We write this down as 3.

3) We add one to the previous one (’1′)…1 + 1 = 2.

This should be thought of as either “200 – 30 – 1” or “200 – 31.”

Example #4: 372962

372962 = 433042

1) Before our “higher numbers” were always grouped together so we had only one “last”, however now we will have 2 in the number “372962”….the ’7′ and the ’96′.

2) The “last” of the first group is 7…so 10 – 7 = 3. We write 3. This takes care of the first group.

3) Add one to the previous one…so 3 + 1 = 4.

4) Now look at the second group, the last is ’6′…10 – 6 = 4. We write 4. The other number, ’9′, we take from 9……9 – 9 = 0.

5) Add one to the previous one…2 + 1 = 3.

This number (433042) represents 403,002 – 30,040.

Here are a few more:

15627283 = 16433323

397968697 = 402031303

37 x 98 = 43 x 102

The great thing about this is that 0 & 1 are twice as likely to show up now, and these numbers are easy to multiply by! Look at the last example…we went from having to multipy by ’9′ and ’8′ which are hard, to ’0′, ’1′ and ’2′ which are much simpler.

Using vinculum numbers in subtraction

Example #5: 347657 – 238294

1) The first thing we do is write these numbers one on top of the other.

347657
238294

2) Now we just start subtracting vertically, and whenever our number is negative we represent it as a vinculum number.

347657
238294
11

3) This next subtraction is 7 – 8 = -1…so we just write this as 1 and continue on.

347657
238294
111443

4) The ‘1‘ and ‘4‘ here are considered as two separate groups (since theres a non-vinculum number in-between them), so each is subtracted from 10. This has the effect of reducing the “previous one” by one.

109363

Example #6: 3251282 – 1896853

3251282
1896853
2645631

Which in turn becomes…

2645631 = 1354429

Using Vinculum numbers in multiplication

Example #7: 78 x 67

78
67

1)Normally this would be somewhat difficult (since we have to multiply by higher numbers and add them), but now we can convert it…

122
133
_____

Now we just do vertically and crosswise.
1) 1 x 1 = 1.

122
133
_____
1

2) (1 x 3) + (1 x 2) = -3 + -2 = -5 or 5. Remember…vinculum numbers represent negative numbers, so 5 is the same as “-5”.

122
133
_____
15

3) (1 x 3) + (1 x 2) + (2 x 3) = -3 + -2 + 6 = 1.

122
133
_____
151

4) (2 x 3) + (2 x 3) = 6 + 6 = 12. We write down the ’2′ and carry the one to the right.

122
133
_____
1522

5) (2 x 3) = 6.

122
133
_____
15226

So our answer is 15226 or 5226.

The practice problems are going to be split up into three groups: converting to vinculum, subtracting, and multiplying.

Practice Problems
Converting to vinculum
1. 937
2. 672
3. 35908795
4. 18978697

Subtraction using vinculum
1. 546328 – 385977
2. 83947 – 63597
3. 475829828 – 63685
4. 28 – 0.67392

Multiplying using vinculum
1. 46 x 23
2. 58 x 19
3. 128 x 121
4. 473 x 398

If you see any errors, let me know!

Anurupyena

Today we’ll be talking about a really useful sutra. The sutra is “anurupyena” which translates to “proportionality.” This sutra allows us to multiply numbers that are near powers of 10, which becomes really useful when doing Nikhilam Multiplication since it allows us to avoid adding/subtracting large numbers.

Normally when we do Nikhilam Multiplication, we would do the following:

Example: 48 x 47
Using Nikhilam Multiplication, we would write down the numbers one above each other. Then write their differences from a base of 10, 100, etc. to the right. Once that happens we cross add/subtract, then we multiply the differences to get the two different parts of the answer. See what happens here…

48  -52
47  -53

When we go to cross add/subtract things become much more difficult…also, the multiplying the differences (52 x 53) is a lot harder. Using “proportionality” though we can choose a more effective base. There are three different ways to find a base to use, I’ll demonstrate all three on this next example.

Example #1: 48 x 47

Method 1: Take the higher multiple base of 10.
In this case, we will treat this as 100/2 = 50. “50″ will be our working base.

1) We write the numbers below each other like before, but this time we write the differences from our working base (50) to the right.

48  -2
47  -3

These numbers are much easier to work with than before!

2) Our next step is to cross add/subtract. So do either (48 – 3) OR (47 – 2) to get 45 for the first part of the answer. (If your really ambitious you could do….(48 + 47) – 50 = 45…but not sure why you would want to when the other two are easier). Please note that I am using ‘/’ to separate the two parts of the answer here.

48  -2
47  -3
45 /

3) The first part of our answer right now is ’45′, but really this is 45 x 50…since our base is 100/2 = 50. In order to correct for this we divide 45 by 2. We get 22 1/2, so the left side will be 22…the 1/2 is carried over to the right as 100 x 1/2 = 50.

48  -2
47  -3
22 / 50

4) The last step we do is multiply the differences and add it to whatever has been carried over (assuming somethings been carried over). So we do (-2) x (-3) = 6, and then 50 + 6 = 56. (Side note: since we started at 100…100/2 = 50, our right-hand side answer can have two digits. However, in Method 2 we are starting with 10 so our right-hand side can only have one digit.)

48  -2
47  -3
22 / 56

Answer: 48 x 47 = 2256

Method 2: Take the higher multiple base of 10
Method 1 and 2 sound the same but are computed differently. In Method 2 we will take 50 as our working base again, but this time we will take it as 5 x 10 = 50 (instead of 100/2 = 50 like before).

1) We write the numbers out, along with their differences from our working base (50).

48  -2
47  -3

2) We cross add/subtract to get 45 like in method 1.

48  -2
47  -3
45 /

3) This time however, since we are treating it as 5 x 10 = 50…we multiply 45 x 5. When we do this we get 225.

48  -2
47  -3
225 / 

4) Since we took this as 5 x 10, our right-hand side can only have one digit (theres only one zero in 10). So we multiply (-2) x (-3) = 6, and this becomes our right-hand side.

48  -2
47  -3
225 / 6

Answer: 48 x 47 = 2256

Method 3: Take the lower multiple base of 10
For this example we will be using 4 x 10 = 40. So our working base is 40.

1) Write the numbers down like before, along with how far they are from the working base (which is 40 this time).

48  +8
47  +7

2) Now we cross add/subtract…[48 + 7] OR [47 + 8] = 55.

48  +8
47  +7
55 /

3) Since we are using 40 as our base (as 4 x 10 = 40), we need to multiply 55 x 4. This gives us…55 x 4 = 220.

48  +8
47  +7
220 /

4) To get the right-hand side we do (8 x 7) = 56. However, remember…we are using 10 as our starting point (10 x 4 = 40) so our right-hand part can only have one digit. In this case we keep the ’6′, and carry the ’5′ over. So the left-hand portion is now 220 + 5 = 225, while the right-hand side is just 6.

48  +8
47  +7
225 / 6

Answer: 48 x 47 = 2256

You will have to decide which method to use when the time comes. I’ll do one more example for you, if you are having trouble with it though let me know in the comments and I’ll add in another example or two.

Example #2: 35 x 24
Here I’ll be using 2 x 10 = 20 as my base. (You could use 40 if you want…or even 30, but with 30 there would be an extra subtraction since one number is above and one is below. I can explain this if you would like)

1) We write down the numbers, along with how far they are from the working base (20).

35  +15
24  +04

2) Now we cross add/subtract….(35 + 4) OR (24 + 15) = 39.

35  +15
24  +04
39 /

3) Since we are using 2 x 10 = 20 as our base, we need to multiply 39 x 2 to get 78. We put ’78′ down on the left-hand side.

35  +15
24  +04
78 /

4) Now we compute the right-hand side…(15 x 4) = 60. We began with (10 x 2 = 20) to find our working base, so since ’10′ only has one zero in it…our right-hand side can only have one digit. In this case we would write down the ’0′ for the right hand side, and carry-over the ’6′. The left-hand side now becomes 78 + 6 = 84.

35  +15
24  +04
84 / 0

Answer: 35 x 24 = 840

In that last example, we also could have taken the base as 100/5 = 20. See if you can work it out using method 1.

Practice Problems
1. 37 x 32
2. 49 x 41
3. 74 x 73
4. 498 x 496
5. 515 x 503

Squaring Numbers Continued

So far I have talked about how to Square Numbers that end in ’5′ and how to Square Numbers near 10, 100, etc. Today, I will explain to you a simple way to square numbers near 5, 50, 500, etc. While I understood this technique for the most part, I would like to thank Raghuthaman from Vedic Matrix for making a post on one of the Orkut Forums that helped clarify some details for me.

When attempting to square numbers close to 5, 50, etc., we must identify what our ‘base number’ is going to be. The base for ’5′ is a little different from ’50′, ’500′, etc., so I will deal with it first.

Squaring Number close to ’5′
First off, what numbers are considered “close” to 5? In this case…3, 4, 6 and 7. I’ll explain afterwards why I don’t consider the other numbers (1, 2, 8, 9) as close to 5. (Note: this is only in the case of ’5′…not for 50, 500, etc.)

Example #1: 4^2
1) We can see that ’4′ is close to ’5′, so we choose 5 as our starting point.
2) Now we must determine what our “base” is. In this case it is 5^2 = 25.
3) Our next step is to determine how far ’4′ is from ’5′ (since 5 is our starting point). We can see that 4-5 = -1.
4) Once we know how far away it is, we have to “add” this to number to the first digit of our “base”. So since our base is 25, the first digit is ’2′. We have to add…2 + (-1) = 1, so ’1′ is the first digit of our base now. The base is now ’15′.
5) Next, we add to the base the square of the deficit. The deficit being how far away the number was from the starting point…so -1 (as we determined in step 3). So we add (-1)^2 to 15…..15 + 1 = 16. This is our answer.

Answer: 4^2 = 16.

Example #2: 7^2
1) It’s close to ’5′ again, so our “base” will once again be 25.
2) To see how far it is from ’5′ we do…7 – 5 = 2.
3) Our next step is to add this difference to the first digit of our base. So we do…2 + 2 = 4. Our first digit of our base is now ’4′, meaning our base has become ’45′.
5) Next we add the difference squared to our base. That means…45 + (2)^2 = 49.

Answer: 7^2 = 49

So why don’t I consider the other numbers (1, 2, 8 and 9) to be “close” to 5? Well for one, the ’1′ and ’2′ wouldn’t work in this case since we’re subtracting from the first digit of the base, but also because those are easy enough to remember. As for ’8′ and ’9′, these two can be more easily solved using the method I showed you for numbers near 10, 100, etc.

The technique for 50, 500, etc. is shown below.

Squaring Numbers near 50, 500, etc.
In order to find the bases for these numbers, all we have to do is divide the number in half. For 50 the base is 25; for 500 the base is 250, for 5000 the base is 2500 (since 5000/2 = 2500). The reason this applies is because of the “proportionality” sutra that I will be talking about in my next post. Now that we know how to find the base, let’s move on to some examples so I can show you how things are different from when we’re dealing with numbers close to 5.

Example #3: 56^2
1) First thing we do is identify what the base will be. We see that 56 is close to 50, so we will be using “25″ as our base (since 50/2 = 25).
2) Now we have to determine how far away ’56′ is from ’50′. We see that 56 – 50 = +6.
3) We add ’6′ to our base. (Note: See the difference here is we are just adding it to the base…not the first digit of the base like when we were dealing with numbers close to ’5′). So we do 25 + 6 = 31. This is the first part of our answer.
4) To get the second part of our answer we square the difference. So (6^2) = 36.
5) Now we have to merge the two parts. So our final answer is 3136. (Note: This step is also different from the case with numbers near ’5′. Here we just “merge” the numbers together…before we were actually adding them.)

Answer: 56^2 = 3136

Example #4: 42^2
1) Yet again, our number is close to 50 so we will be using a “base” of 25.
2) 42 – 50 = -8. Our number is ’8′ below 50.
3) Add the difference to our base….25 + [-8] = 17.
4) Square the difference….[-8]^2 = 64
5) Merge the two parts….1764.

Answer: 42^2 = 1764

Example #5: 521^2
1) This time our number is close to 500, so we will be using a “base” of 250…(500/2 = 250).
2) 521 – 500 = +21. Our difference is 21.
3) Add the difference to our base…250 + 21 = 271.
4) Square the difference…(21)^2 = 441. (Remember…you can always use the “vertically and crosswise” method here if you don’t know the square directly).
5) Merge the two parts….271,441

Answer: 521^2 = 271,441

In this last example I used a number 21 units away 500….so what is considered a reasonable distance? For example…if our base is 50…would you consider 73 to be close? What about for 500…is 436 close? Honestly, it’s up to you. Do what you are comfortable doing. If you can do 250 – 64 (for the case of 436) and find 64 * 64, then by all means do it. If not, then just stick to numbers that are within your range of ability and work your way up.

Ok, now what about when we have a carry-over? The amount of digits you can have in the right-hand part of the answer is equal to the number of digits you are squaring. (Not 100% sure if this is true…but every case I’ve tried has followed this rule, so if someone runs into a problem where it doesn’t obey this rule please let me know in the comments!).

Example #6: 61^2
1) Our number is close to 50, so we use “25″ as our base.
2) 61 – 50 = 11. This is our difference.
3) Add the difference to the base….25 + 11 = 36.
4) Square the difference…(11^2) = 121. (Click here if you don’t know how to square numbers near 10, 100, etc.)
5) We can only have two digits in the right-hand portion of our answer (since the number we are squaring has two-digits). So we keep ’21′ on the right-hand side, and carry the leading ’1′ over to the left. So the left-hand side of our answer now becomes….36 + 1 = 37.
6) Merge the two together to get 3721.

Answer: 61^2 = 3721

What I suggest doing when your learning a new technique:
1) Read over the examples once.
2) Try to do the examples without looking. You can peak if you need to…but try it without looking first.
3) Do the practice problems at the end.
4) Pick random numbers (well numbers that pertain to the sutra at hand), and apply the method. Then check your answer with the calculator.
4) If you were doing it with paper before, try to do it in your head now.

After a while you won’t even need to check your answer with the calculator.

Practice Problems
1. 6^2
2. 53^2
3. 48^2
4. 508^2
5. 4992^2

If there’s any errors in this post, or if you have any questions let me know in the comments.